LECTURE NOTES IN FLUID MACHINERY (1).docx

LECTURE NOTES IN FLUID MACHINERY (ME 413 ) Pump-is a machine used to add energy to a liquid in order to transfer the liquid from one point to another point of higher energy level. A pump is a machine that converts mechanical energy into fluid energy. Function of a pump – is to add to the pressure existing on a liquid an increment sufficient to the required service. This service maybe the production of a velocity or the overcoming of friction or external pressure. Purpose: is to move a quantity of water against a pressure At ordinary temperatures; 1000 kg = 1 m3, or 1 kg= 1 litre, 1 kg/cm2 = 10 m. Note; This could not be used in hot water. Uses of water pumps 1. 2. 3. 4. 5. 6.

Condensing water Condensate Boiler feed Heater drains Booster flows Sump drain and other services

FIGURE:

TYPICAL PUMP INSTALLATION

Upper Reservoir Discharge Pipe

Pressure Gauge Gate Valve Pump Suction Pipe

Check Valve

Foot Valve w/ Strainer Lower Reservoir

CLASSIFICATION OF PUMPS 1. Centrifugal pump- is a machine which the pumping action is accomplished by imparting kinetic energy to the fluid by a high speed revolving impeller with vanes and subsequently converting this kinetic energy into pressure energy either by passing through a volute casing or through diffuser vanes. It is high discharge, low head, high speed, not self priming. 2. Rotary Pump- a positive displacement pump consisting of a fixed casing containing gears, cams, screws, vanes plungers or similar elements actuated by rotation of the drive shaft. It is low discharge, low head, used for pumping viscous liquids like oil. Example; Gear pump, Screw pump, vane Pump 3. Reciprocating pump- is a positive displacement unit wherein the pumping action is accomplished by the forward and backward movement of a piston or plunger provided with valves. It is low discharge, high head, low speed, self- priming.4.0) Deep well pump- is divided into plunger or reciprocating, turbine, ejector-centrifugal types and air lifts. Turbine pumps-are used for pumping water with high suction lift, for pumping condensate. Jet Pump or Injector pump- used for pumping boiler feed water, used as accessory of centrifugal pump. Types of pumps are classified into two major category. 1. Dynamic or kinetic – are types of pumps in which energy is continuously added to the fluid to increase its velocity. Examples; centrifugal, jet and turbine pumps 2. Positive displacement pumps-are types of pumps, in which energy is continuously added by application of force to an enclosed volume of fluid and resulting to a direct increase in its pressure. Examples: Reciprocating, rotary, and diaphragm pumps.

Some other ways of classifying pumps includes the following; 1. Type of flow of fluid at the impeller; radial semi-axial, or axial 2. Type of casing; volute or diffuser pumps 3. Type of design: single stage, multi-stage, single suction, dual suction, horizontal, vertical pump, submersible. 4. Type of application: fire, dredge, slurry, deep well, irrigation, drainage, circulating, boiler feed, condensate, clogless. CENTRIFUGAL PUMP A centrifugal pump consist of stationary casing and an impeller connected in rotating shaft. Liquid enters the centre of the rotating impeller and leaves at a high velocity and passes to a stationary volute casing which transforms kinetic energy into pressure. The term “centrifugal” came from the centrifugal force created as the water move outward from the centre of impeller rotation. A centrifugal pump whose drive unit is supported by the pump having its suction and discharge flanges on approximately the same order is called in line pump. A centrifugal pump characterized by a housing which is split parallel to the shaft is called horizontal split case pump. A centrifugal pump with one or more impellers discharging into one or more bowls and a vertical ejector or column pipe used to connect the bowls to the discharge heads on which the pump driver is mounted is called vertical shaft turbine pump. FIGURE: Impeller

Casing Vane or Blade

Advantages of centrifugal pumps 1. Simple and compact 2. Easy to maintain 3. Adaptability with motor with high rpm 4. Little vibrations 5. Flow can be controlled from full to non-discharge without shutting the pump Disadvantages 1. 2. 3. 4. 5. 6.

Poor suction power Usually needs priming Cavitation may develop during operation Needs multi-stage to increase discharge pressure Cannot handle very viscous fluid Check valve is required to avoid back flow.

One of the disadvantages of a centrifugal pump is that the liquid is delivered at a relatively low pressure. This due to the high velocity acquired by the liquid as it leaves the impeller. Therefore, liquids are delivered at low pressure but at a high velocity. In order, to increase the discharge pressure of liquids, a decrease in its velocity at the impeller discharge is required. This is usually done by providing additional impellers to the pump. The liquid exits the first impeller at high velocity then passes thru a series of impellers thus, additional pressure is added. Parts of centrifugal pump 1. The shaft- is a piece of metal where the internal parts of the pump are mounted. Its function is to transmit torque from the motor to the rotating parts of the pump. Pump shaft has also a shaft sleeve, it is a metal cylinder that fits over the shaft to protect it from corrosion, erosion, and wear. 2. The impellers of a centrifugal pumps are the most critical part of the pump, because the capacity of the pump depends on its diameter and speed of which it turns. Impellers can be open, semi- open and closed type impellers. FIGURE: 3. The pump casing- is the stationary part of a centrifugal pump and it can either be volute or circular. The main function of the volute is to convert the kinetic energy acquired by the fluid into pressure. It could be single and double volute casing. Diffuser volute design has a set of fixed vanes in its casing

4.

5.

6. 7.

that gradually reduces fluid velocity. It gives direction to the flow from the impeller and converts this velocity energy into pressure energy Wear rings-are fitted to the impellers or to the pump casing to protect the actual impeller and casing from wear. Excessive wear in the pump will reduce its efficiency particularly in small pumps used in high speed applications. Wear rings should be replaced periodically to avoid damage in the impeller or in the casing. It keeps internal recirculation down to a minimum. Packing or mechanical seal- the main function of packing or mechanical seal is to protect the pump from leakage in the area where the shaft passes through the casing. The most common means of throttling the leakage between the inside and outside of the casing. Bearings-are used to keep the correct alignment of the shaft with other stationary parts of the pump. It accurately locate shaft and carry radial and thrust loads. Labyrinth and seals and deflectors- the main purpose is to retain oil in the pump housing and to prevent foreign materials entering the pump housing.

HEAD and POWER CALCULATIONS FIGURE: Motor Static Discharge Total Static Head Head

Pump

h Pump Center Line Static Sunction

Discharge-volume flow rate of liquid handled by the pump in m3/s or gal/min Head- total energy developed by the pump, expressed in height of the liquid in meters. Basic Principles: General Flow Equation: Q= A v or v=Q/A H= total head or total dynamic head (TDH) H= ( zd – zs ) +

𝑃𝑑−𝑃𝑠 𝑤

+ (hfs + hfd ) +

𝑉𝑑 2 −𝑉𝑠2 2𝑔

Note: Zs is negative if source is below the pump centre line Ps is negative if it is vacuum Static head- is the height of the surface of the water above the gauge point. Velocity head-is the head required to produce the flow of water. Pressure head= is the static head plus gauge pressure on the water surface plus friction head Dynamic head= is the pressure head plus velocity head. Pump operating head-is the algebraic difference of discharge and suction head. This can also be called as total dynamic head. WATER POWER OR HYDRAULIC POWER- is the theoretical power necessary to raise a given volume of liquid from a lower to a higher elevation.

Water Power= Q w H; kW Where; Q = discharge, m3/s W= specific weight=9.81 kN/m3 for water H= total head, in meters 𝑊𝑎𝑡𝑒𝑟 𝑃𝑜𝑤𝑒𝑟

Brake (Input) Power =𝑃𝑢𝑚𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 Motor efficiency –is the ratio of brake power to input power. 𝐵𝑟𝑎𝑘𝑒 𝑝𝑜𝑤𝑒𝑟

em = 𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟

Overall efficiency – is the ratio of hydraulic power or water power to the input power. e=

𝑊𝑎𝑡𝑒𝑟 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟

Head, as determined from readings of pressure gauges

H=

𝑃𝑑−𝑃𝑠 𝑤

+

𝑉𝑑 2 −𝑉𝑠2 2𝑔

Note: Ps is negative if a vacuum Calculating the friction head Friction head is the head lost by the flow in a stream or conduit due to frictional disturbances set up by the moving liquid and its containing conduit and by intermolecular actions. hf = hf=

𝑓𝐿𝑉 2 , (Darcy equation) 2𝑔𝐷 2 2𝑓𝐿𝑉 , ( Morse equation) 𝑔𝐷

where; hf = friction head loss, m f = coefficient of friction ( should be taken from Morse table if Morse equation is used) L= total length, m (including equivalent lengths of the fittings) V= velocity, m/s g = 9.81 m/s2 D= inside diameter, m Example Problems: 1. Water in the rural areas is often extracted from underground water source whose free surface is 60 m below ground level. The water is to be raised 5m above the ground by a pump. The diameter of the pipe is 10 cm at the inlet and 15 cm at the exit. Neglecting any heat interaction with the surroundings and frictional heating effect, what is the necessary power input for the steady flow of water at the rate of 15 litres/ sec in kW? Solution: FIGURE:

5m 60m

15cm 10cm

Q=15 litres/s = 0.015 m3/s 𝑄 4(0.015) Vs =𝐴𝑠 = 𝛱(0.10)2 0.015

Vd =

𝛱(0.15)2 4

Water Source

= 1.91 m/s

=0.85 m/s

H=( Zd – Zs ) +

𝑉𝑑 2 − 𝑉𝑠2 2𝑔

0.85)2 −(1.91)2 2(9.81)

= 5-(-60 ) +(

=64.85 m

WP= Q w H =0.015 (9.81) (64.85) = 9.54 kW

2. Calculate the drive horsepower for pumping 1703 L/min cold water to a tank suction at 127 mm Hg vacuum, delivery at 5.3 kg/cm2 ga., both measured close to pump, efficiency of the pump is 0.65. Solution:

5.3

𝑘𝑔 𝑐𝑚2 𝑔𝑎𝑔𝑒

Figure: Pump 127mmHgvac

Let; hd= total head at discharge hs =total head at suction h= pump head By Bernoulli’s Equation H= hd – hs =

𝑃𝑑−𝑃𝑠 𝑤

𝑉𝑑 2 −𝑉𝑠2 2𝑔

+

+ (Zd – Zs ) +( hfd + hfs)

Where; Pd= 5.3 kg/cm2ga Ps=-127 mmHg = -0.1727 kg/cm2 H=

(5.3+0.1721)𝑘𝑔𝑝𝑒𝑟 𝑠𝑞.𝑐𝑚(100 𝑐𝑚𝑝𝑒𝑟 𝑠𝑞.𝑚 1000𝑘𝑔 𝑝𝑒𝑟𝑐𝑢.𝑚

= 54.72 m

Pbrake =wQh =1000kg/m3(1703 L/min) (1m3/1000L) (54.72) =93,188.16 kg-m/min ( I kW/6116.3 kg-m/min)=15.24 kW=20.42 hp Drive hp of the pump; Hp=

20.42 0.65

= 31.42 hp

3. Water from a well is to be pumped to a height of 40 meters from the source. The quantity of water to be delivered is 500 L/s, through a 480 mm diameter pipe. It is also desired to maintain a 100 kPa pressure at the summit. The frictional head losses from the suction and discharge of the pump are equivalent to 2.0 m. If the pump speed is 850 rpm, determine the power needed by the pump in hp. Solution: 100kPag Figure:

40m

Water Source

Solving for the water Power Source: suction lift WP = hs + hv + hP + hL hs =40 m Solving for the velocity at discharge; Q= A2 v2 (0.480)2 4

0.5 m3/s = Π

(v2)

v2 = 2.7631 m/s Neglecting the velocity at suction; 𝑣22 −𝑣12 (2.7631)2 −02 = 2(9.81) =0.3891 2𝑔 𝑃 −𝑃 (100−0) hf = 2 𝛶 1 = 9.81 = 10.1937 m

hv =

m

hl = 2.0 m TDH = 40m +0.3891 +10.1937 + 2.0 m = 52.5828 m Thus, WP= 0.5 m3/s (9.81 kN/m3) (52.5828m) = 257.9186 kW WP = 257.9186 kW x 1 hp/0.746 kW =345.7354 hp 4. A water source with an elevation of 10 m is to be pumped by a centrifugal pump to an open reservoir at an elevation of 80 m. The suction pipe and discharge pipe diameter is 255mm and 200 mm, respectively. The pump centreline is at 5m. The head loss at suction pipe is 0.05m and 0.45 at the discharge pipe. The pump delivers water at 15 L/s and has an efficiency of 75% while the electric motor is 80%. Determine: a. Water horsepower of the pump b. Brake horsepower of the pump c. Power input to drive the motor, in kW

d. The pressure gage readings installed before and after the suction and discharge of the pump, kPa. Solution; Figure: B

Reserve Oil

70m

200mm

A Source 80m 10m

2 1

225mm

5m

Solving for WP Source: Suction head WP =QΥTDH; TDH = hs + hv + hf + Hl Solving for the water velocity at the suction and discharge Q= Av= A1v1 = A2v2 (0.255𝑚)2 V1 ; V1 = 0.2937 m/s 4 (0.200𝑚)2 0.015 m3/s = Π v2 ; v2 = 0.4775 m/s 4 2 2 2 𝑣 −𝑣 (0.4775) −(0.2937)2 hv = 22𝑔 1 = =0.0072246 m 2(9.81)

0.015 =Π

Since both the source and the reservoir are open to the atmosphere Hp =0 HL =0.05m + 0.45 m =0.5 m TDH=70m + 7.2246 x 10-3 +0 +0.5 =70.5072 Thus, WP = 0.015 m3/s (9.81)kN/m3 (70.5072 m) = 10.3751 kW ; Thus; Whp =10.3751 kW x

1 𝐻𝑝 0.746 𝑘𝑊

=13.9077 hp

Solving for the brake horsepower of the pump; 𝑊ℎ𝑝 𝐵ℎ𝑝 13.9077 ℎ𝑝 0.75 = 𝐵ℎ𝑝

Np =

Bhp = 18 5436 hp Solving for the power input to drive the motor 𝐵𝑃

nm =𝐸𝑃 x 100 0.80 =

(18.5436) (0.746)𝑘𝑤 𝐸𝑃 1 ℎ𝑝

EP = 17.2919 kW Solving for the gage pressure readings;

Applying energy equation from the water surface level of the source (A) to the suction point of the pump (1), we’ll have ZA + VA2/2g + PA/Υw = Z1 + V12/2g + P1/Υw + HLA-1 Where; PA/Υw = 0 kPa, ( the water is open to the atmosphere) VA = V1 (equal suction pipe diameter) Z1 =0( used as the datum or reference line) 𝑃

