2. Performance Heat Balance and Efficiency Test of a Diesel Electric Power Plant Lecture Copy (2)

NEW ERA UNIVERSITY COLLEGE OF ENGINEERING AND ARCHITECTURE DEPARTMENT OF MECHANICAL ENGINEERING

ME LABORATORY 3 Th; 1:00 PM – 7:00 PM

Performance Heat Balance and Efficiency Test of a Diesel Electric Power Plant

SUBMITTED BY: CORPUZ, RALPH MARWIN U. BSME - V SUBMITTED TO: ENGR. MOSES MABUTE

August 17, 2017

1. The Diesel Electric Power Plant Applications of diesel power plant  Diesel power plant’s is in the range of 2 to 50 MW capacity. They are used as central station for small or medium power supplies.  They can be used as stand-by plants to hydro-electric power plants and steam power plants for emergency services.  They can be used as peak load plants in combinations with thermal or hydro-plants.  They are quite suitable for mobile power generation and are widely used in transportation systems such as automobiles, railways, air planes and ships.  Now-a-days power cut has become a regular feature for industries. The only solution to tide over this difficulty is to install diesel generating sets. Layout diesel engine power plant:

Diesel engine: Diesel engines or compression ignition engines as they are called are generally classified as two stroke engine and four stroke engines. In diesel engine, air admitted into the cylinder is compressed, the compression ratio being 12 to 20. At the end of compression stroke, fuel is injected. It burns and the burning gases expand and do work on the position. The engine is directly coupled to the generator. The gases are then exhausted from the cylinder to atmosphere. Engine starting system: This includes air compressor and starting air tank. The function of this system is to start the engine from cold supplying compressed air. Fuel system: Pump draws diesel from storage tank and supplies it to the small day tank through the filter. Day tank supplies the daily fuel need of engine. The day tan is usually placed high so that diesel flows to engine under gravity. Diesel is again filtered before being injected into the engine by the fuel injection pump. The fuel is supplied to the engine according to the load on the plant.

Air intake system: Air filters are used to remove dust from the incoming air. Air filters may be dry type, which is made up of felt, wool or cloth. In oil bath type filters, the sir is swept over a bath of oil so that dust particles get coated. Exhaust system: In the exhaust system, silencer (muffler) is provided to reduce the noise. Engine cooling system: The temperature of burning gases in the engine cylinder is the order of 1500 to 2000’C. to keep the temperature at the reasonable level, water is circulated inside the engine in water jackets which are passage around the cylinder, piston, combustion chamber etc. hot water leaving the jacket is sent to heat exchanger. Raw water is made to flow through the heat exchanger, where it takes up the heat of jacket water. It is then cooled in the cooling tower and recirculates again. Engine lubrication system: It includes lubricating oil tank, oil pump and cooler. Lubrication is essential to reduce friction and wear of engine parts such as cylinder walls and piston. Lubricating oil which gets heated due to friction of moving parts is cooled before recirculation. The cooling water used in the engine is used for cooling the lubricant also. Advantages of diesel power plant:        

Plant layout is simple. Hence it can be quickly installed and commissioned, while the erection and starting of a steam power plant or hydro-plant takes a fairly long time. Quick starting and easy pick-up of loads are possible in a very short time. Location of the plant is near the load center. The load operation is easy and requires minimum labors. Efficiency at part loads does not fall so much as that of a steam plant. Fuel handling is easier and no problem of ash disposal exists. The plant is smaller in size than steam power plant for same capacity. Diesel plants operate at high overall efficiency than steam.

Disadvantages of diesel power plant:    

Plant capacity is limited to about 50 MW of power. Diesel fuel is much more expensive than coal. The maintenance and lubrication costs are high. Diesel engines are not guaranteed for operation under continuous, while steam can work under 25% of overload continuously.

2. Diesel Engine Power Cycle The ideal thermal cycle of the Diesel engine is illustrated in Figure 1.

Beginning with the working medium at state 1, it is first polytropically compressed to state 2, then heat is added during a limited isobaric expansion, after which a polytropic expansion to the initial volume reduces the pressure to state 4. The ideal work produced by the cycle is represented by its area, and the mean effective pressure is its average height. Polytropic processes 1-2 and 3-4 in the ideal cycle are isentropics with air as the fluid. Thus, for the air standard performance, n = k = 1.4. The ideal air standard efficiency, k 1 r c −1 e=1− k−1 r c −1 kr (1)

[ ]

where QA = Qin = mcp(T3 – T2) rk = rcre rk = compression ratio = V1/V2 rc = cut-off ratio = V3/V2 re = expansion ratio = V4/V3 = V1/V3 Ideal mean effective pressure, pm= p1 r

