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Chap. 4

Domestic hot water system By Dr. Ali Hammoud BAU 2008

Contents • • • • •

Hot Water Consumption and Demands Pipe Sizing and Water Heat Storage Pumping Systems and Performance Boiler types & selection Examples

2

Supply Hot Water Pipe Sizing : In a manner similar to cold water pipe , the hot water Pipe is sized . First of all calculate the total fixture units (0.75 of total FU) then find the corresponding flow in gpm and then refer to pipe flow chart and select the corresponding pipe size. You could hot water pipe sizing. The higher the temperature of the water, the lower would be the limit of flow velocity

Pipe Sizing •Avoid oversizing & undersizing Oversizing • High cost – extra but unnecessary • Delay in getting at outlets • Increase heat loss from distributing piping

• Undersizing • Slow or even no water during peak demand • Variation in temperature & pressure at outlet (obvious in mixer for shower) • High noise level 4

Pipe insulation :

Boilers, storage tanks, calorifiers, chimneys, and hot water pipes lose their heat when exposed to outside air. Hence they should be insulated Insulation material should be non-corrosive, nontoxic, moisture resistant, rot- and vermin-proof, fire-resistant, light, easy to apply, anti-weathering, durable and, in addition possess good insulation properties and low thermal conductivity.

Insulation material could be fiberglass or rubber type insulation like armaflex which comes in the form of tube and the pipe is pulled inside it .

• • • •

Pipe insulation: reasons: Heat conservation. Reduce noise. Control surface condensation. Freeze prevention. The insulating material is the same used in duct insulation (Fiber glass)

Insulation thickness Thicknesses or insulation to be supplied and installed for the various systems shall conform to the following table: (Fiber glass thickness) a. Conditioned air supply and return (ductwork) 1 1/2" b. Refrigerant suction and liquid lines: 1/2" c. Condensate drain pipes 1/2" d. Acoustic duct liner. 1” e: Heating water pipes up to 1" diameter: 1" f. Heating water pipes above 1 ”: 1 1/2" g. Domestic hot water pipes: 1" k. Boiler, water heater 2” L. Boiler breeching & steel chimney 2.1/2”

Domestic hot water system Electrical water heater

Water to water storage Heater Boiler +Water Storage

Instantaneous or semi-instantaneous types of heaters

Electrical Water heater Power 3/4 of the total fixture units are used for hot water Hot water

Thermostat

Drain

1.25 "

4 bar Pressure Releif valve

Inlet water flow ? Electrical power :

Temp.

1.5, 2.4, 4-5, & 9 Kw

Electrical water heater

Size of EWH :

1" Cold water

1"

15, 20 , 30, 40, 50 , 66, 80 &1 20 gallons

3/4 of the total fixture units are used forc old water

Power =

S .hc × Kg × ΔT . 4.2 × 150 × (65 − 10) = = 4.2 Kw Heating time × Efficiency 3 × 3600 × 0.75

Where Shc is the specific heat capacity of water ( 4.2 Kj/kg. ºK), Heating time or recovery period T = 3 hrs, The mass of water 150 Liters = 150 Kg., Temperature rise from 10 to 65 ºC. Insulation efficiency is 75-80 %.

In general, electric water heaters are fully automatic and have a storage tank, one or more electric elements, and operating and safety controls. The heating elements are available in a variety of standard voltages and wattages to meet the specific requirements of the installations. Electrical water heater are fitted with electrical resistance ( heater) 1.5 Kw, 2.4 Kw , 4- 5 Kw, 9 Kw . They could be horizontal or vertical type . EWH are insulated and protected by steel jacketing. EWHs should have pressure /temperature relief valve which will relieve excessive pressure . For high pressure EWH the use of none -return is not recommended. In this case ,the pressure relief valve is connected to nearest floor drain. EWHs are controlled by thermostat which will activate the electrical heater automatically. Available EWHs size are 15, 20 , 30, 40, 50 , 66, 80 &1 20 gallons

Resistance heating element Usually the electrical water heaters have a primary resistance heating element near the bottom , and possible a secondary element located within the upper quarter or third of the tank. Minimum wattage ratings of two element heaters are based as follows: For the upper unit (8 watt / liter) of the tank capacity. For the lower unit (5 watt/liter) of the tank capacity. For fast heating of ( 30-40-50 gal ), we take (26 watt/litter) for each unit . For one single element heater ,we take (26 watt/litter)

Resistance heating element location

12

Hot –water demand (definitions) Demand of water supply : is the rate of flow in pgm furnished by a water supply system to various types of plumbing fixtures and water outlets under normal conditions.

Maximum Demand: is the peak value of the demand. The values Hot of water demand

are shown

previously in table (W-2 ).

Demand factor : is the ratio of the maximum demand of the hot water heating system to the total connected load or the total of the individual requirements of all the fixtures of the system.

Water to water storage heater A hot-water storage tank must meet code requirements that depend on its size and pressure and the authority having jurisdiction. Its capacity should be selected so that 60 to 80 percent of the volume of water in the tank may be drawn off before the temperature drop (caused by the incoming cold water) becomes unacceptable. A value of 70 percent usually is used in design calculations. Heated water from boiler enter the coil of the water storage vessel where it will heat the water , and then returns to boiler at lower temperature . Whereas the domestic Hot water leaves the top of the water tank to the supply distribution piping upon demand for the various fixtures and apparatus. When recalculation of the hot water is provided, the warm water is returned to the bottom of the hot water vessel or tank by a circulate pump. Normally heating water from boiler enters the water tank at 180 º F ( 82 º C ) and return to boiler at 160 º F ( 70 º C), ∆T = 12-15 º C.

Hot water requirement for storage

The hot water storage for Buildings & Hotels will be calculated based on unit hourly demand rates as follows : galon per hour Fixture type Building Hotels Lavatory 2 gph 2 gph bathtub 20 gph 20 gph Shower 30 gph 75 gph Sink 10 gph 30 gph Laundry 20 gph 28 gph Dishwashers 15 gph 50 gph Demand factor 0.3 0.25 Storage factor 1.25 0.8 Domestic hot water temperature will be 60° C.