Thus; ZA = 𝛶𝑤1 +HLA-1 P1 = (ZA –HLA-1) Υw = ( 5m-0.05m)(9.81 kN/m3) = 48.5595 kPa g Applying energy equation from the discharge point of the pump (2) to the water surface level of the reservoir(B) we’ll have, 𝑉2

𝑉2

𝑃

𝑃

2 2 𝐵 Z2 +2𝑔 + 𝛶𝑤 = ZB + 2𝑔 + 𝛶𝑤𝐵 + HL2 – B

𝑃

Where: 𝛶𝑤𝐵 = 0 kPag(the water source is exposed to the atmosphere V2 =VB (equal pipe diameter) Z2 = 0 ( used as the datum or reference line) Therefore:

𝑃2 𝛶𝑤

= ZB + HL2-B

P2 = (ZB – HL2-B ) Υw = (75 m-0.45 m) (9.81 kN/m3) = 731.3355 kPag 5. A pump discharges 150 litres per second of water to a height of 75 m. If the efficiency is 75% and the speed of the pump is 1800 rpm, what is the torque in N-m to which the driveshaft is subjected? Solution: Figure: 150𝐿⁄𝑠 75m

Let ; ep = pump efficiency 2𝛱 𝑇𝑛 P= 60

147.15

𝛶𝑤𝑄𝐻 = 9.81(0.150) 𝑒𝑝 2𝛱𝑇𝑛 2𝛱 𝑇 1800 = 60 = ; 60

=

(75)/(0.75) =147.15 kW, Thus;

T= 0.781 kN-m = 781 N-m Problem Solving: 1. A centrifugal pump delivers 227 m3/hr of water from a source 4 meters below the pump centre line to a pressure tank whose pressure is 2.8 kg/cm2. Friction loss estimates are 2 meters in the suction line and 1 metre in the discharge line. The diameter of the suction pipe is 250 mm and the discharge pipe is 200 mm. Find: a) The water horsepower b) The kW rating of the driving motor if the pump efficiency is 70% (Ans. a. 29 hp, b. 31 kW ) 2. A pump is to deliver 80 galloons/min of water 600C with discharge pressure of 1000 kPag. Suction pressure indicates 50 mmHg vacuum. The diameter of suction and discharge pipes are 5 inches and 4 inches, respectively. If the pump has an efficiency of 70%. Determine the brake horsepower of the pump. (Ans. 9.732 hp) 3. An acceptance test was conducted on a centrifugal pump having a suction pipe 25.4 cm in diameter and a discharge pipe 12.7 cm in diameter. Flow was 186 m3/hr of clear cold water. Pressure at suction was 114.3 mmHg vac and discharge pressure was 107 kPag at a point 91 cm above the point where the suction pressure was measured. Input to the pump was 15 hp. a) Determine the pump efficiency b) If the pump runs at 1750 rpm, what net flow, head, and brake hp would be developed and required if the pump speed were increased to 3500 rpm? Assume constant efficiency. ( Ans. a. 64.4%, b. 372 m3/hr, 56.8 m, 120 hp)

4. A motor driven pump draws water from an open reservoir A and lifts to an open reservoir B. Suction and discharge pipes are 150 mm pipe and 100 mm inside diameter respectively. The loss of head in the suction line is 3 times the velocity head in the 150 mm pipe and the loss of head in the discharge line is 20 times the velocity head in the 100 mm pipeline.. Water level at reservoir A is at elevation 6 meters and that of reservoir B at elevation 75 m. Pump centre line is at elevation 2 m. Overall efficiency of the system is 73%. Discharge is 10 litres/sec. Determine the following; a. Power input of the motor b. Reading in kPa of the pressure gauges installed just at the outlet and inlet of the pump. (Ans. a. 9.51 kW, b. Po =732.34 kPag, Pi = 38.76 kPag ) 5. A boiler feed pump receives 40 litres per second at 1800C. It operates against a total head of 900 meters with an efficiency of 60%. Determine: a. The enthalpy leaving the pump in kJ/kg b. Power output of the driving motor in kW c. Discharge pressure in kPa if suction pressure of 4 MPa (Ans. a. 773.57 kJ/kg, b. 523.3 kW, c. 11,850.3 kPa) 6. A plant has installed a single suction centrifugal pump with a discharge of 68 m 3/hr under 60 m head and running at 1200 rpm. It is proposed to install another pump with double suction but of the same type to operate at 30 m head and deliver 90 m3/hr a. Determine the speed of the proposed pump. b. What must be the impeller diameter of the proposed pump if the diameter of the existing pump is 150 mm? (Ans. a. 877 rpm, b. 145 mm ) 7. A 40 m3/hr pump delivers water to a pressure tank. At the start, the gage reads 138 kPa until it reads 276 kPa and then the pump will shut off. The volume of the tank is 160 litres. At 276 kPa the water occupied 2/3 of the tank volume. a) Determine the volume of water that can be taken out until the gage reads 138 kPa. b) If 1 m3/hr of water is constantly used, in how many minutes from 138 kPa will the pump run until the gage reads 276 kPa? Ans. a) 30.75 litres, b) 0. min. 8. A pump with a 400 mm diameter suction pipe and 350 mm diameter discharge pipe is to deliver 20,000 litres per minute of 15.60C water. Calculate the pump head in metres if suction gage is 7.5 cm below the pump centreline and reads 127 mmHg vacuum and the discharge gage is 45 cm above the pump centre line and reads 75 kPa. Ans. H=10.14 m 9. Water from an open reservoir A at 8m elevation is drawn by a motor driven pump to an open reservoir B at 70 m elevation. The inside diameter of the suction pipe is 200 mm and 150 mm for the discharge pipe. The suction line has a loss of head 3 times that of the velocity head in the in the 200 mm pipe. The discharge line has a loss of head 20 times that of the head in the discharge pipeline. The pump centreline is at 4 m. Overall efficiency of the system is 78% For a discharge rate of 10 li/s, find the power input to the motor and the pressure gages reading installed just at the outlet and inlet of the pump in kPag. Ans. P input = 7.825 kW, Ps= 39.14 kPa, Pd= 650.80 kPa 10. The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and reads 68.95 kPa gage;the discharge static gage is 0.61 m below the pump centre line and reads 344.75 kPagage. The gages are located close to the pump as much as possible. The area of the intake and discharge pipes are; 0.093 m2 and 0.069 m2 respectively. The pump efficiency is 74%. Take density of water equals 1000 kg/m 3. What is the hydraulic power in kW. Ans. P=17.09 kW 11. A pump delivers 20 cfm of water having a density of 62 lb/ft3. The suction and discharge gage reads 5 in.Hg vacuum and 30 psi respectively. The discharge gage is 5 ft above the suction gage. If pump efficiency is 70%, what is the motor power? Ans. P=4.31 hp 12. In a test of a centrifugal pump driven by an electric motor, the suction pipe is 10 in. in diameter and its gage indicates a partial vacuum of 2.5 ft of water. The discharge pipe is 5 in. in diameter, is 2 ft. higher than the suction gage and shows pressure of 50 ft of water. If the pump is discharging 1.6 ft3/s and the electrical power input is 12 kW, what pump efficiency is indicated assuming motor efficiency of 85%. Ans. 74.04% 13. Oil is being pumped from a truck to a tank 10 ft higher than the truck through a 2 in. galvanized pipeline 100 ft long. If the pressure of the discharge side of the pump is 15 psi, at what rate in gpm is oil flowing through the pipe?. The oil has an specific gravity of 0.92 at the temperature in the pipe. Ans. 542.44 gal/min

14. A centrifugal pump delivers 80 litres per second of water on test suction gage reads 10mmHg vacuum and 1.2 m below pump centreline. Power input is 70 kW. Find the total dynamic head in metres. Use np (pump efficiency) = 74%. Ans. H=66m Characteristics of Centrifugal Pump 1. Specific Speed –is the speed at geometrically similar impeller of a pump would run to discharge 1 gpm at 1 foot head. Specific speed of centrifugal pump impeller is the rotative speed at which a geometrically similar impeller would run if it were of such size as to raise 75 kg of water per second against one metre head. 𝑁√𝑄

Ns =

3

; rpm, in English System

𝐻4

Where; Ns = specific speed in rpm N= rotational speed, in rpm Q= discharge, in gpm H= head, in feet 0.0149𝑁√𝑄

Ns =

3

rpm, in

Metric system

𝐻4

Where: N= pump shaft speed in rpm H , or TDH =head in metre per stage Q = is the discharge in litres/min Note; For double suction pumps Q is divided by two and for multi-stage pumps, H is divided by the number of stages. In Figure1.13 shows the relationship of pump efficiency, specific speed, geometry of pump and capacity. As the specific speed of the pump increases, the ratio of impeller outer diameter, D1, to the impeller centre diameter, D2 , decreases. The ratio becomes 1.0 for a propeller type impeller. The observation from the Figure that low specific speed can be designated as” low capacity”, that means the head is developed mainly through centrifugal force and less of axial force. Also high specific speed can be designated as” high capacity”, which indicates that head is developed mainly through axial force and less of centrifugal force. By using Figure 1.14 Pump specific speed when N=3500 rpm, H=120 ft and Q=1200 gpm, the Ns =3350 rpm. It is important to note that specific speed is always calculated at the Best Efficiency Point (BEP) of the pump at maximum impeller diameter. Example Problem; A double suction, single stage centrifugal pump delivers 4000 gpm of water from a well where the water level varies from 2.5 metres from high tide to low tide. The pump centre line is measured to be 3 m above the water level at high tide. The pump discharges the water to an open surface condenser located 4.5 m above the pump centreline. The head loss due to friction in the suction and discharge pipe is 1 m and 2.5 m, respectively. The pump is directly coupled in a motor with 1800 rpm. Determine: a) Total suction head, m b) Total discharge head, m c) Pump specific speed d) Impeller diameter in mm Solution: FIGURE;

Solving for the total suction head Total suction head= 2.5 + 3 + 1 = 6.5 m Total discharge head = 4.5 + 2.5 = 7.0 m

Solving for pump specific speed Ns =

𝑁√𝑄 𝐻43

, where, H=13.5 m (

4000 𝑔𝑝𝑚 2 3 (44.28 𝑓𝑡)4

3.28 𝑓𝑡 )= 1𝑚

44.28 ft

1800 𝑟𝑝𝑚√

Ns =

=4689.5567 rpm, Using Figure 1.13, the specific speed suggests a mixed flow type of

impeller. Solving for the pump impeller diameter, V=

(𝑐𝑖𝑟𝑐𝑢𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒)𝑁 60

=

𝛱𝐷𝑁 60

V=√2𝑔𝐻 = √2(9.81𝑚)(13.5)𝑚 =16.2748 m/s Substituting: 𝛱𝐷(1800𝑟𝑝𝑚) 60

16.2748 m/s =

D= 0.1727 m = 172.6810 mm Example 02] Calculate the impeller diameter of the centrifugal pump that requires 15 m head to deliver water if pump speed is 1500 rpm. Solution V=𝛱𝐷𝑁 Solving for velocity V=√2𝑔ℎ

=√2(9.81)15 = 17.15 m/s

Thus; 17.15 = ΠD( 1500/60) D= 0.21843 m = 218.43 mm Characteristic Curve and Best Efficiency Point Pump performance curves are graphical representation of the characteristic curve of a certain pump model which shows the pump capacity ,(gpm), impeller diameter (inches), total dynamic head, (feet) brake horsepower input (bhp) and pump efficiency in percent. A typical pump performance curve is shown below: The highest point on the efficiency curve is the (Best efficiency Point) FIGURE: H-Q Characteristics and Efficiency

Example; Required; Head, horsepower, and efficiency at 160 gpm, Solution ; From Figure, bhp = 7bhp; head =130 ft, best efficiency is at 160 gpm NET POSITIVE SUCTION HEAD

Pumps are designed to operate continuously for long period of time. To achieve this, the pump utilizes the fluid that it is pumping for its lubrication and cooling. Damage in the pump may occur when circulation of liquid stops for long period of time while the pump is operating. Damage can also occur when the pressure at any point inside the pump drops below the vapour pressure corresponding to the temperature of the liquid. Because at this condition, the liquid will vaporize and eventually forms vapour bubbles as it enters the inlet of the pump. This vapour bubbles then collapse or implodes at the surface of the impeller creating tremendous physical shock to the edges of the impeller. This process of vapour formation is cavitation. Another undesirable result of cavitation is adverse noise accompanied by heavy vibrations. Since cavitation occurs mainly at the inlet of the pump, we can consequently say that it is related to the pump suction head conditions. Net positive Suction Head, NPSH, is an index where the pump may operate without cavitation. NPSH is the difference between actual suction pressure and saturation vapour pressure of the liquid. Two kinds of NPSH, 1.0) NPSHr 2.0) NPSHa NPSHr for a particular pump is experimentally determined and provided by the manufacturer and is a function of pump design. NPSHa is determined by plant engineer during the design and proposed installation of the pump and is a function of the system where the pump will operate. To avoid cavitation, it is necessary that NPSHa is equal or greater than NPSHr. To avoid cavitation, NPSHa>=NPSHr If NPSHa is less than or equal to NPSHr 1.0) Decrease the suction lift by changing the plant layout and raising the source on which the pump draws water 2.0) Reduce the suction head by using pumps with larger capacity but operating it in partial loads or speeds. Cavitation- is defined as the formation of cavities of water vapour in the suction side of a pump due to low suction pressure. Causes of Cavitation 1.0) Low suction pressure 2.0) Low atmospheric pressure 3.0) High liquid temperature 4.0) High velocity 5.0) Rough surface and edges 6.0) Sharp bends Bad Effects of Cavitation 1.0) Drop in capacity and efficiency 2.0) Noise and vibration 3.0) Corrosion and pitting

2.0) Similar Pumps a.) b.)

