A chilled water air conditioning system is to be designed fo

A chilled water air conditioning system is to be designed for a three office, as outlined in Fig 2. The chilled water flow rate is 40 gal/min. The pressure drop across the air handling unit (AHU) of each office is 10 psi, and the pressure drop across the chiller is 25 psi. Neglecting the losses of any elbows or tees, all pipes Schedule 40 have a C value of 120: The pipe lengths, pipe size, are shown in the table. Specify the following What is the flow rate in all pipes? Delta P required for the pump Pump power requirements for eta = 0.85.

Solution

Solution:

Let\'s say flow rate through AHU 2 is X gal/min, so flow rate through B-C pipe is (40 - X) gal/min.

Again consider flow rate through AHU 3 is Y gal/min, so flow rate through AHU 1 is (40 - X - Y) gal/min.

Loss in pipe formula : Hazen-Williams formula is used.

The formula is pressure drop per foot of pipe length, P_d = (4.52 x Q^1.85)/ (C^1.85 x d^4.87)

Where Q is in gal/min, C is a constant = 120 for schedule 40 pipe (given), d is diameter in inches.

Pressure drop across chiller = 25 psi.

Pressure drop across AHU = 10 psi.

Pressure drop in E-A-P-B pipe = 50 x {(4.52 x 40^1.85)/(120^1.85 x 2.5^4.87)} = 0.34 psi.

Pressure drop in B-D = Pressure drop in B-C-D

Or, 10 + 60 x {(4.52 x X^1.85)/(120^1.85 x 1.5^4.87)} = 35 x {(4.52 x (40-X)^1.85)/(120^1.85 x 2^4.87)} +10 + 60 x {(4.52 x (40-X-Y)^1.85)/(120^1.85 x 1.5^4.87)}

After simplification, 12X^1.85 - 1.724(40-X)^1.85 + 12(40-X-Y)^1.85 = 0 ................................i)

Pressure drop in C-D-E = Pressure drop in C-E

Or, 10 +  60 x {(4.52 x (40-X-Y)^1.85)/(120^1.85 x 1.5^4.87)} +  40 x {(4.52 x (40-Y)^1.85)/(120^1.85 x 2^4.87)} = 10 +  60 x {(4.52 x Y^1.85)/(120^1.85 x 1.5^4.87)}

After simplification, 3Y^1.85 - 0.985(40-Y)^1.85 - 3(40-X-Y)^1.85 = 0 ...........................................ii)

From equation i) and ii), after iteration we get, X = 13.573 gal/min and Y = 16.539 gal/min.

a) Flow rate in pipes : E-A-P-B = 40 gal/min (given)

B-D = 13.573 gal/min.

B-C = 26.427 gal/min.

C-D = 9.888 gal/min.

C-E = 16.539 gal/min.

D-E = 23.461 gal/min.

b) Pressure drop for B-D-E line = 10 +  60 x {(4.52 x (13.573)^1.85)/(120^1.85 x 1.5^4.87)} +  40 x {(4.52 x (23.461)^1.85)/(120^1.85 x 2^4.87)}

= 10 + 0.667 + 0.302 = 10.968 psi

So, total pressure drop across pump = 10.968 + 25 + 0.34 = 36.308 psi.

c) Pump power in hp = (GPM x psi) / (1714 x Efficiency) = (40 x 36.308)/(1714 x 0.85) = 0.9968 or 1 hp