Ref [2]

Table Hw-1

Table Hw-2

Ref [2]

Procedure for estimating the heating capacity (recovery capacity) of a hot-water heating system

having a storage tank.

Step(1) Tabulate the number of fixtures of each type in the building. Step(2) Then multiply the number of fixtures of each type by the probable demand for each type of fixture. Step(3) Obtain the maximum demand by taking the sum of products of step (2) Step(4) Then obtain the hourly heating capacity by multiplying the maximum demand in step3 by the demand factor obtained from table 2. Step(5) multiply the hourly heating capacity of step 4 by the storage capacity factor given in table HW-1 for the appropriate type of building to obtain the required capacity of the storage tank. N.B. Not to be used for instantaneous or semi-instantaneous types of heaters

Example Suppose you have to calculate , the hot water storage capacity , the boiler power of an apartment house (building) having the following data: 60 Lavatories , 40 kitchen sinks ,and 10 laundry 60 showers, and 40 dishwasher . Step1& 2. The probable water demand are as follows: From table HW-1; Lavatories =

60 x 2 gph = 120 gph

Kitchen sink = 40 x 10 gph = 400 gph . Laundry

= 10 x 20 gph = 200 gph .

Shower

= 60 x 30 gph = 1800 gph.

Dishwasher

=

40 x 15 gph = 600 gph .

Step 3 : The Maximum demand of the hot water demand is: =((60×2gph)+ (60×30gph)+ (40×10gph) + (40 x 15 gph) +(10x 20 gph ))= 3120 gph. Step 4 Building demand factor = 0.3 (from table HW-1 ). Hourly heating capacity = (3120×0.3) = (936 gph)= 15.6 gpm.= 0.985 liter /sec or 3546 L/hr. Step 5 The required capacity of the storage tank is: Usable capacity = 936 x 1.25 = 1170 gal i.e. (4429 liters) [1.25 is the storage factor from table Hw-1]

Safety storage: Since only 70% of the tank is usable, so the actual

tank capacity = (1170 / 0.7 gallon) = (1670 gallon) that is (6327 liters).

Boiler power Calculation

From the basic equation

Q = m× C × ΔT

In U.S.units Boiler Power Q (BTU/h) = GPM × (60min/hou r) × ∆T1 × (8.3 lb/gal) Q (boiler in BTU/h) = 500 × gpm × ∆T1 Where

gpm = is the calculated water demand in gpm (flow rate ) ; ∆T = is the temperatur e difference between initial & Fianal [ 101 DF]

From the previous example, we have 936 gph = 15.6 gpm of water to be heated , temperature rise ∆T=101 ºF. QT= 500 x 15.6 x 101 = 787800 BTU/hr =230 Kw. In SI units: Power KW =

0.985 Kg / sec . × 4.2 Kj / Kg .C × (55) C = 227.5 Kw

Acceptable Temperature of domestic Hot water

Ref [2]

Fº = 1.8 Cº + 32 , Cº = 0.55 (Fº - 32)

Boiler power 1 Kw = 860 Kcal/hr. = 3413 BTU/ hr. 1boiler hp = 9.81 KW.

The required gross boiler power for heating water + overcome the heat loss from pipe and boiler + the heat needed to rise the initial water temperature ( Pick up ) is given by: QBoiler = QT [1+ a + b] QT = Calculated “boiler power’ ,a = additional heat coefficient to overcome the heat loss in the pipe systems and boiler. [0.1] b = additional heat coefficient to overcome the pick up period [0.1 to 0.2] . The required gross output is then; QBoiler = [1.2 or 1.3 ] QT As mentioned before : for QT= 230 Kw ,

The required boiler power = 1.2 x QT = 1.2 x230 =276 Kw

T=82 °C

T=10 °C T=72 °C

Hot Water Pumps Pumps used in hot water primary and secondary distribution systems are used mainly for maintaining or increasing the rate circulation. Pumps are constructed to withstand the high temperature of water. They have a cast-iron body and a gunmetal impeller. Impellers made of other materials, such as bronze , stainless steel and cast iron are also used. Each pump has a valve on the suction and delivery side and a check valve on the delivery side. A bypass on the lines enables the removal of pumps for maintenance and repairs. In well-designed systems, frictional loss in pipe-lines during recirculation is quite low and the pump horse power is also small .Hot water recirculation pumps should never be used as booster pumps to increase the pressure in the hot water system, as this creates imbalance in the pressures of hot and cold water supply.

Hot water & Boiler Circulated pumps There are two centrifugal circulated pumps used in Hot water system: Pump1: is used to circulate the hot water system (Located in the returned pipe ). Pump2 : Is used to circulate the hot water from boiler to storage tank (closed system). Circulating pump is controlled by an immersion thermostat (in the return line) set to start and stop the pump over a 11 ºC). However for continuous hot water supply the thermostat is eliminated.

Hot water circulating pump ( inside the apartment) If water heater is located far away from plumbing fixtures (more than 30 m) hot water circulator might be provided in order to have hot water in the piping system all the time and not to wait for a long time to have hot water. To size the hot water circulator 1st calculate the total hot water fixture units (as mentioned for cold water ). Normally hot water fixture unit is 0.75 of total fixture unit.

Suppose we have a large flat having the following plumbing fixtures estimate the circulated pump discharge in gpm. :

4 4 2 4 2

showers lavatories bath tubes bidet sinks

x 2 x 0.75 = 6 x 1 x 0.75 = 3 x 2 x 0.75 = 3 x 2 x 0.75 = 6 x 2 x 0.75 = 3 _________ Total = 21 FU For every 20 FU provide 1 gpm of circulation 21 FU/ 20 FU = 1.05 gpm .

Pump head is calculated by multiplying pipe effective length by the pressure drop per 100 ft as discussed earlier including pipe fittings.

For hot water systems in which piping from the heater to the fixture or appliance is short [(30 m), or less], circulating systems are not generally used. But it is common practice to provide circulating pump in all hot water supply systems in which it is desirable to have hot water available continuously at the fixtures. Sizing of hot water circulating pump is simplified by .