𝑁1 √𝑄1 3 𝐻14 𝑄1 𝑁1 𝐷13

=

𝑁2 √𝑄2 3

𝐻2 4

=

𝑄2 𝑁1 𝐷23

; where; D= impeller diameter

3.0 Same Pump (Affinity Laws) Pump affinity laws are rules that express the relationship of pump capacity head, and bhp when the speed or impeller diameter is changed. Assuming that the efficiency is the same for both conditions. a) Constant impeller diameter, variable speed 𝑄1 𝑄2

𝑁

= 𝑁1 2

𝐻1 𝐻2

𝑁

= [𝑁1 ]2 2

𝑃1 𝑃2

𝑁

=[𝑁1 ]3

P= Power

2

b) Constant speed, variable impeller diameter 𝑄1 𝑄2

𝐷

𝐻1 𝐻2

= 𝐷1 ; 2

𝐷

𝑃1 𝑃2

= [𝐷1 ]2 2

𝐷

= [𝐷1 ]3 2

Similarity Laws of Pumps 1. Q is directly proportional to D3N 𝐷3

𝑄2 𝑄1

𝑁

= 𝐷23 x 𝑁2 1

1

2. H is directly proportional to N2 𝐻2 𝐻1

𝑁

𝐷

=( 𝑁2 )2 x (𝐷2 ) 1

1

3. 𝑃𝑜𝑤𝑒𝑟 𝑖𝑠 𝑑𝑖𝑟𝑒𝑐𝑡𝑙𝑦 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎 𝑡𝑜 𝑁3D5 𝑃2 𝑃1

𝑁

𝐷

=( 𝑁2 )3(𝐷2 )5 1

1

𝑁𝑜𝑡𝑒: 𝐹𝑜𝑟 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑝𝑢𝑚𝑝𝑠 𝑇ℎ𝑒 𝑁𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙

Example Problem: 1.0) A pump delivers 500 gpm against a total head of 200 ft and operating at 1770 rpm. Changes have increased the total head to 375 ft. At what rpm should the pump be operated to achieve the new head at the same efficiency. Solution: 𝑯𝟏 𝑯𝟐

𝑵 𝑵𝟐

= [ 𝟏]2 ;

𝟐𝟎𝟎 𝟑𝟕𝟓

𝟏𝟕𝟕𝟎 2 ] 𝑵𝟐

=[

; N2 = 2423.67 rpm

2.0) It is desired to deliver 5 gpm at a head of 640 ft in a single stage pump having a specific speed not to exceed 40. If the speed not to exceed 1352 rpm, how many stages are required? Solution: Ns =

𝑁√𝑄 3 𝐻4

= 40 =

1352√5 (

640 )𝐫𝐚𝐢𝐬𝐞𝐭𝐨 𝟑/𝟒 𝑛

; n= 2 stages

3.0) The power output is 30 hp to a centrifugal pump that is discharging 900 gpm and which operates at 1800 rpm against a head H= 120 ft, 220V, 3 phase, 60 hertz. If this pump is modified to operate 1200 rpm, assuming its efficiency remains constant, determine its discharge in gpm, the theoretical head it imparts to the liquid and the power input to the pump. Solution: 𝑄1 𝑁1 = 𝑄2 𝑁2 900 1800 =[ ] ; Q2 = 600 𝑄2 1200 𝐻1 𝑁 = [𝑁1 ]2 𝐻2 2 120 1800 = [1200]2 𝐻2

gpm.

𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡ℎ𝑒 𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 ℎ𝑒𝑎𝑑 , 𝐻2

𝑯𝟐 = = 53.33 ft 𝐒𝐨𝐥𝐯𝐢𝐧𝐠 𝐟𝐨𝐫 𝐭𝐡𝐞 𝐩𝐨𝐰𝐞𝐫 𝐢𝐧𝐩𝐮𝐭 , P2 𝒑𝟏 𝒑𝟐 𝟑𝟎 𝑷𝟐

𝑵

=

[𝑵𝟏 ]3

=

[𝟏𝟐𝟎𝟎]3

𝟐

𝟏𝟖𝟎𝟎

P2 = 8.89 hp Problem Solving:

1.0) A pump operating at 1750 rpm delivering 500 gpm against a total head of 150 ft. Changes in the piping system have increased the total head to 360 ft. At what rpm should the pump be operated to achieve this new head at the same efficiency? Ans. 2711.09 rpm 2.0) A DC driven pump running at 100 rpm delivers 30 litres per second of water at 400C against a total pumping head of 27 m with a pump efficiency of 60%. Barometer pressure is 758 mmHg. What pump speed and capacity, would result if pump rpm were increase to produce a pumping head of 36 m assuming no change in efficiency. Ans, 115.47 rpm, 34.64 L/s 3.0) A centrifugal pump discharge 20 L/s against a head of 17 m when the speed is 1500 rpm. The diameter of the impeller was 30 cm and the brake horsepower was 6.0. A geometrically similar pump 40 cm in diameter is to run at 1750 rpm, Assuming equal efficiencies, what brake horsepower is required? Ans. Bhp = 40.14 bhp 4.0) A two stage centrifugal pump delivers 15,000 kg/hr of 1100C water against 76 m head at 3500 rpm. What is the specific speed of the pump? Ans. Ns = 780.39 rpm 5.0) A pump running at 1000 rpm delivers water against a head of 300 m. If the pump speed will increased to 1500 rpm, what is the change in head? Ans. ΔH= 375 m 6.0) A test on a centrifugal pump operating at 1150 rpm showed a total head of 37.6 ft at a capacity of 800 gpm. Estimate the total head and capacity if the pump were operated at 1750 rpm. Assume normal operation at point of maximum efficiency in each case. Ans. H2 = 87.07 ft. , Q2 =1217.4 gpm 7.0) A double- suction centrifugal pump delivers 3 m3/s of water at a head of 15 m and running at 1200 rpm. Calculate the specific speed of the pump. Ans. Ns = 9958.56 rpm 8.0) Determine the performance of a centrifugal pump with initial flow of 150 gpm, at an initial head of 120 ft, and initial power input of 5 bhp. The impeller diameter is changed from 9.5 inches to 8.0 inches. Ans. Q2 = 126.3158 gpm; H2 =85.0970 ft. ; BHP2 =2.9859 bhp 9.0) Determine the performance of a centrifugal pump with initial flow of 150 gpm, at an initial head of 120 ft, initial power input of 5 bhp. The speed is changed from1700 rpm to 3000 rpm. Ans. Q 2 =264.7059 gpm, H2 = 373.7024 ft, BHP2 = 27.4781 bhp 10.0) A centrifugal pump operating at 1150 rpm showed a total head of 40 ft at a capacity of 600 gpm. The impeller diameter is 10.5 in. Estimate the total head and capacity of a geometrically similar pump at 1150 rpm with an impeller diameter of 10 inches. Ans. H2= 36.28 ft, Q2 =518.3 gpm 11.0) A double suction, single stage, centrifugal pump delivers 900 m3/hr of sea water(S.G. =1.03) from a source where the water level varies 2 m from high tide to low tide level. The pump centreline is located 2.6 m above the surface at high tide level. The pump discharges into a surface condenser, 3 m above the pump centreline. Loss of head due to friction in the suction pipe is 0.80 m and that in the discharge side is 3 m. Pump is directly coupled to a 11750 rpm, 460 Volts, 3 phase, 60 hertz motor. Calculate the specific speed of pump in rpm Ans. Ns = 5149.20 rpm . Ns =5149.20 rpm.

Net Positive Suction Head Available, NPSHa Formula: NPSHa= hp( + or -) hSL -hV - HL Where: Hp = absolute pressure head on the surface of the liquid source in meters. This will be the atmospheric pressure corresponding to its altitude when the liquid surface is open. HLS = the height of the liquid surface from the pump centreline , designated as positive when suction head and negative when suction lift, in meters. HV = head corresponding to the vapour pressure of the liquid at liquid temperature, can be determined using steam tables, in meters HL= head loss due to friction and turbulence, in meters. Example Problems: 1.0) Determine the NPSH available with following water conditions: Water from a well with a temperature 270C at sea level, with head loss of 0.45m and a suction lift of 3 m. Solution; Figure;

NPSHa = hp + or – hSL –hv - HL From Steam tables: specific volume of water at 270C is; vf = 1.0035 x 10-3m3/kg and the specific weight is ;ɤ = 9.7758 kN/m3 Since the water source is open to the atmosphere, hp=101.325 kPa 𝑃𝑝

101.325 𝑘𝑃𝑎

Hp = ɤ = 9.7758 𝑘𝑁/ 𝑓

𝑐𝑢.𝑚

= 10.3649 m

HSL = -3 m ( since it is suction lift) The vapour pressure of water at 270C is 3.567 kPa 𝑃

Hv = ɤ𝑣 = 𝑓

3.567 𝑘𝑃𝑎 𝑘𝑁 𝑐𝑢.𝑚

9.7758

= 0.3649 m

HL = 0.45 m Thus; NPSHa= (10.3649-3-0.3649-0.45) = 6.55 m

Example Problems on NPSH 2.0)A closed tank contains liquefied butane gas whose specific gravity is 0.60. The tank pressure is 1.6 MPa gage which is also the equilibrium vapour pressure of butane at the pumping temperature. Suction line losses is 1.5 m of gas and the static elevation gain is 4 m. What is the net positive suction head available (NPSH)? Solution: Use (+) for static elevation gain NPSH =

𝑃+𝑃𝑎 −𝑃𝑣 9.81(𝑆𝐺)

+ S – Hf

= 0 + 4 -1.5 = 2.5 m available Problem Solving: 1.0) A condensate pump draws water from a condenser which maintains a pressure head of 724 mmHg vacuum. The friction losses at the piping system between the pump and the condenser is measured to be 1.5 m. Determine the minimum height of water in the condenser that needs to be maintained if the NPSH r for the pump is 3.5 m. Ans. hLS = 4.7 m. 2.0) A condensate pump at sea level take water from a surface condenser where the vacuum is 15 in.Hg. The friction and turbulence in the piping in the condenser hot well and the pump suction flange is assumed to be 6.5 ft. If the condensate pump to be installed has a required head of 9 ft, what would be the minimum height of water level in the hot well that must be maintained above the centreline of the pump to avoid cavitation? Ans. S= 15.5 ft= hLS 3.0) A dearator heater supplies 150 l/min of dearated feed water into a booster pump at 115 0C pumping temperature. The heater pressure is maintained at 100 kPag by bled steam. Pump centreline is located 1 m above the floor level. Suction line losses is 0.60 m. Determine the minimum height of water level in the heater that must be maintained above the centreline of the pump to avoid cavitation, if the pump to be installed has a required suction head of 5.8 m. Ans. S= 3.25 m Pumps in Parallel ( to increase discharge at the same head) FIGURE:

PUMPS in SERIES

(to increase head with same discharge} FIGURE

PUMP SELECTION The following are important items to consider in selecting a pump. 1.0) The desired flow rate 2.0) The suction lift available (NPSHa) 3.0) The total dynamic head in which the pump will operate, TDH 4.0) Nature of liquid handled 5.0) Suction conditions 6.0) Type of drive, motor or engine 7.0) Type of service, continuous or intermittent 8.0) Number of units 9.0) Climate conditions

RECIPROCATING PUMP Reciprocating Pump is a positive displacement Pump wherein the pumping action is accomplished by forward and backward movement of a piston or plunger inside a cylinder usually provided with valves. A POSITIVE DISPLACEMENT PUMP makes a fluid move by trapping a fixed amount and forcing (displacing) that trapped volume into the discharge pipe. Some positive displacement pumps use an expanding cavity on the suction side and a decreasing cavity on the discharge side. Liquid flows into the pump as the cavity on the suction side expands and the liquid flows out of the discharge as the cavity collapses. The volume is constant through each cycle of operation. They are also called constant flow machines. A positive displacement pump must not operate against a closed valve on the discharge side of the pump, because it has no shutoff head like centrifugal pump. A positive displacement pump operating against a closed discharge valve continues to produce flow and the pressure in the discharge line increases until the line bursts, the pump is severely damaged, or both. A relief or safety valve on the discharge side of the positive displacement pump is necessary. Piston types are used for low pressures, light duty or intermittent service. Less expensive than plunger design but cannot handled gritty liquids. Plunger types are used for high pressure, heavy duty or continuous service. Suitable for gritty and foreign material service and more expensive than the piston design. FIGURE:

Types of Reciprocating Pumps 1. Direct acting steam pump__ This type has a steam cylinder and a common piston rod. As there is no lap, the steam is admitted throughout the length of the stroke, hence the pressure volume diagram of the steam end is a rectangle. Consequently the water end flow diagram will also be a rectangle. With the discharge flow constant throughout the length of the stroke and going down to zero value at the instant or reverse at the end of each stroke 2. Cam and flywheel Reciprocating Pump_ this type is driven by cross compound or triple expansion steam engines. 3. Power driven Pump_ this type receives its forward and backward motion of the piston and plunger by means of a crank and connecting rod. Note: Reciprocating pump can be single acting or double acting, simplex, duplex, triplex depending on the number of water cylinder on the machine. Most built in double acting. Pump Material The pump material can be Stainless Steel (SS 316 or SS 3040, CAST IRON ETC. It depends on the specification of the pump. In water industry and for pharma applications SS 316 is normally used, as stainless steel gives better results at high temperatures.

Parts of Reciprocating Pump 1. An air chamber- is a metal box installed in the discharge side of the pump in which air is maintained to cushion the flow of the water from the discharge of the pump as that the final flow will be more or less continuous. 2. Air pressure valve- this should be installed on the discharge side between the pump and any other valve.

3. Foot valve and strainer—This should be installed at the end of suction pipe. Foot valve should be of a size at least equal to the size of the suction pipe.. The area of the strainer should be at least 3 times the area of the suction pipe in order to minimize head loss at this point.

Characteristics of Reciprocating Pump 1. Piston displacement If piston rod neglected 𝛱

Vd= 2( 4 ) D2LN If piston rod considered 𝛱

𝛱

Vd= 4 D2LN + 4 (D2-d2 ) LN; where; d = diameter of piston rod 2. Q = AV 3. Slip = Vd- Q % Slip =[

𝑉𝑑−𝑄 ] 𝑉𝑑

x 100 𝑄

𝑉

4. Volumetric efficiency = 𝑉𝑑 = ηv = 1 + c – c(𝑉1) 2

Example Problem A 25 x 15 x 25 cm direct- acting duplex pump with 4 cm piston rods makes 35 double stroke per cylinder per minute and delivers 510 litres per minute of 380C water against a total head of 140 metres. The steam is supplied to the steam end at a pressure of 1034 kPa gage and exhaust to the atmosphere. Amount of steam supplied is 1725 kg/hr @ 0.98 quality. Barometric reading of 752 mm Hg. Assume that the piston makes full stroke; find: a. b. c. d.

