11

An example calculation of chiller-cooling tower pipe sizing and pump duty.

Figure (1) shows a cooling tower configuration used to cool three condenser units connected in parallel .The flow rate in each condenser , pipe lengths in feet , pressure drop in the condensers ,and tower nozzle are shown in the drawing. The nozzle pressure drop in the tower is taken as 15 ft. The pipes are commercial steel pipe . The system is an open-re circulating circuit. It is required:

Size all the pipes , base your calculation on the maximum allowable flow velocity limitation of 5-6 ft/s. Specify the pumping requirements.

Decide which condenser circuits need to be balanced.

Calculate the required pump power in hp, if its allowable efficiency is not less than 60% .

2

Open reOpen re-- circulated Circuit circulated Circuit

23

Solution:Solution:This is a prototype example of the method by which chiller pipesThis is a prototype example of the method by which chiller pipes, cooling , cooling tower pipes & pump duty are sized. The first step in our calcultower pipes & pump duty are sized. The first step in our calculation is to ation is to select criteria for sizing the pipes . This type of problem callselect criteria for sizing the pipes . This type of problem called (type III ed (type III [Ref 1] ), cannot be solved directly since the pipe diameter & t[Ref 1] ), cannot be solved directly since the pipe diameter & the flow he flow velocity are not known. The solution requires a trial and error velocity are not known. The solution requires a trial and error method. method. Iterate for Iterate for ff" since we cannot solve for "" since we cannot solve for "DD" in term of "" in term of "ff ", the iteration ", the iteration must proceed by assuming a value for must proceed by assuming a value for ff and calculate and calculate DD , ", "/ D/ D and and ""Reynolds numberReynolds number . Finally evaluate . Finally evaluate ff from Moody diagram and compare from Moody diagram and compare it with the estimated it with the estimated ff until the two values of until the two values of ff are nearly equal. This are nearly equal. This way of solving the problem is not practical. way of solving the problem is not practical. AshraeAshrae [Ref2] & Career[Ref3] presented pipe flow charts based on Hazen[Ref2] & Career[Ref3] presented pipe flow charts based on HazenWilliams equation which are used to determine the chiller pipe dWilliams equation which are used to determine the chiller pipe diameters. iameters. Furthermore ,two types of pipe flow charts are available , one fFurthermore ,two types of pipe flow charts are available , one for open reor open re--circulating circuit ( such as chiller cooling tower systems ) ancirculating circuit ( such as chiller cooling tower systems ) and the second d the second closed reclosed re-- circulating circuit( such as radiators heating systems ) . Ref[circulating circuit( such as radiators heating systems ) . Ref[2] 2] & Ref[3] also presented velocity and pressure limitations for si& Ref[3] also presented velocity and pressure limitations for sizing the zing the pipes. In our example we assumed that ,the maximum flow velocitpipes. In our example we assumed that ,the maximum flow velocity is 5y is 5--6 6 ft/s, and the maximum pressure limitation is no more than 7 ft/1ft/s, and the maximum pressure limitation is no more than 7 ft/100 ft in 00 ft in the main pipes . Higher values may be used in the parallel circuthe main pipes . Higher values may be used in the parallel circuits. On the its. On the other hand ,the minor head loss due to fittings, gates etc.. , iother hand ,the minor head loss due to fittings, gates etc.. , is calculated s calculated using the equivalent length technique instead of the using the equivalent length technique instead of the KK values as values as

d d b R f[3] mm nd d b R f[3]

4

Enter the pipe flow chart along the ordinate with the value of Q=810 gpm and then intersect the line of V= 6 ft/sec. From the intersection ,move downward to the abscissa and read the value of ( hL/100ft ) and the corresponding flow pipe diameter (D).Now it is clear that the intersection between the Q & V lies between 6 and 8 diameters . If the 6 in pipe is used , the flow velocity will be around 8.6 ft/s which is greater than the given value 6 ft/s and the pressure drop is at the maximum limit (7 ft/100ft) This is not recommended . If the 8 pipe is used , the pressure drop will be 1.6 ft/100 ft ,and corresponding flow velocity is about 5.2 ft/s which is acceptable .Similarly we size all the other pipes.