1 gpm for every 20 fixtures units in the system.

Or: 0.5 gpm (0.03161/s) for each 0.75”- or 1” riser; 1 gpm (0.06311/s) for each 1.25” - or 1.5” riser; 2 gpm for each 2” riser.

Calculation of circulating pump -1 capacity

Circulating pump 1

H.W. S R.H.W.

H.W.Storage tank

H.W. S R.H.W. Circulating pump C.W.S

Suppose we have a building containing the following plumbing fixtures estimate the circulate pump-1 flow rate in gpm. : Lavatories =

60 x 1 x 0.75 = 45 Fus.

Kitchen sink = 40 x 2 x 0.75 = 60 Fus . Laundry

= 10 x 2 x 0.75 = 15 Fus .

Shower

= 60 x 2 x 0.75 = 90 Fus.

Dishwasher

= 40 x 1 x 0.75 = 30 Fus .

Total = 240 FU For every 20 FU provide 1 gpm of circulation 240 FU/ 20 FU = 12 gpm This is the discharge of the circulated pump , which circulate the water from (boiler tank) to building.

Calculation of the circulating pump-2 capacity Pump 1

P T

P.R.V.

H.W.Storage tank

H.W. S R.H.W.

BOILER

Boiler Circulating pump

FUEL Supply

Expansion Vessel Pump 2

Circulating pump C.W.S

Estimating circulating pump 2 capacity In U.S.units The GPM of the system Circulating pump ; Q (BTU/h) = GPM × (60min/hour) × ΔT1 × (8.3 lb/gal) Q (gpm) = Q (boiler in BTU/h) / (8.3 × ΔT1 × 60min/hour ) = Q (boiler in BTU/h) /500 × ΔT1

In SI units: Power KW = Kg/sec. × 4.2Kj/Kg.C × (11) C Power Kw 227 Kg/sec = = = 4.9 L/s 4.2 × 11 4.2 × 11 This is the discharge of the circulated pump 2, which circulate the water from boiler-storage tank- Boiler.

Head of the Circulated pump As it is known that , the role of the circulated pump is to overcome loss due to pipe friction & fittings.

h A = hL

•The elevation difference is not included . •The head loss due is determined from Darcy equation as mentioned in chap. 9

Suppose we have to estimate the head required of a circulated pump , assuming the following : The pipe length is 600 ft. and an allowance for fittings on straight pipe of 25 %50 % is to be use. 1- Determine, the total effective length E.L that is: The actual pipe length + Equivalent length (due to fittings and valves etc.) L eff . = L + ∑ L e 2- The total head loss or pressure drop hL is determined as : The head loss per unit of length h1(5-7ftw./100ft ) multiplied by the effective length .

Le ff = L + 25% L = 600 + 150 = 750 ft hL = Leff . × 5 ft / 100 ft = 750 × 0.05 = 37.5 ft that is [ 11 m ] Usually the Pump is oversized by 10 % of head & 5% flow .

Instantaneous or semi-instantaneous types of heaters Instantaneous Type: The instantaneous indirect water heater is used to meet a demand for a steady, continuous supply of hot water. In this type of unit, the water is heated instantaneously as it flows through the tubes of the heating coil. The heating medium (steam or hot boiler water) flows through the steel pipe shell yielding a small ratio of hot water volume to heating medium volume. Instantaneous water heaters are designed to provide sufficient capacity to heat the required quantity of water (usually expressed in gpm (l/s)) at the time the hot water draw occurs. Storage tanks are not usually part of an instantaneous water heater, although a separate storage tank may be used to provide hot water. Since instantaneous heaters are of the

high demand type, a circulating pump should be installed in both the boiler water and domestic water piping circuits.

Semi-Instantaneous Type: They are similar to instantaneous water heater except that, Semi-instantaneous water heaters have limited storage. Storage capacities are determined by the manufacturer to average momentary surges of hot water.

Procedure for estimating the heating capacity for instantaneous and semi- instantaneous water heaters. Step(1) Tabulate the number of plumbing fixtures of each type that use hot water. Step(2) Multiply the number of fixtures of each type by the number of fixture units per fixture ( obtained from table HW-3 ) to obtain the total number of fixture units. Step(3) Using the total number of fixtures units obtained from step 2 , determine the maximum demand in gpm using the appropriate curve given in chart HW-4 . Step(4) To the demand of step 3 , add the demand for hot-water fixtures ( or equipment) that operate continuously ( practically 1 gpm) . Step(5) Select a heater that will provide the required rise in temperature ΔT = 101 °F for the total demand of step 3 & 4.

Example on the calculation of water demand using semi-instantaneous type of heaters . Determine the required capacity in gpm of a semiinstantaneous water heater for a high school in which there are 6 wash fountains , 10 showers, 2 service sinks, 1 pantry sink , and 4 private lavatory basins. Step-1& 2: Tabulate the number of plumbing fixture & Multiply

these numbers by the number of fixture units per fixture ( obtained from table HW-3 ) to obtain the total number of fixture units. 6 circular wash fountain =

6 x 2.5 = 15 Fus.

Service sink

= 2 x 2.5 = 5 Fus .

Pantry sink

= 1 x 2.5 = 2.5 Fus .