Piston displacement of the pump in cm3/hr Pump slip Water power delivered Pump overall efficiency

Solution: Figure:

a).Solving for the piston displacement of the pump Piston displacement = displacement at the head end + displacement at crank end. Vd= VHE + VCE 𝜋

𝜋

𝜋

Vd= 4 D2LN + 4 ( D2-d2) LN = 4 LN [ D2 + ( D2 – d2) Where; L= 25 cm = 0.25 m N= 35 x 2 = 70

𝑐𝑦𝑐𝑙𝑒𝑠 60 𝑚𝑖𝑛 x 𝑚𝑖𝑛𝑢𝑡𝑒 ℎ𝑟

= 4200 cycles/hr

D= 15 cm = 0.15 m d = 4 cm = 0-.04 m 𝜋

Vd= 4 (0.25) (4200) [(0.15)2+ (0.15)2- (0.04)2] = 35.79 m3/hr Solving for the pump slip

𝑄

Vol. efficiency = 𝑉 x 100% 𝑑

Q= 510 litres/min = 0.510 m3/min 𝑚3

𝑚3

Vd = 35.79 ℎ𝑟 = 0.5965𝑚𝑖𝑛 Vol. efficiency =

0.510 0.5965

= 0.85498 x 100% = 85.498%

Pump slip = 100%- 85.498% = 14.502% Solving for power water delivered Fluid Power = ɤQH ɤ of water at 380C = 992.95 kg/m3 = 9.7375 kN/m3 Q= 0.510 m3/min = 0.0085 m3/s H= 140 m FP = 9.7375 (0 0085) (140) = 11.59 kW Solving for the pump Thermal efficiency et =

𝐻𝑒𝑎𝑡 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑝𝑜𝑤𝑒𝑟 𝐻𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑎𝑡 𝑠𝑡𝑒𝑎𝑚 𝑒𝑛𝑑

=

𝑤𝑎𝑡𝑒𝑟 𝑝𝑜𝑤𝑒𝑟 𝑚𝑠( ℎ1 −ℎ𝑓2)

ms = 1725 kg/hr = 0.479167 kg/s 1𝑎𝑡𝑚

P1= 1034 kPa + 752mmHg(760 𝑚𝑚𝐻𝐺) (

101.325 𝑘𝑃𝑎 ) 1 𝑎𝑡𝑚

= 1134.2258 kPa

From steam Tables h1= hf + x hfg = 787.393 + 0.98 ( 1995.4326) = 2742.9162 kJ/kg hf2 @ atm. Pressure atm. Pressure = 752 mm Hg = 0.00258 MPa hf2 = 417.7582 kJ/kg 11.59

et = 0.479167 ( 2742.9162−417.7582) = 1.043% Rotary pump- is a positive displacement pump consisting of a fixed casing containing gears, cams, screws, vanes, plungers or similar elements actuated by rotation of the drive shaft. Types of Rotary pumps 1) Cam and piston pump- type of rotary pump consist an eccentrically bored cam rotated by a shaft concentric cylindrically bored casing, with an aburment or follower so arranged that with each rotation of the driveshaft a positive quantity of liquid is displaced from the space between the cam and the pump casing. 2) Screw pump—is a type of rotary pump consists of two or three screw motors so arranged that as the rotors turn liquid fills the shape between the screw threads and is displaced axially as the rotor threads mesh. 3) Vane pump- a type of rotary pump consist of one rotor in a casing machined eccentrically to the drive shaft. The rotor is fitted with a series of vanes, blades or buckets which follow the bore of the casing thereby displacing the liquid with each revolution of the driveshaft. 4) Gear pump- type of rotary pump, consists two or more gears, operating closely fitted casing so arranged that when the gear teeth unmesh on one side liquid fills the space between the gear teeth and is carried around in the tooth space to the opposite side and displaced as the teeth mesh again.

1. 2. 3. 4.

Special Classification of Pump based on suction lift Shallow well pump—(ordinary centrifugal pump for suction lift up to 25 ft) Deep well Pump___ (centrifugal pump with injector for suction lift up to 120 ft) Turbine pump (multi stage pump, for suction lift up to 300 ft) Submersible pump (multi stage pump driven by submersible motor)

Deep well Pumps may divided into: 1. Plunger or reciprocating 2. Turbine 3. Ejector –centrifugal type 4. Air lifts Plunger types-a ball valve, plunger and check valves are used in this pump. In operation only the plunger moves. When the plunger is raised a vacuum is created below it, a water flows through the check valve in the plunger to be lifted on the next upward stroke of the plunger. Turbine type Pumps- These pumps represents the application of vertical centrifugal pump to dep well service and are built for heads up to 305 metres and capacities up to26295 litres/min. Ejector centrifugal pump- type of deep well pump used for small capacities combines a single stage centrifugal pump at the top of the well and an ejector o jet located down in the water This is best suited when the lifts is 7.6 m or over and capacities up to 190 liters/min net discharge Air lifts- another method of pumping wells is by air lift with compressed air being admitted to the well to lift water to the surface.

CHAPTER 3 COMPRESSORS Compressors are machines that are used to transfer another gasses from one location to another at a higher differential pressure ranges. There are various types of compressors and they are classified in two major categories: (1) dynamic or also known as continuous flow or turbo compressor and (2) positive displacement or intermittent compressor. Dynamic or continuous flow are types of compressor in which the flow of gas is accelerated thru high speed rotating element, converting velocity head into pressure head. Continuous flow includes centrifugal and axial flow compressors. Positive displacement or intermittent flow type compressor, however, are those units that confine gas in a closed space, reducing its volume and discharging it at a higher pressure. Intermittent flow includes reciprocating, screw, and root types of compressors. Some other ways of classifying compressors includes the following: a) b) c) d) e)

Number of stages: single, two, three-stages or multi-stage Reciprocating compressor element: single acting or double acting Cylinder arrangement: vertical, horizontal, V-type Cooling system: water cooled, air cooled Mounting condition: portable, stationary

RECIPROCATING COMPRESSORS Reciprocating compressors are positive displacement, intermittent flow machines that are capable of delivering air from a pressure of 35 psig up to 250 psig. They are widely used in industrial applications because of their overall efficiency, wide range of capacity simplicity of use and compactness. Some of the disadvantages of reciprocating compressors are the excessive vibrations due to the forces exhibited by the reciprocating movement of the piston. This requires strong machine foundation and higher maintenance cost compared with other types of compressor. Figure 3.1 shows a typical reciprocating compressor with its parts.

FIGURE 3.1: Portable Air Compressor

A reciprocating compressor resembles automotive engines and can be air-cooled or liquid-cooled. Generally, low capacity compressor are air-cooled and they have air fins cast around the cylinder to dissipate heat. On the other hand, liquid-cooled compressors have cooling fins and water jackets cast around the cylinder to circulate water. Reciprocating compressors needs lubrication to reduce wear, provide cooling and to act as a sealant between moving parts. Lubricant is distributed thru a splash system crank and connecting rods with oil in the crankcase.

Non-lubricating reciprocating compressors use special design piston and non-metallic piston rings without oil in the crankcase. They have higher maintenance cost because valves and piston rings wear faster than in lubricated reciprocating compressor.

FIGURE 3.2: Schematic view of air-cooled single-stage, single-acting reciprocating compressor

Single-acting Reciprocating Compressor The cylinders of reciprocating compressors can be classified as single-acting and double –acting. Single-acting compressors as shown in Figure 3.3, compresses gas in one side of the piston only. They are commonly air cooled and widely used for small capacity applications.

FIGURE 3.3: Single-acting Reciprocating Air compressor and P-v diagram

Compression stroke starts at point 1. As the compressor compress the air from point 1 to 2, the suction valves immediately closed, shutting off the suction line of the cylinder. At point 2, the discharge valves opens and compressed air is pushed out of the cylinder into the discharge line of the cylinder. The discharge stroke is completed at point 3. At this point, the piston returns to point 4, where the pressure drops and closing the discharge valve. The clearance between the end of the cylinder and the top dead center (TDC) is known as the “clearance volume”. The piston expands from point 3 to 4. And the intake stroke occurs from point 4 to 1, which is also known as the “compressor capacity”. The complete cycle is shown in the P-v diagram in figure 3.3.

Piston Speed, v Piston speed is measured as the total distance travel by the piston multiplied by the angular speed. N

v = 2L 60 Where: v = piston speed, m/s L = length of stroke, m N = compressor angular speed, rpm

Volume Displacement, VD VD = V1 – V3 N

VD = AL 60 VD = (π

D2 LN ) ∗ 4 60

(No. of cylinder) ∗ (No. of piston action)

Where: D = piston diameter, m L = stroke, m N = compressor angular speed, rpm No. of piston action = 1 for single-acting; 2 for double-acting

Double-acting Reciprocating Compressor Double-acting compressors as shown in the figure 3.4, compresses gas in both sides of the piston. For the same motor speed, double-acting compressors deliver twice of that single-acting compressors. They are water-cooled and mostly used in large capacity applications.

FIGURE 3.4: Schematic

view of double-acting reciprocating compressor

The P-v diagram of a double-acting reciprocating compressor is shown in Figure 3.5. The cycle occurs at the opposite side of the piston.

FIGURE 3.5: Double-acting Reciprocating Air compressor and P-v diagram

COMPRESSOR FUNDAMENTALS Compressor Work, Wc a) Work for Polytropic Compression n−1 n

nP1 V1 ′ P2 Wc = [( ) n − 1 P1

− 1]

b) Work for Isentropic Compression k−1 k

kP1 V1 ′ P2 Wc = [( ) k − 1 P1

− 1]

c) Work for Isothermal Compression P2 Wc = P1 V1 ′ ln ( ) P1 Note: Pressure must be in absolute values Compressor Capacity, V1’ Compressor capacity is the actual volume of air drawn in by the compressor as measured at the intake pressure and temperature.

V1 ′ = V1 − V4 V1 ′ =

ma RT1 P1

Where: R = ideal gas constant, Rair = 0.287 KJ/kg-K for air PVT Relationship 𝐧−𝟏

𝐓𝟐 𝐓𝟏

=

𝐏 𝒏 (𝐏𝟐) 𝟏

𝐕

𝐧−𝟏

= (𝐕𝟏) 𝟐

Where: 1 ˂ n ˂ 1.4 = polytropic compression n = k = 1.4 = isentropic compression n = 1 =isothermal compression Volumetric Efficiency, ƞv Volumetric efficiency describes how efficient air is being drawn into the cylinder of the aircompressor. It is the ratio of the amount of air drawn into the cylinder of the air-compressor. It is the ratio of the amount of air drawn in divided by the volume displacement or the maximum possible amount of air can be drawn in. actual volume

ƞv = volume displacement ∗ 100%

ƞv =

V1 ′ VD

∗ 100% =

And;

V1 −V4 VD

∗ 100%

V1 = V3 + VD = cVD + VD Where: c = percent clearance ranges 3% to 10% V n−1 (V4 ) 3

V4 =

n−1

=

P n (P3 ) 4

; P3 = P2, P4 = P1

1

1

P n V3 (P2 ) 1

P n cVD (P2 ) 1

=

Substituting; 1

′

ƞv =

V1 VD

∗ 100%

P (cVD +VD )−cVD ( 2 )n P1

VD

∗ 100%

Therefore; ƞv =

V1 ′ VD

∗ 100%

1

ƞv = 1 + c

P n − c (P2 ) 1

∗ 100%

It is evident that as the clearance becomes smaller, the volumetric efficiency increases. Compressor Efficiency The compressor efficiency is the ratio of the compressor power output over the brake power input to the compressor. Consider the compressor shown in Figure 3.6:

FIGURE 3.6: The compressor as the system

ƞc =

Wc BP

∗ 100%

Example: A single-acting reciprocating air compressor with a clearance of 5% receives air at 100 kPa and 30°C and is delivered at 450 kPa. The bore and stroke are 350 mm and 390 mm, respectively when operating at 1000 rpm. Determine: a) b) c) d) e) f)

Volume displacement, in m3/s Volumetric efficiency, in percent Compressor capacity, in m3/s Compressor work, in kW Compressor efficiency, in percent, if the brake power input to the compressor is 120 kW. Temperature of air at the discharge, in °C

Assume no pressure drop in the intake and discharge port of compressor and take the compression and expansion process to be PV1.3 = C.

Solving for the volume displacement, VD VD = (π VD = π

D2 LN ) ∗ 4 60

(0.350 m)2 4

𝐕𝐃 = 𝟎. 𝟔𝟐𝟓𝟒

(No. of cylinder) ∗ (No. of piston action) 1000 rpm ) (1)(1) 60

(0.390 m) (

𝐦𝟑 𝐬

Solving for the volumetric efficiency, ƞv 1

ƞv = 1 + c

P n − c (P2 ) 1

∗ 100% 1

ƞv = 1 + 0.05 − ƞ𝐯 = 𝟖𝟗. 𝟏𝟎%

450 kPa 1.3 0.05 (100 kPa)

∗ 100%

Solving for the compressor capacity, V1’ V1 ′ VD

ƞv =

∗ 100% V1 ′

0.8910 =

0.6254

𝐕𝟏 ′ = 𝟎. 𝟓𝟓𝟕𝟐

m3 s

𝐦𝟑 𝐬

Solving for the compressor work, in kW n−1

Wc =

Wc =

nP1 V1 ′ P2 n [( ) n−1 P1

− 1] m3 ) s

1.3(100 kPa)(0.5572 1.3−1

450 kPa

1.3−1 1.3

[(100 kPa)

− 1]

𝐖𝐜 = 𝟏𝟎𝟎. 𝟏𝟗𝟏𝟐 𝐤𝐖 Solving for the compressor efficiency, ƞc =

Wc BP

ƞc =

100.1912 kW ∗ 120 kW

∗ 100% 100%

ƞ𝐜 = 𝟖𝟑. 𝟒𝟗% Solving for the temperature of air at the discharge, in °C 𝐧−𝟏

𝐓𝟐 𝐓𝟏

=

𝐏 𝐧 (𝐏𝟐) 𝟏 𝟏.𝟑−𝟏

𝐓𝟐 (𝟑𝟎+𝟐𝟕𝟑)𝐊

=

𝟒𝟓𝟎 𝐤𝐏𝐚 𝟏.𝟑 ( ) 𝟏𝟎𝟎 𝐤𝐏𝐚

T2 = 428.7301 K 𝐓𝟐 = 𝟏𝟓𝟓. 𝟕𝟑 °𝐂

Example: A double acting compressor with a volume displacement of 0.432 m3/s, delivers air at 725 kPa at a rate of 0.188 m3/s. The inlet condition of air 100 kPa and 30°C and the angular speed of the compressor is 200 rpm. For a compression and expansion process given by PV1.3 = C, determine: a) The percent clearance of the compressor b) The bore and stroke, in meters, assuming that the stroke is equal with the bore and the volume displacement of the crank end and head end are the same. Solving for the percent clearance of the compressor; ƞv =

V1 ′ VD

∗ 100% =

m3 s m3

0.188 0.432

∗ 100%

s

ƞv = 43.52% Then; 1 n

P

ƞv = 1 + c − c (P2 ) ∗ 100% 1

1

0.4532 = 1 + c −

725 kPa 1.3 c (100 kPa)

𝐜 = 𝟏𝟓. 𝟕𝟑% Solving for the bore and stroke, in mm; VD = (π 0.432

m3 s

D2 ) LN 4

=π

∗ (No. of cylinder) ∗ (No. of piston action)

(D)2 4

200 rpm ) (1)(2) 60

(D) (

𝐃 = 𝟎. 𝟒𝟑𝟓𝟑 𝐦 𝐋 = 𝟎. 𝟒𝟑𝟓𝟑 𝐦

Actual Compressor Cycle An increase in area occurs in actual compressor cycle because of the fluid losses thru the inlet and discharge ports of the compressor which cause pressure drops in the cycle. Other contributors to the pressure drop in the compressor are the friction losses and fluid slippage past the piston rings, intake and discharge valve. A larger area covered by the cycle in the P-v diagram, as shown in Figure 3.7, means larger horsepower input required to the compressor.