Step Step --1 Sizing the pipes1 Sizing the pipes

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As mentioned previously for Q= 810 gpm, the nearest pipe diameter is D = 8 and the corresponding velocity is 5.2 ft / s .The friction factor is 1.6 ft / 100 ft .

The length of the pipe is L = 4 + 37 + 63 + 7 + 6 = 117 ft

From Table (1) The equivalent length of fittings and valves are :

4 * 90 long radius elbow Le1 = 4x13 ft

one gate valve Le2 = 1 x 9 ft

Le= 61 ft The effective length is: Leffective = L + Le = 178 ft

hL = 178 x 1.6/100 = 2.848 ft

Due to aging effect the pipe head loss becomes :

hL(AB) = 2.848 x 1.15 = 3.3 ft

Step Step --2 Calculate the total head loss in the suction 2 Calculate the total head loss in the suction side of the pump side of the pump Line ALine A--BB

6

Since the system is open system the following pipe flow chart will be used

Selected Pipe Selected Pipe diameterdiameter

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Table (1)Table (1)

8

table 2

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The head loss in Line CThe head loss in Line C--D :D :Since it is the same flow rate 810 gpm and the same velocity & pressure limitations, then the pipe diameter , velocity and head loss per 100 Ft are the same as line AB ,that is D=8 diameter and velocity 5.2 ft / s, and the head loss is 1.6 ft / 100 ft .

The geometrical length of the pipe is L = 29 ft

From table (1 & 2) the equivalent length of the fittings and valves :

2 x 90 long radius elbow = 2x13

One gate valve = 9 ft

Tee- Through = 13 ft

Check valve = 80 ft

Leffective = = L + Le = 29+128 = 157ft

hL = 157 x 1.6/100 = 2.512 ft

Due to aging effect

hL(CD) = 2.512 x 1.15 = 2.9 ft

Le= 128 ft

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The head loss in Line DThe head loss in Line D--E:E:Condenser (a) requires 300 gpm, therefore the flow through DE will be 810-300 = 510 gpm. Now for Q = 510 gpm , and V= 6 ft/mi. the nearest pipe diameter is D = 6 and the corresponding flow velocity is 5.75 ft / s

The pipe head loss /100 ft is 3 ft / 100 ft .

The length of the pipe is L = 52 ft

the equivalent length of fittings and valves :

2x Tee (flow through )= 2 x13 ft Le= 26 ft Leffective = L + Le = 52+26= 78 ft

The head loss becomes hL = 78 x 3/100 = 2.34 ft

Aging effect :

hL(DE) = 2.34 x1.15 = 2.7 ft

611

The head loss in branch DDThe head loss in branch DDI:I:Similarly for Q = 300 gpm , we get D = 5 and the velocity is 4.9 ft / s

The corresponding friction factor is 2.7 ft / 100 ft .

the length of the pipe is L = 6+2+4+8= 20ft

Equivalent length for fittings and valves :

2 x 90 long radius elbow = 2x 8.2ft. Two gate valves = 2x 6 ft.

2x Tees ( branch ) = 2x25 ft. Le= 70.2ft

Leffe. = L + Le = 20+ 70.2 = 90.2 ft

Pipe head loss hL = 90.2 x 2.7/100 = 2.4 ft

Aging effect : hL(DDI) = 2.4 x 1.15 = 2.8 ft

Pressure drop in the condenser (a) = 21 ft.

The total head loss in branch DDI including condenser loss:

The head loss in branch DDThe head loss in branch DDI: I: =2.8+21 =23.8 ft.

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The head loss in branch EEThe head loss in branch EEH :H :Similarly for Q = 60 gpm , we get D =2. and the velocity is 6 ft / s ,the corresponding friction factor is 13 ft / 100 ft .

the length of the pipe is L = 6+1+3+8= 18 ft

the pressure drop due to fittings and valves :

2 x 90 long radius elbow = 2x 3.3 ft. Two gate valves = 2x 2.3 ft.