Showers = 30 x 1.5 = 45 Fus. Private lavatory basins = 4 x 0.75 = 3Fus , Total Fixture units = 70.5 FUs

Step-3 & 4 : Using the total number of fixtures units obtained from step 2 , determine the maximum demand in gpm using the appropriate curve given in chart HW-4 . Refer to figure HW-4 ( the enlarged section) for schools, and read the corresponding value for 70.5 FUs , which is 15 gpm. This is the hot water demand for fixtures that operate intermittently . Now assume at least one fixture operates continuously , and it needs a demand of 1 gpm. The total water flow rate becomes 16 gpm ,this is the capacity of the semi-instantaneous Boiler. Select the desired temperature of the water leave & the temperature of cold water enters the boiler in order to calculate the boiler power. Q (boiler in BTU/h) = 500 × gpm × ΔT Q (boiler ) = 500 × 16 × 101 F = 808 000 BTU/h = 236.6 Kw

HW-3

Ref [2]

HW-4

Another way to determine gpm

HW-4 (b)

15 gpm

Ref [2]

Drawing of Water Distribution Systems

Boiler Selection And Specifications

BOILER SPECIFICATIONS

Qualities of a good boiler are: 1-It should be capable of quick start-up. 2-Should meet large load fluctuations. 3-Occupy less floor space. 4- Should afford easy maintenance and inspection. 5-Should essentially possess the capacity of producing maximum steam with minimum fuel consumption. i.e. high thermal efficiency 6- Simple in construction. 7-Tubes should be sufficiently strong to resist wear and corrosion. 8- Mud and other deposits should not collect on heated plates. 9-The velocity of water and that of flue gas should be a minimum.

Selection of a boiler The selection criteria of a boiler depends very much on the purpose of the boiler i.e. the load requirement. Boiler may be either used to produce steam to a steam turbine, or for heating process. If steam is required for power Generation then superheated steam with the pressure of inlet to the turbine is essential. On the other hand, if the boiler is required for a heating process like an industrial load other applications like hospitals, hotels, kitchens steam or hot water boilers must be considered. For power generation we need essentially a water tube boiler. On the other hand, for Heating process all types are possible. Hot water are usually produced around 100°C And pressure from 2 ~ 8 atm, and high-temp hot water HTHW from 121°C~260°C and pressure over 10.8 atm

High efficiency Cast Iron Boilers

Diesel Fuel

Cast Iron Boilers

Gas boilers 25 to 200 horsepower Water

Boiler for light commercial heating

Cast iron boilers (gas) Cast iron boilers are limited to low-pressure steam or hot water applications, and typically range in size from 25 to 200 horsepower. An example of gas cast iron boilers is the AtmoGas LN is available with an output range from 29 to 51kW. The Atmo-Gas LN is a cast iron boiler supplied for use with natural gas or propane. The boiler is suitable for both central heating and indirect hot water supply for working pressures up to 4.0 Bar. An integral draught diverter is provided for reduced boiler height aiding plant room access.

Diesel fuel boilers

Fire tube boilers • Scotch Marine Fire tube boilers are available for low pressure hot water applications, in which the product of combustion( gases) pass through the tubes which are surrounded by water. • Fire tube boiler has a flame inside the furnace and the combustion gases inside the tubes. The furnace and tubes are within a large vessel which contains water.

• This type of boiler possesses some characteristics that differ from other types, because of it’s large quantity of water (high stored energy) , making it reliable to respond for load changes, if it required an instantaneous load demand where a large quantity of water is needed for a short period of time , so it is preferable to choose this type of boiler to meet instantaneous water requirements.

Fire tube boiler

15 to 1500 horsepower

Burner

• Burner is used in order to burn fuel in more efficient way to give a complete combustion ,prevent the formation of toxic contaminations (NOx, SOx, HC ….). The burner uses air in order to atomize the fuel into small droplets to become more easily to vaporize and combust.

1 Kw = 960 Kcal/hr. = 3413 BTU/ hr. 1boiler hp = 9.81 KW.

References 1- Mechanical & electrical equipment for buildings –by Stein/Reynolds, Ninth edition, John Wiley, 2000. 2-Practical Plumbing Engineering, Cyril M.Harris, ASPE,1998. 3- Building Services & equipment (I/II/III), F.Hall, Third edition, 1994. 4- Upland engineering, Mechanical consulting office, Dr. Ali Hammoud. 5- Applied hydraulics Part I & II .”Lecture notes." by A. Hammoud BAU- 1995 6- Pumps with practical applications, .”Lecture notes.” by A. Hammoud BAU- 1999. 7- Lowara catalogue 8- Plumber’s & pipefilter’s , Calculations Manual by R. Dodge Woodson. 9- Plumbing Design & practice by S G Deolalikar 10-

fluidedesign – Jacques Chaurette

12- Internet web sides

Expansion of Pipe material

ΔL = λ × L × ΔT ΔL =Amount of change in pipe length, mm λ = Coefficient of linear expansion, mm/m. C° ( for (API ) PPr λ=0.15 mm/m ° C) ΔT= Temperature difference , C° L = is the original length of pipe m For example : A 6 m copper pipe is subjected to temperature difference of 50° C Calculate the pipe expansion ΔL : ΔL = 0.15 x 6 x 50 =45 mm

Calculate the length of the Arm Ls ,assuming that D= 25 mm & ΔL =45 mm.

Ls = 30 × 25 × 45 = 10 cm

Calculation of the Arm Length Ls

Determination of the Width between the arms B

B

B= 200 +2 x ΔL =200+ 2x45 = 290 mm.

Domestic Hot Water-Return Pipe Sizing [large systems]. Usually for small installation, a 0.5 or 0.75 in hot water return will be satisfactory. However for large installation, the heat loss from the return line becomes a major consideration. The following method is used to size the return pipe. A. Determine the approximate total length of all hot water supply + return piping. B. Multiply this total length by 30 Btu/Ft (28.8 W/m), for insulated pipe and 60 Btu/Ft (57.6 W/m) for un-insulated pipe to obtain the approximate heat loss.

C. Divide the total heat loss by 10,000 to obtain the total pump capacity in GPM or by 40000 to obtain the pump capacity in L/s. [ 1Kg water /liter x 3600 sec/hr x 11 °C=40 000] [ 1lb water /gal x 60 min/hr x 20 °F=10 000] where 11°C is the allowable temperature drop. D. Select a circulating pump to provide the total required GPM and obtain from pump curves the head created at this flow. E. Multiply the head required by 100 (30.5) and divide by the total length of the longest run of the hot water return piping to determine the allowable friction loss per 100 feet of pipe. F. Determine the required GPM (L/s ) in each circulating loop and size the hot water return pipe based on this GPM and the allowable friction loss as determined above step E.