FIGURE 3.7: Actual Compressor

P-v diagram as given by an indicator card

Example: A single acting reciprocating compressor receives air at 100 kPa and 30°C and delivered at 700 kPa. The clearance volume is taken to be 10% and the compression and expansion process to have a polytropic exponent of 1.3. The compressor piston displacement is 450 cm3 and operating at a speed of 850 rpm. Determine the mass of air compressed, in kg/hr and the required compressor power, in kilowatts when the pressure drop at the suction and discharge port are 10 kPa and 15 kPa, respectively.

Solving for the mass of air compressed, in kg/hr P1 V1 = ma RT1 ma =

P1 V1 ′ RT1

Solving for V1’, form volumetric efficiency, ƞv; V1 ′ VD

ƞv =

1

P n P1

= 1 + c − c ( 2)

Solving for the volume displacement, VD, im m3/s N

850rpm ) 60

VD = AL 60 = (4.5 × 10−4 m3 ) ( VD = 6.375 × 10−3

m3 s

Substituting to solve for the actual volume of air taken in by the compressor, V1; 1

V1 ′ (6.375×10−3

m3 ) s

= 1 + 0.10 −

715 kPa 1.3 0.10 ( 90kPa )

V1 ′ = (3.8732 × m3 3600 s ) ( 1 hr ) s m3 13.9434 hr

10−3

=

For mass of air; ma = (90kPa)(13.9434 (0.287

m3 ) hr

kJ )(30+273)K kg−K

kg

ma = 14.4307 hr

The mass taken in by the compressor was reduced by 10% because of the pressure drop at the suction and discharge port of the compressor. Solving for the compressor power required, in kW n−1 n

nP1 V1 ′ P2 Wc = [( ) n − 1 P1 Wc =

− 1]

(1.3)(90kPa)(3.8732 ×10−3

m3 ) s

1.3−1

715 kPa

1.3−1 1.3

[( 90kPa )

− 1]

𝐖𝐜 = 𝟎. 𝟗𝟐𝟔𝟒 𝐤𝐖 In general, single compressors are generally used for pressures from 35 to 100 psig, and a two-stage compressor are used from 100 to 250 psig. Multi-stage Reciprocating Compressor Two-stage compressor In multistage compressor, the gas is delivered by the low pressure cylinder at intercooler pressure to succeeding stages. In this way, the low pressure cylinder determines the volumetric efficiency of the whole machine because whatever the low pressure cylinder passes to the succeeding stages must be discharged. Figure 3.8 shows the P-v diagram of a two-stage compressor.

FIGURE 3.8: Two-stage reciprocating compressor

By using multistage compressor, the power input to the compressor is lessen, the gas discharge temperature and pressure differential are decreased. Intermediate pressure, PHX, for two-stage compressor can be theoretically approximated by: PHX = √P1 P4 Where: P1 = pressure intake at the first stage P4 = pressure at the second stage discharge The compressor work for two-stage, WC1−2 is given by; n−1

WC1−2 =

2nP1 V1 ′ PHX n [( P ) n−1 1

− 1]

The heat rejected by the intercooler; Q HX = ma Cp (T1 − THX ) Three-stage compressor Figure 3.9 shows the P-v diagram of a three-stage compressor.

FIGURE 3.9: Three-stage reciprocating compressor

For two stage compressor, intermediate pressure, PHX and PHY, can be theoretically approximated by: 3

PHX = √P1 2 P6

3

PHY = √P1 P6 2 Where: P1 = pressure intake at the first stage P6 = pressure at the third stage discharge

The compressor work for three-stage, WC1−2−3 is given by; n−1

WC1−2−3 =

3nP1 V1 ′ PHX n [( P ) n−1 1

− 1]

For ideal conditions: PHX P1

P

P

= PHY = P 6 HX

HY

Example: A single acting, two stage, reciprocating compressor receives air at 100 kPa and air capacity of 0.05 m 3 per second and delivered at 400 kPa. The clearance volume is taken to be 8%. The compression and expansion process is isentropic with compressor mean piston speed of 150 m/min. Assuming that each piston has the same amount of stroke, no pressure drops at each suction and discharge ports of the compressor and perfect intercooling, determine: a) The piston diameter, in cm b) The total power required, in kW c) The heat loss at the intercooler, in kW

a) Solving for the diameter of the piston at the first stage, d1; π VD1 = ALN = ( d1 2 ) LN 4 Piston speed:

V = 2 LN 150 m 60 s

= 2 LN

LN = 1.25

m s

Solving for the volume displacement, VD1 at the first stage; ƞ𝐯 =

V1 ′ VD1

1

=1+c−

P k c ( PHX ) 1

Intermediate pressure, PHX at the intercooler; PHX = √P1 P4 = √(100 kPa)(400 kPa) PHX = 200 kPa Solving for the volume displacement of the first stage, VD1; ƞ𝐯 =

V1 ′ VD1

1

= 1 + 0.08 −

200 kPa 1.4 0.08 (100 kPa)

= 0.9487 = 94.87%

V ′

ƞ𝐯 = V 1

D1

0.9487 =

m3 s

0.05

VD1

VD1 = 0,05270

m3 s

Substituting; π

VD1 = ( 4 d1 2 ) LN 0.05270

m3 s

π

m

= ( 4 d1 2 ) (1.25 s ) 100 cm ) 1m

d1 = (0.231 m) ( 𝐝𝟏 = 𝟐𝟑. 𝟏𝟕 𝐜𝐦

Solving for the diameter of the piston at the second stage, d2; π VD2 = ALN = ( d2 2 ) LN 4 Solving for the volume displacement of the second stage, VD2; and knowing that the volumetric efficiency is equal at the first stage and the second stage of compressor; V ′

ƞ𝐯 = V 3

D2

Solving for V3’; At the suction of the first stage:

P1 V1 ; = mRT1

At the suction of the second stage:

P3 V3 ; = mRT3

Therefore; P3 V3 ; = P1 V1 ; (200 kPa)V3 ; = (100 kPa) (0.05 V3 ; = 0.025

m3 ) s

m3 s

Substituting to solve for the volume displacement of the second stage; V ′

ƞ𝐯 = V 3

D2

0.9487 =

0.025

m3 s

VD2

VD2 = 0.02635

m3 s

Substituting; 0.02635

m3 s

π

m

= ( 4 d2 2 ) (1.25 s ) 100 cm ) 1m

d2 = (0.1638 m) ( 𝐝𝟐 = 𝟏𝟔. 𝟑𝟖 𝐜𝐦

In summary: 𝐝𝟏 = 𝟐𝟑. 𝟏𝟕 𝐜𝐦 𝐝𝟐 = 𝟏𝟔. 𝟑𝟖 𝐜𝐦 b) Solving for the total power required, WC1−2

WC1−2

k−1 k

2kP1 V1 ′ PHX = [( ) k−1 P1

− 1]

2(1.4)(100 kPa) (0.05 WC1−2 =

m3 ) s

1.4 − 1

1.4−1 1.4

200 kPa ) [( 100 kPa

− 1]

𝐖𝐂 𝟏−𝟐 = 𝟕. 𝟔𝟔𝟓𝟓 𝐤𝐖 c) Solving for the heat loss in the intercooler, QHX; From the intercooler: Q HX = ma CP (T1 − THX ) For THX; THX T1

k−1 k

P

= ( PHX ) 1

1.4−1

THX T1

=

200 kPa 1.4 (100 kPa)

THX = 1.2190T1 For ma; P1 V1 ; = ma RT1 (100 kPa) (0.05 ma =

m3 ) s

kJ

= ma (0.287 kg−K) (T1 )

17.4216 T1

Substituting; 17.4216 kJ ) (1.0062 kg−K) (T1 T1

− 1.2190T1 )

Q HX = (

𝐐𝐇𝐗 = −𝟑. 𝟖𝟑𝟗𝟎

𝐤𝐉 𝐬

Example: A reciprocating compressor receives air at 100 kPa and 0.20 m3 of air per second and delivered at 1000 kPa. Assumimg conditions are ideal, and the compression and expansion has a polytropic exponent of 1.3, determine the savings in compressor work kW, due to (a) two staging and (b) three staging. 1. Solving for the work required by the compressor for single stage, WC1; WC1

WC1 =

n−1 n

nP1 V1 ′ P2 = [( ) n − 1 P1

(1.3)(100 kPa)(0.20

m3 ) s

− 1] 1.3−1

1000 kPa 1.3 [( 100 kPa )

1.3−1

− 1]

𝐖𝐂𝟏 = 𝟔𝟎. 𝟕𝟕𝟓𝟒 𝐤𝐖 Solving for the work required by the compressor for two stage, WC1−2 ; n−1

WC1−2 =

2nP1 V1 ′ PHX n [( ) n−1 P1

− 1]

PHX = √P1 P4 = √(100 kPa)(1000 kPa) = 316.2278 kPa WC1−2 =

m3 ) s

2(1.3)(100 kPa)(0.20 1.3−1

WC1−2 = 52.7490 kW

1.3−1

316.2278 kPa 1.3 [( 100 kPa )

− 1]

Solving for the compressor work savings, in kW: Work savings = (60.7754 – 52.7490) kW Work savings = 8.0264 kW

2. Solving for the work required by the compressor for three stage, WC1−2−3 ; WC1−2−3

n−1 n

3nP1 V1 ′ PHX = [( ) n−1 P1

− 1]

3

3 PHX = √P1 2 P6 = √(100 kPa)2 (1000 kPa) = 215.4435 kPa

WC1−2−3 =

m3 ) s

3(1.3)(100 kPa)(0.20 1.3−1

1.3−1

215.4435 kPa 1.3 [( 100 kPa )

− 1]

WC1−2−3 = 50.3818 kW Solving for the compressor work savings, in kW: Work savings = (60.7754 – 50.3818) kW Work savings = 10.3936 kW We can say that by having multiple compressor stages decreases the required work of the compressor, thus, increasing the power savings, in kW. General Formula for Multi-stages Compressor,(“m” number of stages) The compressor work, WC m is given by; n−1

WC m =

mnP1 V1 ′ PHX n [( P ) n−1 1

− 1]

The intermediate pressure, PHX can be theoretically approximated by: m

PHX = √(P1 )m−1 PF Where: P1 = suction pressure PF= final or discharge pressure

COMPRESSOR SELECTION The following are important items to consider in selecting a compressor:     

The desired capacity, V1’ The required discharge pressure The foundation required The characteristics of gas to be handled Type of control required

Other things to be considered in the selection are the space requirements, availability of intercooler cooling water, maintenance costs, power source and economics. Reciprocating compressors are generally driven by a belt. Similar with belt driven fans, one advantage is that the belts tend to slip that reduce load in the motor during start up. But on the other hand, maintenance cost could be high in belt driven compressors because belts are designed to break to some degree of usage.

COMPRESSOR INSTALLATION

The following are some of the guidelines and precautions in installing a compressor. It is also important to consult the compressor manufacturer for additional recommendation and further information regarding the compressor to be used. These outlines are similar with the other fluid machineries installation described in previous chapters.      

It is necessary to have a good foundation for the compressor unit in order to stabilize the vibration that occurs during its operation. Inlet filter should be installed in the suction of the compressor to protect the unit from dust, foreign object, moisture and corrosive elements present that may damage the compressor. To reduce friction and turbulence, all compressed air piping system lines should be short and straight as possible, with minimum elbows, valves and fittings. One of the challenges in maintaining compressed air piping system is the collection of water in the pipelines. A suitable water drain or collector should be available at each point in the line Install the compressor in a secure location and at the same time, available for inspection and maintenance. Silencers may be used if noise level is a primary consideration

Air Receivers It is necessary for all reciprocating compressor to have an air receiver to eliminate the pulsation of air delivered. It is also used to store compressed air, and to condense some of its moisture content by cooling it down. Air receiver size, VT can be theoretically calculated from; VT =

V1 ′ P1 P2

Air receiver size depends on the capacity of the compressor, pipeline network and pressure drop, air consumption of the plant and switching cycle per unit-time of the compressor operation.