2x Tees ( branch ) =2x10 ft.

Le= 2x3.3+2x2.3+20 = 31.2 ft Leffe. = L + Le = 18+ 31.2=49.2 ft

Pipe head loss hL = 49.2 x 13/100 = 6.4 ft

Aging effect : hL(EEH ) = 6.4 x 1.15 = 7.3 ft

The head loss including the condenser pressure drop (b) , 30 ft(b) , 30 ft becomes :

The head loss in branch EFHThe head loss in branch EFH = 7.3+30 =37.36 ft.

713

The head loss in branch EFHThe head loss in branch EFH ::Similarly for Q = 450 gpm , we get D = 6 and the velocity is 4.9 ft / s , and the corresponding head loss factor is 2.5 ft / 100 ft .

the length of the pipe is L = 28+6+2+4+8+28= 76 ft

the fittings and valves :

4 x 90 long radius elbow = 4x 10ft. Two gate valves = 2x 7 ft.

2x Tees ( Thru ) = 10 x2 = 20 ft. Le= 40+14+20 = 74 ft

Leffe. = L + Le =76 + 74= 150 ft

The head loss becomes hL = 150 x 2.5/100 = 3.75 ft

Aging effect : hL(EFH) = 3.75 x 1.15 = 4.3 ft

The head loss including the condenser pressure drop (c) , 27 ft(c) , 27 ft becomes :

The head loss in branch EFHThe head loss in branch EFH = 4.3+27=31.3 ft.

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3.888IKIK

2.766HIHI

31.36 6 EFHEFH

37.3622EEEEHH

Pipe name Diameter Total head (ft)

23.8 55DDDDII

Summary of pressure drop in ft Summary of pressure drop in ft

815

Since the Line AB and KI are common , the head loss in circuits ( DDI ) , ( DEEHI) and (DFI) are calculated.hL (DDI) = 23.8 ftft

hL (DEEHI) = hL(DE) + hL(EEH)+ hL (HI)

= 2.7 + 37.36 + 2.7 = 2.7 + 37.36 + 2.7 = 42.76 ft42.76 ft is the longest runis the longest run..

hL (DFI) = hL(DE) + hL (EFH) + hL (HI)

= 2.7 + 31.3 + 2.7 2.7 + 31.3 + 2.7 = 36.7 ft.

The highest head loss is in branch bThe highest head loss is in branch b

Therefore both controlling valves Therefore both controlling valves aa & & cc need to be adjusted in need to be adjusted in order to balance the circuit or we may increase the pipe order to balance the circuit or we may increase the pipe diameter in branch diameter in branch bb from 2from 2 to 3to 3 in order to reduce the in order to reduce the pressure drop per 100 ft from 13 ft/100 ft to 1.6 ft/100 ft. pressure drop per 100 ft from 13 ft/100 ft to 1.6 ft/100 ft.

The head loss in the longest run is determined as follows: The head loss in the longest run is determined as follows:

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The pumps head is :hA= hL + hL equipemeny +V2/(2g) + ( ZKZ1 )Where the total head loss in the longest circuit is:hL= [hLAB + hLCD +hL (IK)] = [ 3.3+2.9+3.8]= 10ft ,hL equipemeny = pressure drop condenser branch (b)+ pressure

drop in nozzle (15 ft). Elevation difference =9 ft, Kinetic energy =0.413 fthA = 10+42.76+15+0.413 +9 = 77 ft = 23.15 m

For the pump head of hA=18.6 m ,the flow rate of 810 gpm( Qv=0.0511 m3/s) , and the pump efficiency of = 60% we get; Pumps power = ( 9.81 x0.0511x23.15 ) /0.60

= 19 KW= 25.8 hp

gg

/)hQ(Power'Pump Av =s

917

References

1- Manson & Young Fluid MechanicsJohn Wiley & son 2000

2- Carrier 1987 System design manual 2- Ashrae 1997 Fundamental