End of part one

73

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Domestic hot water system By Dr. Ali Hammoud BAU 2008

Contents • • • • •

Hot Water Consumption and Demands Pipe Sizing and Water Heat Storage Pumping Systems and Performance Boiler types & selection Examples

2

Supply Hot Water Pipe Sizing : In a manner similar to cold water pipe , the hot water Pipe is sized . First of all calculate the total fixture units (0.75 of total FU) then find the corresponding flow in gpm and then refer to pipe flow chart and select the corresponding pipe size. You could hot water pipe sizing. The higher the temperature of the water, the lower would be the limit of flow velocity

Pipe Sizing •Avoid oversizing & undersizing Oversizing • High cost – extra but unnecessary • Delay in getting at outlets • Increase heat loss from distributing piping

• Undersizing • Slow or even no water during peak demand • Variation in temperature & pressure at outlet (obvious in mixer for shower) • High noise level 4

Pipe insulation :

Boilers, storage tanks, calorifiers, chimneys, and hot water pipes lose their heat when exposed to outside air. Hence they should be insulated Insulation material should be non-corrosive, nontoxic, moisture resistant, rot- and vermin-proof, fire-resistant, light, easy to apply, anti-weathering, durable and, in addition possess good insulation properties and low thermal conductivity.

Insulation material could be fiberglass or rubber type insulation like armaflex which comes in the form of tube and the pipe is pulled inside it .

• • • •

Pipe insulation: reasons: Heat conservation. Reduce noise. Control surface condensation. Freeze prevention. The insulating material is the same used in duct insulation (Fiber glass)

Insulation thickness Thicknesses or insulation to be supplied and installed for the various systems shall conform to the following table: (Fiber glass thickness) a. Conditioned air supply and return (ductwork) 1 1/2" b. Refrigerant suction and liquid lines: 1/2" c. Condensate drain pipes 1/2" d. Acoustic duct liner. 1” e: Heating water pipes up to 1" diameter: 1" f. Heating water pipes above 1 ”: 1 1/2" g. Domestic hot water pipes: 1" k. Boiler, water heater 2” L. Boiler breeching & steel chimney 2.1/2”

Domestic hot water system Electrical water heater

Water to water storage Heater Boiler +Water Storage

Instantaneous or semi-instantaneous types of heaters

Electrical Water heater Power 3/4 of the total fixture units are used for hot water Hot water

Thermostat

Drain

1.25 "

4 bar Pressure Releif valve

Inlet water flow ? Electrical power :

Temp.

1.5, 2.4, 4-5, & 9 Kw

Electrical water heater

Size of EWH :

1" Cold water

1"

15, 20 , 30, 40, 50 , 66, 80 &1 20 gallons

3/4 of the total fixture units are used forc old water

Power =

S .hc × Kg × ΔT . 4.2 × 150 × (65 − 10) = = 4.2 Kw Heating time × Efficiency 3 × 3600 × 0.75

Where Shc is the specific heat capacity of water ( 4.2 Kj/kg. ºK), Heating time or recovery period T = 3 hrs, The mass of water 150 Liters = 150 Kg., Temperature rise from 10 to 65 ºC. Insulation efficiency is 75-80 %.

In general, electric water heaters are fully automatic and have a storage tank, one or more electric elements, and operating and safety controls. The heating elements are available in a variety of standard voltages and wattages to meet the specific requirements of the installations. Electrical water heater are fitted with electrical resistance ( heater) 1.5 Kw, 2.4 Kw , 4- 5 Kw, 9 Kw . They could be horizontal or vertical type . EWH are insulated and protected by steel jacketing. EWHs should have pressure /temperature relief valve which will relieve excessive pressure . For high pressure EWH the use of none -return is not recommended. In this case ,the pressure relief valve is connected to nearest floor drain. EWHs are controlled by thermostat which will activate the electrical heater automatically. Available EWHs size are 15, 20 , 30, 40, 50 , 66, 80 &1 20 gallons

Resistance heating element Usually the electrical water heaters have a primary resistance heating element near the bottom , and possible a secondary element located within the upper quarter or third of the tank. Minimum wattage ratings of two element heaters are based as follows: For the upper unit (8 watt / liter) of the tank capacity. For the lower unit (5 watt/liter) of the tank capacity. For fast heating of ( 30-40-50 gal ), we take (26 watt/litter) for each unit . For one single element heater ,we take (26 watt/litter)

Resistance heating element location

12

Hot –water demand (definitions) Demand of water supply : is the rate of flow in pgm furnished by a water supply system to various types of plumbing fixtures and water outlets under normal conditions.

Maximum Demand: is the peak value of the demand. The values Hot of water demand

are shown

previously in table (W-2 ).

Demand factor : is the ratio of the maximum demand of the hot water heating system to the total connected load or the total of the individual requirements of all the fixtures of the system.

Water to water storage heater A hot-water storage tank must meet code requirements that depend on its size and pressure and the authority having jurisdiction. Its capacity should be selected so that 60 to 80 percent of the volume of water in the tank may be drawn off before the temperature drop (caused by the incoming cold water) becomes unacceptable. A value of 70 percent usually is used in design calculations. Heated water from boiler enter the coil of the water storage vessel where it will heat the water , and then returns to boiler at lower temperature . Whereas the domestic Hot water leaves the top of the water tank to the supply distribution piping upon demand for the various fixtures and apparatus. When recalculation of the hot water is provided, the warm water is returned to the bottom of the hot water vessel or tank by a circulate pump. Normally heating water from boiler enters the water tank at 180 º F ( 82 º C ) and return to boiler at 160 º F ( 70 º C), ∆T = 12-15 º C.

Hot water requirement for storage

The hot water storage for Buildings & Hotels will be calculated based on unit hourly demand rates as follows : galon per hour Fixture type Building Hotels Lavatory 2 gph 2 gph bathtub 20 gph 20 gph Shower 30 gph 75 gph Sink 10 gph 30 gph Laundry 20 gph 28 gph Dishwashers 15 gph 50 gph Demand factor 0.3 0.25 Storage factor 1.25 0.8 Domestic hot water temperature will be 60° C.