PROBLEM SET: 1. An air compressor delivers air at a flow rate of 0.25 m3/s with initial pressure of 100 and discharge pressure of 680 kPa. Determine the power of the compressor. (Ans. 63.8) 2. A double acting compressor with 225 mm x 380 mm cylinder runs at 480 rpm with a clearance of 8%, compressor air 7 times its initial pressure. Compute the compressor capacity in cubic feet per minute. (Ans. 388.3 ft2/min)

3. A single stage compressor with initial pressure of 100 kPa and discharge pressure of 850 kPa has a suction volume of 0.3 m3/s. Determine the percent decrease in power due to two staging if the compressor process is PV1.33 = C. (Ans. 13.2%) 4. An air compressor takes air at 98 kPa at a rate of 0.4 m3/s and delivers it at a pressure of 620 kPa. If the power input to the compressor is 135 kW, find the heat loss in the compressor. (Ans. 39.8 kW) 5. A single stage compressor with a suction pressure of 15 psi discharge ar at a pressure of 75 psi. If the suction volume is 110 ft3/min, determine the horsepower capacity of the motor needed to drive the compressor if the compressor efficiency is 80%. (Ans. 183.9 hp) 6. A two stage compressor has a suction volume of 720 m3/hr at 100 kPa and 26°C, the discharge is 700 kPa, determine the heat rejected at the intercooler. (Ans. 22.5 kW) 7. An air compressor has a suction volume of 720 m3/s and discharges to 650 kPa. If the power input to the compressor is 120 kW, find the heat loss in the compressor. (Ans. 5.3 kW) 8. A two stage compressor takes air at 98 kPa and 24°C with a volume flow rate of 0.122 m3/s and discharges to 680 kPa. What is the amount of heat rejected at the intercooler? (Ans. 13.4 kW) 9. The piston speed of an air compressor is 155 m/min and the displacement volume is 0.25 m3/s. Find the diameter of the compressor cylinder. (Ans. 504.6 mm) 10. Determine the volume displacement of a double acting compressor that has cylinder dimension of 52 cm x 65 cm and runs at 660 rpm. (Ans. 3.04 m3/s)

11. An air compressor has suction condition of 98 kPa, 28°C and 0.22 m3/s. If the surrounding air 100 kPa and 23°C, calculate the free air capacity in m3/s. (Ans. 0.212 m3/s) 12. An air compressor has a suction volume of 0.35 m3/s at 28°C and 101.325 kPa and discharges to 680 kPa. Determine the amount of power saved by the compressor in two staging? (Ans. 14.3 kW) 13. The initial condition of air is 100 kPa and 24°C and compressed to 550 kPa. The bore and stroke are 365 mm and 380 mm, respectively. If the percent clearance is 8% and runs at 320 rpm, determine the mass flow rate of air at the suction, (Ans. 0.20 kg/s) 14. A 15 hp motor is used to drive an air compressor. The compressor efficiency is 75% and the air is available at 29%. Find the mass of air needed if the pressure compresses to 7 times initial pressure. (Ans. 2.23 kg/min) 15. The discharge pressure of an air compressor is 6 times the suction pressure. If the volume flow at the suction is 0.6 m3/s, find the power required by the compressor operating at PV1.33 = C with a suction pressure of 100 kPa. (Ans. 135.4 kW)

CHAPTER 2 Fans and Blowers Fans and blowers are machines that move air or gases under moderate pressures. They are significant and important component in all industrial plant specifically if there is a need in supplying, circulating and removing air in a space in order to provide comfort and safety. They are also utilized for heating, ventilating, air conditioning and pollution control needs for industrial plants.

FIGURE 2.1 Industrial Fans (Picture courtesy of Pollrich DLK Fan Factories)

Although fans and blowers, including compressors (to be discussed in the next chapter) performs a similar function of adding energy and moving gasses, they all differ in terms of their pressure ranges. Generally, fans works at low pressures from 1 psig, while blowers (also known as turbo-blowers) works up to 35 psig and compressors starts at 35 psig.

TYPE OF FANS Two general types of fans are Centrifugal Fans and Axial Fans 1. Centrifugal Fans Centrifugal fans moves air radially thru the impellers. Air drawn at the center of a revolving wheel, connected to a rotating shaft, and then enters the spaces of the wheel blades. The air is then thrown out perpendicularly to the impeller’s axis of rotation at high velocity and static pressure. They are widely used in comfort applications because it’s quite, efficient and operates at relative high pressure heads. Centrifugal are classified in terms of wheel blades: (1) Forward- curved, (2) backward curved and (3) radial

FIGURE 2.2 Centrifugal Fan

2. Axial Fans Axial fans moves air in a helical-type flow pattern that is parallel thru the impeller. This type of fan moves large amount of fluid and develops small pressure heads. They are widely used in applications where noise levels are of secondary. Axial fans are classified as: (1) Propeller, (2) tubeaxial and (3) vaneaxial

TYPES OF CENTRIFUGAL FANS Forward-Curved Fan

A forward-curved fan or also known as “squirrel cage fan” is generally used for high-volume air flow applications and low to medium static pressures applications. Widely used in heating and ventilating work because of its quite operation, fan should be operated at relatively clean air because contaminants might clog its blades.

A typical performance curve of forward-curved fan is shown in Figure 2.3. As we can observe, the horsepower is at minimum value at no delivery and the highest efficiency occurs when the fan is operating at 35-45% of free delivery capacity. Furthermore, the behaviour of the horsepower input curve increases as the air delivered increases, while the static pressure decreases. Forward-curved fan is “overloading fan” because the motor may overload when the static pressure is below its design value.

FIGURE 2.3: Forward-Curved Blade Performance Curve and Blade Profile

An example of this is, let’s say, a fan is designed to operate at 40% of free delivery capacity and at 95% static pressure. The motor installed for this fan to operate properly is say, with maximum capacity of 4 bhp and is operating at 3.5 bhp. If fir instance, the fan access door is removed, the static pressure or the system resistance will drop, say at 30% and at the same time the free air delivered will increase to 80%, at this point the motor must operate at a higher rate, say a 5 bhp, which is greater than the power capacity of the installed motor, in this case the motor will stop on overload.

Backward-Curved Fan Backward-curved fan or also known as “limit loading fan” because of the backward profile of its blade and is widely used for high-static pressure applications. This type of fan is more efficient than forward-curved with efficiency between 50-70% of free delivery capacity, and is quieter than any other types of fan. A typical performance curve of a backward-curve fan is shown in Figure 2.4. It has a minimum horsepower at no delivery. The horsepower curve gradually increases at the range of maximum efficiency and then gradually decreases. Backward-inclined and airfoil blade fan are modifications of backward-curve fan. Airfoil blade fan has a high efficiency because of its aerodynamic blade profile which allows smooth flow of air thru its blades.

FIGURE 2.4: Backward-Curved Blade Performance Curve and Blade Profiles and Modifications

Radial Fan

that require high velocities and high pressure heads. This type of high dust loading in gasses because of its self-cleaning wheel suited for high temperature applications.

Radial (straight) fans are widely used in waste collection systems fan is preferred for design and well

A typical performance curve of a backward-curve fan is shown in Figure 2.5. The horsepower input curve increases as the air delivered increases, while the static pressure decreases. This fan is an overloading fan and the least efficient of all centrifugal fans.

`

FIGURE 2.5: Radial Blade Performance Curve and Blade Profile

TYPES OF AXIAL FANS Propeller Fan Propeller fans consist of two or more blades and are widely used for general ventilation systems in which require large volume of air at low pressure heads. It is used ad exhaust system for indoor applications and for air-cooled condensers and cooling towers for outdoor applications. Propeller fans are relatively noisy because of the turbulence created when air is discharged in a circular or helical pattern. This type of fan has a simple ring enclosure for its housing and no ducting system is required. A propeller fan operates generally at 0 to 1 inch of water gage or less.

FIGURE 2.6: Propeller Fan

A typical performance curve of a backward-curve fan is shown in Figure 2.7. The horsepower decreases as the percent of free delivery increases and the lowest at maximum percent of free delivery. Horsepower also decreases as the static pressure decreases. The highest efficiency of propeller fans occur when it is delivering 50%-70% free delivery capacity.

FIGURE 2.7: Propeller Performance Curve

Tubeaxial Fan Tubeaxial fans are axial fans that are fabricated in a tubular casing. They are widely used in ducted ventilating applications that require medium pressure heads and where air flow pattern in downstream of the fan are not of primary consideration, such as in fume exhaust system, drying ovens and paint spray bottles. The blades of tubeaxial fans are similar to propeller fans, except that it usually has 4 to 8 blades and is designed for heavy duty applications.

FIGURE 2.8 Tube Axial Fan

Typical performance curve of axial fans are presented in Figure 2.9. The efficiency and static pressure curve patterns of tubeaxial fans has a significant improvement compared with propeller fans.

FIGURE 2.9: Axial Flow Performance Curve

Vaneaxial Fan Vaneaxial fans are tubeaxial fans with guide vanes. They operate in medium to high pressure heads and are widely used when good stream air distribution is needed.

FIGURE 2.10 Vaneaxial Fan

The general advantages of axial fans are in terms of its simplicity and its installation, small space requirements and economy. They are widely used in applications where high volume of air is needed against a low pressure heads and noise is not a primary consideration.

FAN FUNDAMENTALS Air Power, AP

Air power is the energy added by the impeller of the fan to air or gasses to move it against a particular pressure. Air power is referred to as “total air power” when total pressure head, HT is used, and as “static air power”, when static pressure head, hs is used in the equation.

Total Air Power AP = Qγa HT

g

γa = ρa [go ] = c

ma go [ ] va gc

g

AP = ma [go ] HT c

Where: AP = Total air power, kW or hp 1 hp = 0.746 kW Q = v = volume flow rate, m3/s or ft3/sec γa = specific weight of air For air at standard condition @ 21°C and 101.325 kPa: SI units: 0.0118 kN/m3 Eng units: 0.075 lbf/ft3 ma = mass flow rate of air,

kg or lbm/sec s

HT = total pressure head of the fan meters of air or feet of air

Static Air Power

APstatic = Qγa hs

g

APstatic = ma [ga ] hs c

Where: APstatic = Static air pressure, kW or hp hs ͌ = static pressure head of the fan meters of air or feet of air

The Total Fan Pressure Head, HT Total fan head, the total amount of work needed by the fan (usually measured in meter or feet) per specific weight of air flowing through the fan. It is the sum of the velocity head and static pressure head. hw(Water Gage, WG)

Manometer Fluid: Water SG = 1

AIR FLOW

AIR FLOW

FIGURE 2.11 Fan configuration showing velocity and static pressure head

Consider a fan configuration in Figure, applying conservation of energy: [Ein = Eout]

PE1 + KE1 + Wf1 + U1 + AP = PE2 + KE2 + Wf2 + U2 AP = (PE2 – PE1) + (KE2 – KE1) + (Wf2 – Wf1) + (U2 – U1)

If we assume that the temperature and elevation from the suction and discharge of the pump are almost equal, t1 ≈ t2 and z1 ≈ z2, then we can say that the change in internal energy and potential energy are negligible, ∆U ≈ 0 and ∆PE ≈ 0

AP =

1ma (v2 2 2gc

− v1 2 ) + Va (P2 − P1 )

g

Multiplying both sides of the equations by m cg a

gc AP ma g

=

1(v2 2 −v1 2 ) 2(go )

+

(P2 −P1 ) ma go [ ] va gc

Therefore: HT =

(P −P ) 1 (v2 2 −v1 2 ) + 2 1 2 go γa

HT = (velocity head) + (static pressure head) Or; HT = hv + hs

Velocity Head, hv Velocity head is the energy possessed by air because of its velocity. This has a significant effect in fan performance and must be considered in the calculation. Velocity head is given by the equation: 1 (v2 2 −v1 2 ) go

hv = 2

Static Pressure Head, hs Static pressure head is the energy required to overcome resistance. Static pressure head becomes irrelevant when the fan operates against no resistance (e.g. no ducting systems or dampers) and when the fan outlet velocity is high. To determine the capacity of the fan to be installed, it is necessary to know the resistance characteristics of the system. A fan operating in a system with narrow ducting systems, multiple short elbows, bends and twists will require more power to overcome system resistance, compared with a fan operating with a larger ducting system and minimum elbows, bends and twists. Static pressure head is given by the equation: hs =

(P2 −P1 ) γa

=

γw hw γa

Therefore, we can say that static pressure head is also equivalent to:

Or;

hs = hw

γw γa

hs = hw

ρw ρa

Where: ρw = density of water For water at standard condition SI units: 1000 kg/m3 Eng units: 62.4 lbm/ft3 Ρa = density of air For air standard condition @ 21°C and 101.325 kPa SI units: 1.2 kg/m3 Eng units: 0.075 lbm/ft3 Example: Air enters a fan at an initial velocity of 7 m/s and a static vacuum pressure head of 22.5 mm of water. The air is delivered at a velocity of 12 m/s and a static pressure head of 81 mm of water. The fan has a capacity of 9.5 m3/s and is coupled to a 15 kW motor. Determine the total air power if the density of air is 1.2 kg/m3. Solving for the total air power, AP; AP = Qγa HT Solving for the total pressure head; Ht = hv - hs 1 (v2 2 −v1 2 ) hv= 2 go

hs =

m2

=

2 2 1 (12 −7 ) s2 2 9.81 hs

= 4.841997961 m of air

m

(P2 −P1 ) γa

P2 = γw hw2 = (9.81

kN ) (0.081m) m3

P1 = γw hw1 = (9.81

kN ) (−0.0225m) m3

g

hs =

(0.79461+0.220725)kPa kN

0.011772 3 m

= −0.220725 kPa

m

9.81 2 s

kg

γa = ρa [ga ] = (1.02 m3 ) ( c

= 0.74461 kPa

kg−m kN−s2

1000

kN

) = 0.011772 m3

= 86.25 m of air

HT = (4.8420 + 86.0424) m of air HT = 90.8844 m of air Substituting HT, to solve for AP; AP = Qγa HT = (9.5 𝐀𝐏 = 𝟏𝟎. 𝟏𝟖𝟕 𝐤𝐖

m3 kN ) (0.011772 m3 ) (91.091999 s

m)

Example: Air is flowing with an initial and final velocity of 1.2 m/s and 8.3 m/s, respectively, against a static pressure of 2.25 cm water gage. The duct diameter is 1.5 cm and the condition of air is at 99.5 kPa and 30°C. Determine the total pressure in which the fan must operate (a) in meters of air and (b) in centimeters of water. (a) Solving for the total pressure in meters of air HT = hv + hs hv =

ms

2 2 1 (v2 2 −v1 2 ) 1 (8.3 −1.2 ) s2 [ ] m 2 go 2 9.81 2

= 3.4378 m of air

s

hs = hw ρa =

ρw ρa

m v

P

= RT =

99.5 kPa KJ (0.287 )(30+273)K kg−K

kg

= 1.1442 m3

kg

hs = (0.0225m)

(1000 3) m kg

(1.1442 3) m

= 19.6644 m of air

HT = (3.4378 + 19.6644) m of air 𝐇𝐓 = 𝟐𝟑. 𝟏𝟎𝟐𝟐 𝐦 𝐨𝐟 𝐚𝐢𝐫 (b) Solving for the total pressure in centimeters of water HT = hv + hs kg

hv = (3.4378 m of air)

(1.1442 3) m kg

(1000 3) m

= 3.9335 × 10−3 m of water

hv = 0.39335 cm of water Hs = 2.25 cm of water 𝐇𝐓 = 𝟐. 𝟔𝟒𝟑𝟒 𝐜𝐦 𝐨𝐟 𝐰𝐚𝐭𝐞𝐫

Fan Efficiency The fan efficiency is the ratio of the air power output of the fan over the brake power input to the fan. It shows how effective the fan in converting brake power input into air power. Consider the fan shown in Figure 2.12:

FIGURE 2.12 The as the system

There are two ways on how manufacturer’s describes the fan efficiency: the (1.) fan total or the mechanical efficiency and (2.) fan static efficiency. The fan efficiency is known as “Fan total efficiency,” if total air power is used, and as “Fan static efficiency,” if static air power I used. Fan total efficiency: AP

ƞF = BP ∗ 100% =

Qγa HT BP

∗ 100%

Fan static efficiency: ƞFstatic =

APstatic BP

∗ 100% =

Qγa hs BP

∗ 100%

Example: Determine the brake horse power required for a fan with a static efficiency of 50% and delivers air at 1200 cfm. The static pressure developed is 2.5 inches of water. Air is at 60°F and barometric pressure of 30 inches mercury.