[

k kr k−1 ( r c−1 ) −( r c −1 ) k

( r k −1 ) ( k −1 )

]

(2) These equations show that high efficiency is promoted by high rk and low rc, but that engine size (which may be considered proportional to pmi) is increased as rc decreases. The requirement of adequate fuel combustion imposes a limitation on maximum rc that can be used. Since Diesels are load-governed by varying point of cutoff, the ideal efficiency increases at part load. This partially offsets other tendencies, and a fairly flat efficiency-load characteristic result. Real engines have cylinder cooling for mechanical reasons; also they work on an open cycle, meaning that the fluid at state 1 is not the same as that which completed the previous cycle. The

products of combustion of each working cycle are discharged as exhaust gas, and fresh air is inducted for use in the following cycle. Instead of heat being transferred between states 2 and 3, fuel is injected into the air and its heat of combustion provides the energy input. The real engine will have nonisentropic compression and expansion processes; n = 1.35 is a fair average in practice. Although Equation (1) has its uses in Diesel engine studies, the actual thermal efficiencies are considerably less than those of the air standard. Example No. 1 An air standard Diesel cycle will be analyzed for state of the working fluid and performance. Using the nomenclature of Figure 1, state 1 is at 0.9 kg/cm 2 ab and 27 C. The volume quantity is that of a single-cylinder engine with 25 cm bore and 38 cm stroke. It will be assumed that, after a compression sufficient to produce 538 C, heat is added during the first 10% of the working stroke.

Given: State 1 is at 0.9 kg/cm2 ab and 27 C Single-cylinder engine with 25 cm bore and 38 cm stroke. Produce 538 C Heat added during the first 10% of the working stroke. Required: Analysis of the cycle Solution:

Ratio of compression, T1 = 27 + 273 = 300 K T2 = 538 + 273 = 811 K

v1 T 2 rk = = v2 T 1

()

1 k−1

,

p2 v 1 = p1 v 2

k

()

p1 = 0.9 kg/cm2 ab k= 1.4

811 rk = 300

( ) ()

1 1.4−1 =12.0

k

p2 v 1 = =( r k ) k p1 v 2 1.4

p2 =( 0 .9 )( 12 ) =29. 2 kg /cm2 ab 2

The piston displacement = D = 25 cm = 0.25 m L = 38 cm = 0.38 m

v 1 −v2 =

πD L 4

2

π ( 0 . 25 ) ( 0 . 38 ) v 1 −v2 = =0 . 01865 m3 4 Clearance volume

v 1 −v2 =rv 2 −v 2 v −v v2= 1 2 r−1 0 .01865 v 2= =0. 00170 m3 12−1 v 1 =( v 1−v 2 ) + v 2

v 1 =0 . 01865+0 . 00170=0 . 02035 m v 4 =v 1 =0 . 02035 m3

3

v 3 =v 2 +10 ( v 1 −v 2 )

v 3 =0 . 00170+0 . 10 ( 0. 01865 ) =0.003565 m3 T 3 v 3 0 .003565 = = =2. 097=R T 2 v 2 0 .00170 T 3 =( 811 ) ( 2. 097 ) =1701 K

p 4 v 3 k 0 .003565 = = p3 v4 0 .02035

()(

1. 4

)

=0 . 0873

2

p3 =p 2 =29. 2 kg /cm ab 2 p4 =( 29 . 2 )( 0. 0873 ) =2. 55 kg /cm ab

T 4 v 3 k −1 0. 003565 1. 4−1 = = =0. 498 T3 v4 0. 02035 T 4 =( 1701 ) ( 0. 498 )=847 K

() (

)

Principal physical states of the cycle are summarized:

State Pressure, kg/cm2 ab Volume, m3 Temperature, K Temperature, C

1 0.9

2 29.2

3 29.2

4 2.55

0.02035 300 27

0.00170 811 538

0.003565 1701 1428

0.02035 847 574

Ideal mean effective pressure

[

pm= p1 r

k kr k−1 ( r c−1 ) −( r c −1 ) k

( r k −1 ) ( k −1 )

pm=( 0 . 9 ) ( 12 )

[

] ]

( 1 . 4 )( 12 )1. 4−1 ( 2 .097−1 )−( 2. 0971 . 4 −1 ) ( 12−1 ) ( 1. 4−1 ) 2

pm=5 . 72 kg/cm ab

(answer)

Net work done per cycle

2

100 cm ( 0.01865 m3) = pm ( v 1−v 2 ) =( 5.72 kg/cm ) 1m =1067 kg⋅m (answer) 2

(

)