Ref [2]

Table Hw-1

Table Hw-2

Ref [2]

Procedure for estimating the heating capacity (recovery capacity) of a hot-water heating system

having a storage tank.

Step(1) Tabulate the number of fixtures of each type in the building. Step(2) Then multiply the number of fixtures of each type by the probable demand for each type of fixture. Step(3) Obtain the maximum demand by taking the sum of products of step (2) Step(4) Then obtain the hourly heating capacity by multiplying the maximum demand in step3 by the demand factor obtained from table 2. Step(5) multiply the hourly heating capacity of step 4 by the storage capacity factor given in table HW-1 for the appropriate type of building to obtain the required capacity of the storage tank. N.B. Not to be used for instantaneous or semi-instantaneous types of heaters

Example Suppose you have to calculate , the hot water storage capacity , the boiler power of an apartment house (building) having the following data: 60 Lavatories , 40 kitchen sinks ,and 10 laundry 60 showers, and 40 dishwasher . Step1& 2. The probable water demand are as follows: From table HW-1; Lavatories =

60 x 2 gph = 120 gph

Kitchen sink = 40 x 10 gph = 400 gph . Laundry

= 10 x 20 gph = 200 gph .

Shower

= 60 x 30 gph = 1800 gph.

Dishwasher

=

40 x 15 gph = 600 gph .

Step 3 : The Maximum demand of the hot water demand is: =((60×2gph)+ (60×30gph)+ (40×10gph) + (40 x 15 gph) +(10x 20 gph ))= 3120 gph. Step 4 Building demand factor = 0.3 (from table HW-1 ). Hourly heating capacity = (3120×0.3) = (936 gph)= 15.6 gpm.= 0.985 liter /sec or 3546 L/hr. Step 5 The required capacity of the storage tank is: Usable capacity = 936 x 1.25 = 1170 gal i.e. (4429 liters) [1.25 is the storage factor from table Hw-1]

Safety storage: Since only 70% of the tank is usable, so the actual

tank capacity = (1170 / 0.7 gallon) = (1670 gallon) that is (6327 liters).

Boiler power Calculation

From the basic equation

Q = m× C × ΔT

In U.S.units Boiler Power Q (BTU/h) = GPM × (60min/hou r) × ∆T1 × (8.3 lb/gal) Q (boiler in BTU/h) = 500 × gpm × ∆T1 Where

gpm = is the calculated water demand in gpm (flow rate ) ; ∆T = is the temperatur e difference between initial & Fianal [ 101 DF]

From the previous example, we have 936 gph = 15.6 gpm of water to be heated , temperature rise ∆T=101 ºF. QT= 500 x 15.6 x 101 = 787800 BTU/hr =230 Kw. In SI units: Power KW =

0.985 Kg / sec . × 4.2 Kj / Kg .C × (55) C = 227.5 Kw

Acceptable Temperature of domestic Hot water

Ref [2]

Fº = 1.8 Cº + 32 , Cº = 0.55 (Fº - 32)

Boiler power 1 Kw = 860 Kcal/hr. = 3413 BTU/ hr. 1boiler hp = 9.81 KW.

The required gross boiler power for heating water + overcome the heat loss from pipe and boiler + the heat needed to rise the initial water temperature ( Pick up ) is given by: QBoiler = QT [1+ a + b] QT = Calculated “boiler power’ ,a = additional heat coefficient to overcome the heat loss in the pipe systems and boiler. [0.1] b = additional heat coefficient to overcome the pick up period [0.1 to 0.2] . The required gross output is then; QBoiler = [1.2 or 1.3 ] QT As mentioned before : for QT= 230 Kw ,

The required boiler power = 1.2 x QT = 1.2 x230 =276 Kw

T=82 °C

T=10 °C T=72 °C

Hot Water Pumps Pumps used in hot water primary and secondary distribution systems are used mainly for maintaining or increasing the rate circulation. Pumps are constructed to withstand the high temperature of water. They have a cast-iron body and a gunmetal impeller. Impellers made of other materials, such as bronze , stainless steel and cast iron are also used. Each pump has a valve on the suction and delivery side and a check valve on the delivery side. A bypass on the lines enables the removal of pumps for maintenance and repairs. In well-designed systems, frictional loss in pipe-lines during recirculation is quite low and the pump horse power is also small .Hot water recirculation pumps should never be used as booster pumps to increase the pressure in the hot water system, as this creates imbalance in the pressures of hot and cold water supply.

Hot water & Boiler Circulated pumps There are two centrifugal circulated pumps used in Hot water system: Pump1: is used to circulate the hot water system (Located in the returned pipe ). Pump2 : Is used to circulate the hot water from boiler to storage tank (closed system). Circulating pump is controlled by an immersion thermostat (in the return line) set to start and stop the pump over a 11 ºC). However for continuous hot water supply the thermostat is eliminated.

Hot water circulating pump ( inside the apartment) If water heater is located far away from plumbing fixtures (more than 30 m) hot water circulator might be provided in order to have hot water in the piping system all the time and not to wait for a long time to have hot water. To size the hot water circulator 1st calculate the total hot water fixture units (as mentioned for cold water ). Normally hot water fixture unit is 0.75 of total fixture unit.

Suppose we have a large flat having the following plumbing fixtures estimate the circulated pump discharge in gpm. :

4 4 2 4 2

showers lavatories bath tubes bidet sinks

x 2 x 0.75 = 6 x 1 x 0.75 = 3 x 2 x 0.75 = 3 x 2 x 0.75 = 6 x 2 x 0.75 = 3 _________ Total = 21 FU For every 20 FU provide 1 gpm of circulation 21 FU/ 20 FU = 1.05 gpm .

Pump head is calculated by multiplying pipe effective length by the pressure drop per 100 ft as discussed earlier including pipe fittings.

For hot water systems in which piping from the heater to the fixture or appliance is short [(30 m), or less], circulating systems are not generally used. But it is common practice to provide circulating pump in all hot water supply systems in which it is desirable to have hot water available continuously at the fixtures. Sizing of hot water circulating pump is simplified by .