Solving for BHP; ƞFstatic =

APstatic BHP

∗ 100%

APstatic = Qγa hs Specific weight of air, hs = hw

γw γa

1200 ft3 lb 2.5 ) (62.4 ft3f) ( 12 ft) 60 sec

APstatic = Qγw hw = ( APstatic = (260

lbf −ft 1hp ) ( lbf−ft) sec 550 sec

= 0.4727 hp

Therefore; 0.50 =

0.4727 hp BHP

BHP = 0.9455 hp

Example: A fan with a capacity of 1.5 m3/s draws air at a static pressure of 3.0 cm of water through a duct. The air drawn is measured at 27°C and 760 mmHg. If the inlet and outlet duct is 320 mm and 280 mm, respectively, determine the fan static efficiency when the total fan efficiency is 80%. Solving for fan static efficiency, ƞFstatic =

APstatic BP

∗ 100% γ

APstatic = Qγa hs and hs = hw γw , hence; a

APstatic = Qγw hw = (1.5

m3 s

kN

) (9.81 m3 ) (0.03 m)

APstatic = 0.4415 kW

Solving for the brake power input, BP, from total fan efficiency; AP

ȠF = BP ∗ 100% AP = Qγa HT HT = hv + hs 1 v2 2 −v1 2 ) go

hv = 2 (

Solving for the velocity of air at the inlet and exit ducts, using; Q = A1 v1 1.5

m3 s

=𝜋

(0.320 m)2 4

v1

v1 = 18.6510 m/s Q = A2 v2 1.5

m3 s

=𝜋

(0.280 m)2 4

v2

v2 = 24.3605 m/s

Solving velocity head, hv m2

hv =

2 2 1 (24.3605 −18.6510 ) s2 2 m 2 9.81

= 12.5165 m of air

s2

Solving for the static pressure head, hs hs = hw

γw γa

ρa =

m v

P

γa =

g ρa [go ] c

= RT =

101.325 kPa

kg

kJ (0.287 )(27+273)K kg−K

= 1.1768 m3

m

=

9.81 2 kg s 1.1768 m3 [ kg−m ] 1000 kN−s2

kN

hs = (0.03 m) (

9.81 3 m kN

0.01154 3 m

) = 25.5026 m of air

The total fan pressure head; HT HT = (12.5165 + 25.5026) m of air

kN

= 0.01154 m3

HT = 38.0191 m of air For total air power, AP; AP = (1.5

m3 kN ) (0.01154 3 ) (38.0191 m) s m

AP = 0.6581 kW Substituting to the total fan efficiency, to solve for the brake power input, BP; 0.08 =

0.6581 kW ∗ BP

100%

BP = 0.8226 kW Then for the fan static efficiency; ƞFstatic =

0.4415 kW ∗ 0.8226 kW

100%

ƞ𝐅𝐬𝐭𝐚𝐭𝐢𝐜 = 𝟓𝟑. 𝟔𝟕%

FAN SPECIFIC SPEED Fan specific speed is a dimensionless parameter used to describe the applications of different types of fans. It is defined as the revolution per minute at which a given geometrically similar impeller of a fan would operate if reduced proportionally in size so as to deliver a rated capacity of 1 cubic feet per minute against a differential head of 1-foot of air under standard conditions. Ns =

N√Q 3 hs ⁄4

Where: Ns = specific speed of the fan, rpm Q = capacity of the fan, cfm hs = fan static pressure head, ft. of air N = impeller speed, in revolution per minute FAN TYPE AND SPECIFIC SPEED Figure shows the specific speed of centrifugal and axial fans based in speed, fan capacity and static pressure. It shows that forward-curved fans generally operate at low speeds while propeller fans operate at high speeds to attain its peak efficiencies.

FIGURE 2.13: Specific speed ranges (ref. Handbook of Air conditioning System Design Carrier Air Conditioning Design Company)

FAN AFFINITY LAWS Fan affinity laws are rules that express the relationship of fan capacity, head (either static or total), BHP input and air density when one of each parameter is changed. In applying the following equations, we consider that the efficiency is the same for both conditions.

Capacity

Variation in speed N1 ≠ N2;ρ1 = ρ2

Variation in density N1 = N2;ρ1 ≠ ρ2

Variation in density N1 ≠ N2;ρ1 ≠ ρ2

Q1 N1 = Q 2 N2

Q1 = Q 2

Q1 N1 = Q 2 N2

Head

h1 N1 2 =( ) h2 N2

h1 ρ1 = h 2 ρ2

h1 ρ1 N1 2 = ( ) h 2 ρ2 N 2

BHP

BHP1 N1 3 =( ) BHP2 N2

BHP1 ρ1 = BHP2 ρ2

BHP1 ρ1 N1 3 = ( ) BHP2 ρ2 N2

Where: Q = Capacity, in gpm h = Total head (static or total), in feet BHP = Brake horsepower, in hp N = Fan speed, rpm Subscript 1 refers to initial condition and subscript 2 refers to the new condition. Example: Determine the performance of a fan with a static pressure head of 6.20 cm of water, speed of 410 rpm and initial power input of 3 kW. The capacity is changed from 5.5 m3/sec to 8.6 m3/sec. In this example problem, fan capacity (which is a function of motor speed) is changed while maintaining a constant air density; Q1 Q2

N

= N1 2

5.5 8.6

m3 s m3 s

=

410 rpm N2

N2 = 641.0909 rpm h1 h2

N 2 N2

= ( 1)

6.2 cm of water H2

410 rpm

2

= (614.0909 rpm)

h2 = 15.1587 cm of water BP1 BP2

N 3 N2

= ( 1)

3 kW BP2

410 rpm

3

= (614.0909 rpm)

BP2 = 11.4691 kW Example: A fan delivering air through a ducting system that has a capacity of 100m3/s, static pressure head of 30 cm of water, speed of 420 rpm and initial power output of 600 kW. Determine the performance of the fan if the air is to be delivered at 85°C and a barometric pressure 734 mmHg. In this example problem, the fan speed and capacity remains constant while the air density is changed because of the increase in temperature from standard air temperature of 21°C to 85°C. Q1 = Q 2 Q1 = 100 m3/s h1 h2

=

ρ1 ρ2 kg

ρ1 = 1.2 m3 at standard condition of 101.325 kPa and 21°C. ρ2 =

m v

=

P2 RT2

=

kg

ρ2 = 0.9524 m3 30 cm of water h2

kg

=

1.2 3 m kg

0.9524 3 m

(734 mmHg)(

101.325 kPa ) 760 mmHg

KJ )(85+273)K Kg−K

(0.287

h2 = 23.81 cm of water BP1 BP2

ρ

= ρ1 2

600 kW BP2

kg

=

1.2 3 m kg

0.9524 3 m

BP2 = 476.20 kW Example: A fan has a capacity of 80 m3/s, static pressure head of 22 cm of water, speed of 380 rpm and initial power input of 400 kW. Determine the performance of the fan when the air is to be delivered at 70°C and at the speed of 410 rpm. In this example problem, the fan speed and air density is changed. Q1 Q2

N

= N1 2

80

m3 s

Q2

380 rpm

= 410 rpm

Q2 = 86.3158 m3/s h1 h2

2

N

= (N1 ) 2

kg

ρ1 = 1.2 m3 at standard condition of 101.325 kPa and 21°C. m v

ρ2 =

P

= RT2 = 2

101.325 kPa (0.287

KJ )(70+273)K Kg−K

kg

ρ2 = 1.0293 m3 22 cm of water h2

kg

=

1.2 3 m kg 1.0293 3 m

380 rpm 2

(410 rpm)

h2 = 21.9677 cm of water BP1 BP2

ρ

3

N

= ρ1 (N1 ) 2

400 BP2

2

kg

=(

1.2 3 m kg 1.0293 3 m

380 rpm 3

) (410 rpm)

BP2 = 430.9447 kW

FAN SELECTION The following are important items to consider in selecting a fan:  The desired flow rate, Q  The static pressure head, hs  The characteristic of air to be handled (e.g. dust concentration)  Operating temperature  The prevailing noise level of the space to be served Other things to be considered in the selection are the space requirements, maintenance costs, space requirements, power source and economics. Fans can be directly coupled to a motor or driven by a belt. One advantage of belt driven fans is that the belts tend to slip that reduce load in the motor during start up. Maintenance cost could be high in belt driven fans because belts are designed to break to some degree of usage. Sometimes it is necessary to vary the flow rate of air to balance the flow rate of air delivered in the system. This can be accomplished by varying the speed of the motor. Or, if the fan is belt driven, a change in the drive pulley can also be made. Another method is by installing dampers in the duct system, which increases the static pressure of the fan.

FAN INSTALLATION The following are some of the guidelines and precautions in installing a fan. It is also important to consult the fan manufacturer for additional recommendations and further information regarding the fan to be used.

These outline are similar with the outline given for pump installation. The only difference is the type of fluid they handle and the Net Positive Suction Head, NPSH considerations are critical for pump installation.       

To reduce friction and turbulence, all ducting lines should be short and straight as possible, with minimum elbows, valves and fittings. Elbows in ductings should have small radius. Both the suction and discharge lines should be supported independently to protect the casing from strains that may cause distortion to the fan. Flexible ducting system can also be used to isolate the vibration of the fan. Install the fan away from heat source if the fan takes air from the atmosphere so that the air will be cooler. The suction of the fan should be protected from dust, foreign object, moisture and corrosive elements present that may damage the fan. Silencers may be used if the noise level is primary consideration. The inlet damper positioning should be checked regularly to satisfy the closed and fully open condition of the fan. Install the fan in a secure inspection and maintenance.

FANS IN SERIES AND PARALLEL Similar with pumps, fans can also be installed in series and or in parallel operation. Fans in Series Fans in series are done by staging multiple fans as shown in Figure 2.14. The total pressure is increased at a given capacity as shown in the performance curve.

TWO FANS IN SERIES

FIGURE 2.14: fans installed in series

Fans in Parallel Fans in parallel are the result of installing multiple fan side by side as shown in Figure 2.15. The capacity is doubled while maintaining the total pressure.

TWO FANS IN PARALLEL

FIGURE 2.15: Fans installed in parallel

PROBLEM SET: 1. The power output of a fan is 120 kW with efficiency of 80%. Determine the horsepower output required by the motor to drive the fan. (Ans. 201 hp) 2. The static head of a fan is measured to be 160 mm of water gage at an air velocity of 25 m/s. Find the air power at an air condition of 28°C and 98 kPa with a volume flow rate of 5 m3/s. (Ans. 9.6 kW) 3. Determine the horsepower required for a fan delivering 35 ft/sec of air through a 2.3 ft x 3.5 ft duct with a total pressure of 3.5 in water gage. Take density of air to be 0.075 lb/ft3. (Ans. 9.3 hp) 4. A fan operating at a standard air condition registered a total static head of 230 mm of water gage. If the static efficiency is 65% and the fan efficiency is 80% determined the velocity of air if the volume of air delivered is 6 m3/s. (Let the velocity head to be 35% of the static head). (Ans. 34.6 m/s) 5. A fan initially operating at a speed of 380 rpm at an air temperature of 26°C. If the speed is increased to 460 rpm with 55°C, determine the new head in mm of water gage for an initial head 190 mm of water gauge. (Ans. 253.8 mm) 6. What is the horsepower required for a fan to deliver 230 ft3/sec of air through a 2.5 ft x 4.5 ft duct under a total pressure of 4.2-in water gage? (Ans. 0.35 hp) 7. The rating of a fan is 610 m3/min when running at 360 rpm and requires 8 kW motor to drive it. If the fan speed is increased to 620 rpm and the air handled becomes 60°C instead of 30°C, determine the power in kW. (Ans. 37.1 kW) 8. A fan has a static head of 110 m at a pressure of 1.5 kg/cm 2 and 75°F. If the air velocity is 18 m/s, determine the total equivalent head in mm of water gage. (Ans. 145.61 mmWG)

HYDRAULIC TURBINES Fluid machines are those machines which are used to convert mechanical energy into fluid energy or vice versa. Machines that coverts fluid energy into mechanical energy are known as turbines. Machines that converts mechanical energy into fluid energy are known as pumps. Water turbines. Turbines are devices to convert energy of water into mechanical energy which can be used for running electricity generators.

Types of Hydraulic Turbines 1) Reaction turbines 2) Impulse turbines

In reaction turbines, pressure head of water is converted into velocity head as water flows through the turbine. Reaction turbines run full of water and hence the turbine maybe entirely submerged below the tail race or may discharged into the atmosphere. It may also discharge into a suction or draught tube when placed 9.14 m above the foot of fall. In these turbines water must enter over the whole circumference. In impulse turbine, the energy of water is converted into velocity before entering the wheel at atmospheric pressure and this will not permit the turbine tube to be flooded. The water is passed through nozzle or guide vanes for converting all its energy into velocity. Impulse turbine must be placed at the foot and above the tail race. Hence these turbines can be inspected easily. The water maybe entered over entire circumference or part of the circumference. If H is the head at which water is available and leaves the turbine at velocity V1, then the energy available from turbine or energy of water absorbed by turbine will be; H=

𝑉12 2𝑔

However, the impulse turbine, this energy H will first be converted into velocity or kinetic energy at velocity V. Both reaction and impulse turbines can be further classified depending upon direction of flow. The flow is axial in axial flow turbines and radial and radial flow turbines. Mixed flow turbines are also in used. The radial flow turbine may have flow from centre to circumference or vice-versa. Commonly used turbines are: a) b) c) d)

Kaplan turbine/Propeller type, reaction turbine suitable for very low heads. Francis turbine. A reaction turbine suitable for medium heads. Pelton wheel. An axial flow impulse turbine for high heads Turgo wheel. An impulse turbine suitable for medium heads.