Ideal Thermal Efficiency k 1 r c −1 e=1− k−1 r c −1 kr

[ ] [

( 2 . 097 )1. 4 −1 1 e=1− ( 1 . 4 ) ( 12 )1. 4−1 2 . 097−1 e=0 . 56=56 (answer)

]

3. Combustion Here the special features of combustion as carried out in the Diesel engine cylinder are to receive attention. However, first it appears desirable to repeat and summarize the equations pertaining to fuel oil. Density scales:

140 −130 [email protected]/15.6 141.5 ° API= −131.5 [email protected]/15.6 '

° Be =

(3) (4)

Ignition quality:

Diesel index=0.018×° API×tap +0.32×° API

(5)

Heating value:

Qh =41130+139. 6×° API kJ /kg

(6)

2

Qh=51716−8793.8 ( S.G . ) kJ /kg

(7)

Q L=Qh −2442.7×9 H 2

(8)

Hydrogen content: H 2 =26−15 ( S . G. ) percent by weight(9) where tap = aniline point in C. Combustion in the Diesel engine cylinder begins theoretically at the instant injection starts and continues, at constant pressure, until injection ceases. The distillate fuel used may be considered to have an average chemical formula of C16H32 for which the ideal air quantity is found as follows:

C16 H 32+24O2=16CO 2+16 H 2 O

Considering the numerical prefixes to be mols, the equation of combining weights is written as follows:

224 kg C16 H 32+24×32 kg O2=16×44 kg CO 2 +16×18 kg H 2 O

Since 1 kg air provides 0.232 kg O2,

Air per kg C16 H 32=

24×32 =14. 8 kg 0. 232×224

Early fuel cutoff is necessary to good thermal efficiency, but early cutoff is not possible with the ideal A:F ratio of 14.8. This is due to the need for limiting maximum temperature of the cycle for mechanical and thermal reasons, under circumstances as set forth in the following example. Example No. 2 The ideal maximum temperature of combustion of a fuel of 24 oAPI is calculated, on the assumption of 427 C compression temperature and 14.8 kg air per kg fuel. Given: Fuel of 24 oAPI, 427 C compression temperature, 14.8 kg air per kg fuel. Required: Ideal maximum temperature Solution: Specific gravity

141.5 131.5+° API 141 .5 S .G .= =0.91 131 .5+24

S .G .=

Heating value:

2

Qh =51 ,716−8793 . 8 ( S. G. ) kJ /kg

Qh =51 ,716−8793 . 8 ( 0. 91 )2 =44 , 434 kJ/kg Q L=Qh −2442.7×9 H 2 Q L=44 , 434−2442. 7 ( 9 ) ( 0 . 26−0 . 15×0. 91 ) Q L=41 , 719 kJ /kg During isobaric combustion, using cp = 1.00 kJ/kg.K, the sensible heat, QL, will raise the products t degrees, according to the relation

Q L=wc p Δt w=14 .8 kg air+1 kg fuel=15 .8 kg

Δt=

QL wc p

=

41 ,719 =2640 C ( 15 . 8 ) (1 . 00 )

Maximum temperature t3 = t2 + t = 427 + 2640 = 3067 C. Since T3/T2 = V3/V2 = rc during an isobaric process, T 3067+273 rc = 3 = =4 . 8 T 2 427+273 Not only is the calculated t3 higher than can be considered practical in the engine cylinder, it is high enough so that thermal dissociation of the products would have prevented its attainment. Furthermore, an R of 4.8 is large enough to impair ideal efficiency seriously. The solution is to use A:F ratio, higher than the chemical ideal. Assume that 1650 C is the limit of t3. Then T 1650+273 rc = 3 = =2 . 75 T 2 427+273 This is compatible with good efficiency.

w=

QL

c p Δt

=

41, 719 =34 . 1 kg ( 1. 00 ) ( 1650−427 )

A : F ratio=34.1−1=33 .1 Although this example is based on an ideal engine, the actual engine is similar. 4. Engine Performance The rapid cyclic action in an actual engine modifies the ideal cycle of Figure (1). The diagram corners are no longer sharp, fuel must be injected ahead of dead-center, etc. The forms of “indicator cards,” which are instrument-drawn p-v’s from running engines, are shown in Figure (2) for both 2and 4- cycle types.