1 gpm for every 20 fixtures units in the system.

Or: 0.5 gpm (0.03161/s) for each 0.75”- or 1” riser; 1 gpm (0.06311/s) for each 1.25” - or 1.5” riser; 2 gpm for each 2” riser.

Calculation of circulating pump -1 capacity

Circulating pump 1

H.W. S R.H.W.

H.W.Storage tank

H.W. S R.H.W. Circulating pump C.W.S

Suppose we have a building containing the following plumbing fixtures estimate the circulate pump-1 flow rate in gpm. : Lavatories =

60 x 1 x 0.75 = 45 Fus.

Kitchen sink = 40 x 2 x 0.75 = 60 Fus . Laundry

= 10 x 2 x 0.75 = 15 Fus .

Shower

= 60 x 2 x 0.75 = 90 Fus.

Dishwasher

= 40 x 1 x 0.75 = 30 Fus .

Total = 240 FU For every 20 FU provide 1 gpm of circulation 240 FU/ 20 FU = 12 gpm This is the discharge of the circulated pump , which circulate the water from (boiler tank) to building.

Calculation of the circulating pump-2 capacity Pump 1

P T

P.R.V.

H.W.Storage tank

H.W. S R.H.W.

BOILER

Boiler Circulating pump

FUEL Supply

Expansion Vessel Pump 2

Circulating pump C.W.S

Estimating circulating pump 2 capacity In U.S.units The GPM of the system Circulating pump ; Q (BTU/h) = GPM × (60min/hour) × ΔT1 × (8.3 lb/gal) Q (gpm) = Q (boiler in BTU/h) / (8.3 × ΔT1 × 60min/hour ) = Q (boiler in BTU/h) /500 × ΔT1

In SI units: Power KW = Kg/sec. × 4.2Kj/Kg.C × (11) C Power Kw 227 Kg/sec = = = 4.9 L/s 4.2 × 11 4.2 × 11 This is the discharge of the circulated pump 2, which circulate the water from boiler-storage tank- Boiler.

Head of the Circulated pump As it is known that , the role of the circulated pump is to overcome loss due to pipe friction & fittings.

h A = hL

•The elevation difference is not included . •The head loss due is determined from Darcy equation as mentioned in chap. 9

Suppose we have to estimate the head required of a circulated pump , assuming the following : The pipe length is 600 ft. and an allowance for fittings on straight pipe of 25 %50 % is to be use. 1- Determine, the total effective length E.L that is: The actual pipe length + Equivalent length (due to fittings and valves etc.) L eff . = L + ∑ L e 2- The total head loss or pressure drop hL is determined as : The head loss per unit of length h1(5-7ftw./100ft ) multiplied by the effective length .

Le ff = L + 25% L = 600 + 150 = 750 ft hL = Leff . × 5 ft / 100 ft = 750 × 0.05 = 37.5 ft that is [ 11 m ] Usually the Pump is oversized by 10 % of head & 5% flow .

Instantaneous or semi-instantaneous types of heaters Instantaneous Type: The instantaneous indirect water heater is used to meet a demand for a steady, continuous supply of hot water. In this type of unit, the water is heated instantaneously as it flows through the tubes of the heating coil. The heating medium (steam or hot boiler water) flows through the steel pipe shell yielding a small ratio of hot water volume to heating medium volume. Instantaneous water heaters are designed to provide sufficient capacity to heat the required quantity of water (usually expressed in gpm (l/s)) at the time the hot water draw occurs. Storage tanks are not usually part of an instantaneous water heater, although a separate storage tank may be used to provide hot water. Since instantaneous heaters are of the

high demand type, a circulating pump should be installed in both the boiler water and domestic water piping circuits.

Semi-Instantaneous Type: They are similar to instantaneous water heater except that, Semi-instantaneous water heaters have limited storage. Storage capacities are determined by the manufacturer to average momentary surges of hot water.

Procedure for estimating the heating capacity for instantaneous and semi- instantaneous water heaters. Step(1) Tabulate the number of plumbing fixtures of each type that use hot water. Step(2) Multiply the number of fixtures of each type by the number of fixture units per fixture ( obtained from table HW-3 ) to obtain the total number of fixture units. Step(3) Using the total number of fixtures units obtained from step 2 , determine the maximum demand in gpm using the appropriate curve given in chart HW-4 . Step(4) To the demand of step 3 , add the demand for hot-water fixtures ( or equipment) that operate continuously ( practically 1 gpm) . Step(5) Select a heater that will provide the required rise in temperature ΔT = 101 °F for the total demand of step 3 & 4.

Example on the calculation of water demand using semi-instantaneous type of heaters . Determine the required capacity in gpm of a semiinstantaneous water heater for a high school in which there are 6 wash fountains , 10 showers, 2 service sinks, 1 pantry sink , and 4 private lavatory basins. Step-1& 2: Tabulate the number of plumbing fixture & Multiply

these numbers by the number of fixture units per fixture ( obtained from table HW-3 ) to obtain the total number of fixture units. 6 circular wash fountain =

6 x 2.5 = 15 Fus.

Service sink

= 2 x 2.5 = 5 Fus .

Pantry sink

= 1 x 2.5 = 2.5 Fus .