In an inward radial flow turbine fixed guide blades surround the revolving blades externally whereas in an outward flow turbine the ring of moving blades surrounds the fixed guide blades The water leaves through the centre in latter case while it enters the turbine though centre in former. In outward flow turbine the relative velocity of water is increased and hence quantity of water passing through the turbine is increased. This tendency would not exists in inward flow turbines. In fact, this case may increase the speed of the wheel will tend to decrease the flow through the wheel and reduce power. In axial flow turbine, the guide blade and rotating blade rings are mounted side by side and water flows from guide blades into moving blades parallel to the axis of turbine. For this reason the turbines are called parallel flow.

Types of Hydraulic turbines 1.0)Impulse turbine also known as Tangential wheel or Pelton wheel It is a Pressureless Turbine __ a turbine that utilizes kinetic energy of high velocity jet which acts upon a small part of the circumference at an instant. __ Movement of the water is tangential

__ Suited for very high heads plants (150 m and above) and low volume of water __ No exact value for critical head, hence heads are given in range __Impulse turbine has no draft tube __ Typical turbine efficiencies are in the range of 80% to 90% __Higher efficiencies are associated with turbines having two or more runners. 2.0) Reaction Turbines (Francis type or Propeller type). It is Pressure Turbine __ a turbine which develops power from the combined action of pressure and velocity of water that completely fills the runner and the water passages __Movement of water for reaction turbines can be radial for Francis type and axial for Propeller/Kaplan type __Reaction turbine has draft tube which keeps the turbine up to 5m( 15 ft) above the tail water __Reaction turbine conversion efficiency is usually higher than that of Impulse turbine __Reaction turbines’ conversion efficiency is about 85% to 95%

Turbine type recommended based on head Net head

Type of turbine

Up to 70 ft

Propeller type

70 to 110 ft

Propeller or Francis type

110 to 800ft

Francis type

800 to 1300 ft

Francis type or Impulse type

1300 and above

Impulse type

Types of reaction turbines A. Francis Type--- for medium head B. Propeller and Kaplan type Reaction turbine for very low head Propeller has fixed blade. A type of reaction turbine with reduced number of fixed blades. The flow is axial. Suited for low head plant and has usual conversion efficiency of 80% Kaplan type has adjustable blades. The flow is inward flow axial. Suited for low headed and large volum e of water and usual conversion efficiency of more or less than 93%.

Performance of Hydroelectric Power Plant 1. Gross head, hg---- is the difference between water and tail water elevations . hg= hhw - htw where: hhw =headwater elevation htw = tail water elevation 2. Friction head loss,hf Friction head loss- is the head lost by the flow in a stream or conduit due to frictional disturbances set up by the moving fluid and its containing conduit and by intermolecular friction. Using Darcy equation 𝑓𝐿𝑉 2

hf= 2𝑔𝐷

Using Morse equation hf =

2𝑓𝐿𝑉 2 𝑔𝐷

where: hf= friction head in meter f = coefficient of friction L = total length in m

g = 9.81 m/s2 D= inside diameter in m Note: Friction head loss is usually expressed as a percentage of the gross head 3. Net head or Effective head is the difference between the gross head and the friction head loss H = hg-hf 4. Penstock efficiency is the ratio of the net head to the gross head ℎ ep = ℎ𝑔 5. Volume flow rate of water, Q –it is the product of the velocity and the cross-sectional area. Q=AV 6. Water Power- is the power generated from an elevated water supply by the use of hydraulic turbines 7. Turbine efficiency—is the ratio of turbine power output to the water power output 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟 𝑃𝑡 et = 𝑊𝑎𝑡𝑒𝑟 𝑃𝑜𝑤𝑒𝑟 = 𝑃𝑤, and Pt = ɤQhet 8. Electrical or generator efficiency is the ratio of the generator output to the turbine power output 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑜𝑢𝑡𝑝𝑢𝑡 𝑃𝑔𝑒𝑛 eG = = ; Pgen =Ptegen = ɤQhetegen 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑜𝑢𝑡𝑝𝑢𝑡

𝑃𝑡 120𝑓 = 𝑃 ,

9. Generator Speed, N where: N= angular frequency, rpm F= frequency (usually 60 hertz), P = no. of poles,(even number) 10. Head of impulse turbine 𝑉2

𝑃

h = ɤ + 2𝑔’ meters of water 11. Head of reaction turbine 𝑉𝐴2 −𝑉𝐵2 2𝑔

𝑃

H= ɤ + Z +

12. Peripheral coefficient It is a ratio of the peripheral velocity(Vp) to the velocity of the jet (Vj) 𝜋𝐷𝑁 𝟇= where: D= diameter of the runner √2𝑔ℎ

N= angular speed, h= net head 13. Specific speed-is a number used to predict the performance of the hydraulic turbines a) In English units Ns=

𝑁√𝐻𝑃 5

, where: N= angular speed in rpm, h= net head in feet

ℎ4

b) In SI Units Ns=

0.2623√𝑘𝑤 ℎ45

14. Hydraulic efficiency is the ratio of the utilized head to the net head ℎ𝑤 Eh = ℎ

Components Parts of a water turbine A. Wheel- commonly known as runner, it is along with the vanes on its periphery, rotates under the action of water gliding on the vanes comes from the gliding apparatus. B. Guiding Apparatus (or mechanism), which guides the water to the vanes of runner. C. In addition, there will be a source of supply from which water will come through a pipe line ( known as penstock) to the guiding mechanism also, there will be a tailrace in which the water, after gliding over the moving blades of the runner and passing out of runner, will ultimately fall from the turbine. Classification of turbine according to the direction of flow A. B. C. D.

Radially inward flow turbine Radially outward flow turbine Axial flow turbine or parallel flow Mixed flow turbine

A turbine, whether Impulse or Reaction may have one of the following settings A. Vertical setting, B. Horizontal setting C. Above the tai race D. Below the tail race A turbine is said to be vertical or with vertical setting when its shaft is vertical and its runner is horizontal, this arrangement gives better efficiency due to the reduced losses. This setting is usually used for reaction turbines, it may also used for pelton wheel having more than two nozzles.The vertical setting is said to be used for low heads. A turbine is said to be horizontal or with horizontal setting when its shaft is horizontal and its runner is vertical. This setting is usually for Pelton wheel with one or two nozzles.

HEADS at Inlet and Outlet of Runner of Reaction Turbine 𝑉2

𝑃

H=𝑤 + 2𝑔, meters of water; Similarly, if V1 and P1 are respectively the velocity and intensity of pressure at the outlet of runner, the total head at outlet runner, 𝑉2

𝑃

H1 = 𝑤1 + 2𝑔 meters of water; where: h= total head at inlet, in m; V= absolute velocity with which water from guide passages, enters the inlet of the runner, m/s; P= intensity of pressure at the inlet of the runner in kg/m2. Applying modified Bernoullis’ Theorem to the outlet of the horizontal runner. H= h1+ total losses between inlet and outlet of the runner = h1 + friction loss(hydraulic loss) in runner + head used due to the work done by water on runner. If hf be the loss of head in guide passages and Z be the elevation of inlet above the tail race, he total head H on turbine 𝑉2

𝑃

H= hf + 𝑤 + 2𝑔 + Z Efficiency and horsepower of reaction turbine The efficiency of a turbine may be a) Hydraulic efficiency, ɳh b) Mechanical efficiency,ɳm c) Overall efficiency, ɳo ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑜𝑢𝑡𝑝𝑢𝑡 𝐻𝑃 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑤𝑎𝑡𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑢𝑛𝑛𝑒𝑟 Hydraulic efficiency = ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑖𝑛𝑝𝑢𝑡 = 𝑊𝐻 𝑊𝐻 = 75

Where;

75

horsepower of water on available water horsepower

𝑘𝑔−𝑚 ℎ𝑝−𝑠𝑒𝑐 𝑊𝑜𝑟𝑘 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚 𝑠ℎ𝑎𝑓𝑡 𝑆𝐻𝑃 𝑜𝑓 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 Mechanical efficiency= 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑜𝑛 𝑟𝑢𝑛𝑛𝑒𝑟 =𝐻𝑃 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑤𝑎𝑡𝑒𝑟 𝑡𝑜 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑆ℎ𝑎𝑓𝑡 ℎ𝑝 𝑜𝑓 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑆𝐻𝑃 Overall efficiency, ɳ0 =ɳm x ɳh = 𝐼𝑛𝑝𝑢𝑡 ℎ𝑃 𝑜𝑓 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 = 𝑊𝐻𝑃

75=1 metric horsepower,

Hydraaulic Turbines (Performance) 1) Specific Speed of turbine- speed of an imaginary or specific turbine( which is a small model of the original turbine) which develops one SHP when working under a net head of one meter. Discharge of turbine, Q= Area of flow times velocity of flow.= πDbVf; taking the turbine to be a reaction turbine , but; 𝜋𝐷𝑁 60

U= tangential velocity =

; D=

𝑈(60) 𝑈 ; D∞𝑁, 𝜋𝑁

Speed ratio=

𝑈

√2𝑔𝐻

of a turbine is a certain constant,

U= constant ( √2𝑔𝐻) : Constant =Cv= coefficient of velocity of nozzle √𝐻

For a turbine, b =fD; therefore b∞ 𝑁 Ns=

𝑁√𝑃 5

𝐻4

Equation for specific speed using 2 jet pelton wheel 𝑁√2𝑃

Ns2=

5

𝐻4

3. Conditions of working of a Turbine 𝐷1 𝑁1 𝐷2 𝑁2 = √𝐻1

√𝐻2

𝑤ℎ𝑒𝑛 𝐷1= 𝐷2 𝑄1 𝑁12 3 𝐻12 𝑃1 𝑁12 5 𝐻12

𝑄2 𝑁22

=

3

,

𝐻22 𝑃2 𝑁22

=

5

𝐻22 𝑁1 √𝑃1

𝑇ℎ𝑢𝑠 ; Under two conditions; Ns =

5 𝐻14

=

𝑁2 √𝑃2 5

𝐻24

Example Problem #01 A reaction turbine develops 500 BHP. Flow through the turbine is 50 cfs. Water enters at 20 fps with a 100 ft pressure head. The elevation of the turbine above the tail water level is 10 ft. Find the effective head. Solution: 𝑝

𝑉𝐴2 −𝑉𝐵2 2𝑔 (20)2 −(0)2 + 10 + 2(32.2)

h= ɤ +z + = 100

h= 116.2 ft

Example #02 A proposed scheme has an available head of 480 m and the 3 single jet pelton wheels to be installed are required to run at 330 rpm. Assuming an over-all efficiency of 85%, determine the total quantity of water required per second. Assume that the specific speed of Pelton is 18. Solution: H= 480 m N= 330 rpm, ɳ0 =85%, Ns = 18 From Formula: Ns= 𝑁𝑠(𝐻)1.25

𝑁√𝑆𝐻𝑃 5

;

𝐻4 18(480)1.25 330

= = 122.54 √𝑆𝐻𝑃 = 𝑁 𝑆𝐻𝑃 = 15018.28093 hp 𝑆𝐻𝑃 𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝐼𝑛𝑝𝑢𝑡 ℎ𝑝 𝑠ℎ𝑝

𝐼𝑁𝑃𝑈𝑇 𝐻𝑃 = 𝑤𝐻𝑃 = 𝑂𝑉𝐸𝑅𝐴𝐿𝐿 𝐸𝐹𝐹𝐼𝐶𝐼𝐸𝑁𝐶𝑌= ɤ𝑄𝐻 𝑤ℎ𝑝 = 75 𝑊𝐻𝑃(75) 17668.56 (75) 𝑄 = ɤ𝐻 = 1000(480)

15018.28093 0.85

= 17,668.56 HP

= 2.76 m3/sec, For single jet Pelton wheel

𝑄 = 2.76(3)= 8.28 m3/sec Example #03 Determine the specific speed of a pelton wheel of which the basis design criteria are as follows: Coefficient of velocity of nozzle, Cv= 0.98, runner velocity = 0.46 x jet velocity, 𝐽𝑒𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑑𝑐 1 = = 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑟𝑢𝑛𝑛𝑒𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

𝐷𝑚𝑒𝑎𝑛

10

Overall efficiency = 82% 𝜋𝐷𝑚𝑒𝑎𝑛𝑁 Solution; for runner , tangential velocity, U= eq. 1 60 Where: Dmean= effective runner diameter,m N= speed of the runner, rpm U= tangential velocity, m/s For water jet, absolute velocity or jet velocity= V= Cv √2𝑔𝐻----eq.2 Where Cv =coefficient of velocity of nozzle H= net head of turbine,m V= jet velocity m/s, But runner velocity, U=0.46 x V----eq.3 Substituting eq.1 and eq.2 to eq.3 𝜋𝐷𝑚𝑒𝑎𝑛 𝑁 = 0.46 x Cv√2𝑔𝐻; Cv= 0.98 (given) 60 𝜋𝐷𝑚𝑒𝑎𝑛𝑁 = 0.46 60 38.135√𝐻 𝑁 = 𝐷𝑚𝑒𝑎𝑛

x 0.98 √2 𝑥 9.81 𝑥 𝐻 𝑆𝐻𝑃

𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = ɳ0= 𝐼𝑛𝑝𝑢𝑡 𝐻𝑃 =

𝑆𝐻𝑃 𝑥 75 𝑊𝐻𝑃

𝑊 = mass of striking water, kg/s ɳ 𝑊 𝑥 𝐻 ɳ0 𝑥 𝑤 𝑥 𝑄 𝑥 𝐻 𝑆𝐻𝑃 = 0 𝑥75 = 75

𝜋𝑑 2 x Cv √2𝑔𝐻 4 0.82 𝑥 1000 𝑥 𝜋 𝑥𝑑 2 𝑥 0.98√2 𝑥 9.81 𝑥 𝐻 4 𝑥 75

𝑄 = 𝐴𝑉 =

SHP= SHP= 37.275dc2x H3/4 Taking the square root of both sides SHP1/2 = [ 37.275 dc2 x H3/2]1/2 = 6.1 dc H3/4 From Formula Ns=

𝑁√𝑆𝐻𝑃 5

H

; N= speed of the runner

𝐻4

Ns= Ns= Ns=

𝟑 38.135(√𝐻 ) 𝑥 6.1 𝑑𝐜 𝑯𝟒 𝐷𝑚𝑒𝑎𝑛 5 𝐻4 1 3 38.135 𝐻 2 𝑥 6.1 𝑥 𝑑𝑐 𝑥 𝐻 4 5 𝐷𝑚𝑒𝑎𝑛 𝑥 𝐻 4

38.135 𝑥 6.1 𝑥 4 5

10 𝑥 𝐻 4

Ns= 23.26 rpm

; but

𝑑𝑐 𝐷𝑚𝑒𝑎𝑛

=

1 10