A Diesel engine is used as a source of power which can utilize a relatively cheap fuel. Therefore its power capacity and thermal efficiency are paramount consideration. The developed cylinder power is the indicated horsepower, abbreviated, ihp. After engine friction and accessory power needs have been satisfied, the net power available at the engine shaft is the brake horsepower, bhp, so called because it can be (but nowadays seldom is) measured by a brake dynamometer. Indicated power is difficult to measure directly on small Diesels with great accuracy, because the volume of indicator lead through the cylinder head appreciably lowers the normal compression ratio. Also, the cylinder heads of small-bore high-speed engines have special chambers, valves, and other equipment, leaving little or no space available for indicator lead. Since none of these difficulties exists with the large stationary Diesel, indicator openings are provided as standard equipment.

A hypothetical pressure, known as brake mean effective pressure, bmep, can be employed to show the magnitude of mean effective pressure. The true pressure, pmep, is higher on account of engine friction losses.

ihp=

p mep LAN p

hp 33 , 000 (10) 2 πWrN bhp= hp 33,000 (11) bhp×33 , 000 bmep= lb/ft 2 LAN p

(12)

in which pmep = Indicated mep, lb per sqft L = Piston stroke, ft A = Piston face area, sqft Np = Number of power strokes per min (N for two-cycle and N/2 for four-cycle) N = Rotative speed, rpm W = Net dynamometer force, lb r = Dynamometer arm length, ft As is true of all prime movers, there are a number of efficiency expressions applying to Diesels. Mechanical efficiency is the ratio, bhp/ihp. Indicated thermal efficiency,

e i=

2545 wi Q

(13) Brake thermal efficiency,

e b=

2545 wbQ

(14)

wi, wb = fuel consumption, lb per hr per ihp or bhp Q = Fuel heating value, Btu/lb, either Qh or QL, according to policy. Example No. 3 A 6-cylinder Diesel engine on dynamometer test was found to use 84 lb of fuel, having Qh = 19,351 Btu/lb, in a one-hour test at steady load. The brake thermal efficiency and the brake mep will be determined from the following test data and measurements. Cylinder is 8.5 in x 10.5 in 4-cycle type. Speed, 600 rpm. Dynamometer torque, 1809 lb-ft.

Given: Wr = dynamometer torque = 1809 lb-ft. Qh = 19,351 Btu/lb N = 600 rpm D = 8.5 in = 0.7083 ft L = 10.5 in = 0.875 ft Fuel weight = 84 lb 4-cycle type. Required: Brake thermal efficiency and the brake mep. Solution:

bhp=

2 πWrN hp 33 ,000 2 π (1809 )( 600 )

bhp=

33 ,000

=206 . 7 hp

Brake thermal efficiency:

Brake thermal efficiency , ηtb =

2545 wbQ

84 =0 . 4064 lb per bhp hr 206 .7 2545 Brake thermal efficiency , ηtb = =0 . 324 or 32 . 4 ( 0 . 4064 ) (19 , 351 )

w b=

Brake mep:

bmep=

bhp×33 , 000 lb/ft 2 LAN p

2

πD ×no . of cylinders 4 N N p = for 4−cycles 2 206 . 7×33 ,000 bmep= =10 , 991 lb/ ft 2 π 600 ( 0 . 875 ) ( 0 . 7083 )2 ( 6 ) 4 2 A=

[

bmep=

10 , 991 =76 .33 psi 144

5. Heat Balance

]( )

The energy supplied to I.C. engines in the form of Qh of the fuel input is generally broken down into the following items for heat-balance purposes. a. Useful work, the actual net shaft output. b. Cooling. Heat absorbed by water jackets. Sometimes this is not separable from some of the friction and exhaust loss due to merging streams of water that cool cylinders, exhaust manifolds, and lube oil heat exchangers. A wct type of loss. c. Exhaust gas loss. Heat carried off as sensible and latent heat in the products of combustion. Sensible heat is wct loss; latent heat is of the form 2442.7 x 9H2 (in kJ/kg units). d. Mechanical friction. Crankshaft and connecting rod bearings, piston, etc. Usually taken to include energy supplied to engine-mounted auxiliaries such as pump, governor. e. Radiation and unaccounted-for. Heat radiated from engine and incomplete combustion loss due to presence of fuel in exhaust gas. TYPICAL FULL-LOAD HEAT BALANCES (%) (Based on Qh) Otto Cycle, Spark Ignition Useful work Cooling Exhaust Friction, radiation, and unaccounted Input; heating value of fuel

25 30 37 8

Diesel Cycle, Compression Ignition 34 30 26 10

100

100

Item d and e are not often separable nor measurable. They are usually combined, then the balance is made by subtracting Items a, b, and c from the heat of the fuel and considering the difference to be Item d and e.