Showers = 30 x 1.5 = 45 Fus. Private lavatory basins = 4 x 0.75 = 3Fus , Total Fixture units = 70.5 FUs

Step-3 & 4 : Using the total number of fixtures units obtained from step 2 , determine the maximum demand in gpm using the appropriate curve given in chart HW-4 . Refer to figure HW-4 ( the enlarged section) for schools, and read the corresponding value for 70.5 FUs , which is 15 gpm. This is the hot water demand for fixtures that operate intermittently . Now assume at least one fixture operates continuously , and it needs a demand of 1 gpm. The total water flow rate becomes 16 gpm ,this is the capacity of the semi-instantaneous Boiler. Select the desired temperature of the water leave & the temperature of cold water enters the boiler in order to calculate the boiler power. Q (boiler in BTU/h) = 500 × gpm × ΔT Q (boiler ) = 500 × 16 × 101 F = 808 000 BTU/h = 236.6 Kw

HW-3

Ref [2]

HW-4

Another way to determine gpm

HW-4 (b)

15 gpm

Ref [2]

Drawing of Water Distribution Systems

Boiler Selection And Specifications

BOILER SPECIFICATIONS

Qualities of a good boiler are: 1-It should be capable of quick start-up. 2-Should meet large load fluctuations. 3-Occupy less floor space. 4- Should afford easy maintenance and inspection. 5-Should essentially possess the capacity of producing maximum steam with minimum fuel consumption. i.e. high thermal efficiency 6- Simple in construction. 7-Tubes should be sufficiently strong to resist wear and corrosion. 8- Mud and other deposits should not collect on heated plates. 9-The velocity of water and that of flue gas should be a minimum.

Selection of a boiler The selection criteria of a boiler depends very much on the purpose of the boiler i.e. the load requirement. Boiler may be either used to produce steam to a steam turbine, or for heating process. If steam is required for power Generation then superheated steam with the pressure of inlet to the turbine is essential. On the other hand, if the boiler is required for a heating process like an industrial load other applications like hospitals, hotels, kitchens steam or hot water boilers must be considered. For power generation we need essentially a water tube boiler. On the other hand, for Heating process all types are possible. Hot water are usually produced around 100°C And pressure from 2 ~ 8 atm, and high-temp hot water HTHW from 121°C~260°C and pressure over 10.8 atm

High efficiency Cast Iron Boilers

Diesel Fuel

Cast Iron Boilers

Gas boilers 25 to 200 horsepower Water

Boiler for light commercial heating

Cast iron boilers (gas) Cast iron boilers are limited to low-pressure steam or hot water applications, and typically range in size from 25 to 200 horsepower. An example of gas cast iron boilers is the AtmoGas LN is available with an output range from 29 to 51kW. The Atmo-Gas LN is a cast iron boiler supplied for use with natural gas or propane. The boiler is suitable for both central heating and indirect hot water supply for working pressures up to 4.0 Bar. An integral draught diverter is provided for reduced boiler height aiding plant room access.

Diesel fuel boilers

Fire tube boilers • Scotch Marine Fire tube boilers are available for low pressure hot water applications, in which the product of combustion( gases) pass through the tubes which are surrounded by water. • Fire tube boiler has a flame inside the furnace and the combustion gases inside the tubes. The furnace and tubes are within a large vessel which contains water.

• This type of boiler possesses some characteristics that differ from other types, because of it’s large quantity of water (high stored energy) , making it reliable to respond for load changes, if it required an instantaneous load demand where a large quantity of water is needed for a short period of time , so it is preferable to choose this type of boiler to meet instantaneous water requirements.

Fire tube boiler

15 to 1500 horsepower

Burner

• Burner is used in order to burn fuel in more efficient way to give a complete combustion ,prevent the formation of toxic contaminations (NOx, SOx, HC ….). The burner uses air in order to atomize the fuel into small droplets to become more easily to vaporize and combust.

1 Kw = 960 Kcal/hr. = 3413 BTU/ hr. 1boiler hp = 9.81 KW.

References 1- Mechanical & electrical equipment for buildings –by Stein/Reynolds, Ninth edition, John Wiley, 2000. 2-Practical Plumbing Engineering, Cyril M.Harris, ASPE,1998. 3- Building Services & equipment (I/II/III), F.Hall, Third edition, 1994. 4- Upland engineering, Mechanical consulting office, Dr. Ali Hammoud. 5- Applied hydraulics Part I & II .”Lecture notes." by A. Hammoud BAU- 1995 6- Pumps with practical applications, .”Lecture notes.” by A. Hammoud BAU- 1999. 7- Lowara catalogue 8- Plumber’s & pipefilter’s , Calculations Manual by R. Dodge Woodson. 9- Plumbing Design & practice by S G Deolalikar 10-

fluidedesign – Jacques Chaurette

12- Internet web sides

Expansion of Pipe material

ΔL = λ × L × ΔT ΔL =Amount of change in pipe length, mm λ = Coefficient of linear expansion, mm/m. C° ( for (API ) PPr λ=0.15 mm/m ° C) ΔT= Temperature difference , C° L = is the original length of pipe m For example : A 6 m copper pipe is subjected to temperature difference of 50° C Calculate the pipe expansion ΔL : ΔL = 0.15 x 6 x 50 =45 mm

Calculate the length of the Arm Ls ,assuming that D= 25 mm & ΔL =45 mm.

Ls = 30 × 25 × 45 = 10 cm

Calculation of the Arm Length Ls

Determination of the Width between the arms B

B

B= 200 +2 x ΔL =200+ 2x45 = 290 mm.

Domestic Hot Water-Return Pipe Sizing [large systems]. Usually for small installation, a 0.5 or 0.75 in hot water return will be satisfactory. However for large installation, the heat loss from the return line becomes a major consideration. The following method is used to size the return pipe. A. Determine the approximate total length of all hot water supply + return piping. B. Multiply this total length by 30 Btu/Ft (28.8 W/m), for insulated pipe and 60 Btu/Ft (57.6 W/m) for un-insulated pipe to obtain the approximate heat loss.

C. Divide the total heat loss by 10,000 to obtain the total pump capacity in GPM or by 40000 to obtain the pump capacity in L/s. [ 1Kg water /liter x 3600 sec/hr x 11 °C=40 000] [ 1lb water /gal x 60 min/hr x 20 °F=10 000] where 11°C is the allowable temperature drop. D. Select a circulating pump to provide the total required GPM and obtain from pump curves the head created at this flow. E. Multiply the head required by 100 (30.5) and divide by the total length of the longest run of the hot water return piping to determine the allowable friction loss per 100 feet of pipe. F. Determine the required GPM (L/s ) in each circulating loop and size the hot water return pipe based on this GPM and the allowable friction loss as determined above step E.

End of part one

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