Domestic hot water system

Chap. 4

By Dr. Ali HammoudBAU 2008

2

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

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.

Supply Hot Water Pipe Sizing :

The higher the temperature of the water, the lower would be the limit of flow velocity

4

Pipe Sizing

•Avoid oversizing & undersizingOversizing• 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

Pipe insulation :Boilers, storage tanks, calorifiers, chimneys, and hot water pipes lose their heat when exposed to outside air. Hence they should be insulatedInsulation material should be non-corrosive, non-toxic, 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)

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”

Insulation thickness

Domestic hot water system

Water to water storage HeaterBoiler +Water Storage

Electrical water heater

Instantaneous or semi-instantaneoustypes of heaters

KwEfficiencytimeHeating

TKghcSPower 2.475.036003

)1065(1502.4..=

××−××

=×

Δ××=

Electrical Water heater Power

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 %.

4 bar

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

3/4 of the total fixture units are used for hot water

Electrical water heater

Cold water

Hot water

1"1"

1.25 "

Inlet water flow ? Pressure Releif valve

Temp.

Drain ThermostatElectrical power :

1.5, 2.4, 4-5, & 9 Kw

Size of EWH :

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

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

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

12

Resistance heating element location

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 fixturesand water outlets under normal conditions.

Maximum Demand:is the peak value of the demand. The values Hot ofwater 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 thetotal of the individual requirements of all the fixturesof 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 hourlydemand rates as follows :

galon per hourFixture type Building Hotels

Lavatory 2 gph 2 gphbathtub 20 gph 20 gphShower 30 gph 75 gphSink 10 gph 30 gphLaundry 20 gph 28 gphDishwashers 15 gph 50 gphDemand factor 0.3 0.25Storage factor 1.25 0.8Domestic hot water temperature will be 60° C.

Ref [2]

Table Hw-1

Table Hw-2

Ref [2]

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 probabledemand for each type of fixture.

Step(3) Obtain the maximum demand by taking the sum of products ofstep (2)

Step(4) Then obtain the hourly heating capacity by multiplying themaximum demand in step3 by the demand factor obtained from table 2.

Step(5) multiply the hourly heating capacity of step 4 by the storagecapacity factor given in table HW-1 for the appropriate type of building toobtain the required capacity of the storage tank.

N.B. Not to be used for instantaneous or semi-instantaneous types ofheaters

Procedure for estimating the heating capacity (recovery capacity) of a hot-water heating system having a storage tank.

Suppose you have to calculate , the hot water storagecapacity , 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 .

Example

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).

From the basic equation Q = m× C × ΔT

Boiler power Calculation

F]101[Fianal &initialbetweendifference etemperatur theis ∆T;)rate(flowgpmindemandwatercalculatedtheisgpmWhere

∆Tgpm 500BTU/h) in (boiler Qlb/gal)(8.3∆Tr)(60min/houGPM(BTU/h) QPowerBoiler

unitsU.S. In

1

1

=

=

××=

×××=

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:KwCCKgKjKgKWPower 5.227)55(./2.4.sec/985.0 =××=

Acceptable Temperature of domestic Hot water

Fº = 1.8 Cº + 32 , Cº = 0.55 (Fº - 32)Ref [2]

Boiler power1 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=72 °CT=10 °C

T=82 °C

Pumps used in hot water primary and secondary distributionsystems are used mainly for maintaining or increasing the ratecirculation. Pumps are constructed to withstand the high temperature of water. They have a cast-iron body and a gun-metal 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 Pumps

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.

4 showers x 2 x 0.75 = 64 lavatories x 1 x 0.75 = 32 bath tubes x 2 x 0.75 = 34 bidet x 2 x 0.75 = 62 sinks 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.

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

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

R.H.W.

H.W. S

Circulating pump

H.W.Storage tank

C.W.S

R.H.W. H.W. S

Circulating pump 1

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 .

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

Total = 240 FUFor 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

H.W.Storage tankTP

Expansion Vessel

BOILER

FUEL Supply

P.R.V.

Circulating pump

H.W. S

R.H.W.

Circulating pumpBoiler

C.W.S

Pump 2

Pump 1

Estimating circulating pump 2 capacity

1

1

1

T/500BTU/h) in(boiler Q )60min/hourT(8.3/BTU/h) in(boiler Q(gpm) Q

lb/gal)8.3(T)60min/hour(GPM(BTU/h) Q;pumpgCirculatinsystem theof GPM The

unitsU.S. In

Δ×=×Δ×=×Δ××=

This is the discharge of the circulated pump 2, which circulate the water from boiler-storage tank- Boiler.

In SI units:

L/s4.9114.2

227114.2KwPowerKg/sec

C)(114.2Kj/Kg.CKg/sec.KWPower

=×

=×

=

××=

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.

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

LA hh =

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.)

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 .

L L Leff e. = +∑

]11[5.3705.0750100/5

750150600%25

. misthatftftftLh

ftLLLe

effL

ff

=×=×=

=+=+=

Usually the Pump is oversized by 10 % of head & 5% flow .

Instantaneous or semi-instantaneoustypes 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.

Step(1) Tabulate the number of plumbing fixtures of each type that usehot water.

Step(2) Multiply the number of fixtures of each type by the number offixture units per fixture ( obtained from table HW-3 ) to obtain the totalnumber 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 inchart 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 temperatureΔT = 101 °F for the total demand of step 3 & 4.

Procedure for estimating the heating capacity for instantaneous and semi- instantaneous water heaters.

Example on the calculation of water demand using semi-instantaneous type of heaters .

Determine the required capacity in gpm of a semi-instantaneous 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.

Kw236.6BTU/h 000808F10116 500)(boiler Q ΔTgpm500BTU/h) in(boiler Q

==××=××=

HW-3

Ref [2]

Another way to determine gpm

HW-4

(b)

HW-4

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 efficiency6- Simple in construction.7-Tubes should be sufficiently strong to resist wear andcorrosion.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 HTHWfrom 121°C~260°C and pressure over 10.8 atm

High efficiency Cast Iron Boilers

Diesel Fuel

Cast Iron Boilers

Boiler for light commercial heating

Gas boilers

Water

25 to 200 horsepower

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 Atmo-Gas 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.

Cast iron boilers (gas)

Diesel fuel 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.

Fire tube boilers

• 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 Burner

15 to 1500 horsepower

• 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 K

w =

960

Kca

l/hr.

= 34

13 B

TU

/ hr.

1boi

ler

hp

= 9.

81 K

W.

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

ΔL =Amount of change in pipe length, mmλ = 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° CCalculate the pipe expansion ΔL :ΔL = 0.15 x 6 x 50 =45 mm

TLL Δ××=Δ λ

Expansion of Pipe material

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

cm10452530Ls =××=

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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