Question

Calculate the head loss if the water is discharged through a 2,000 ft 8” steel pipe and a 1,500 ft 6” cast iron pipe at the discharge rate of 1.63 cfs;

The 8 inch pipe has a square-edged entrance; a swing check valve; fully opened gate valve; contraction (20 degree);

The 6 inch pipe has 6 - 45 deg. elbows; 2 - 90 deg. bends; 1 r=1 ft; 1 r=2ft.

Use Hazen-Williams equation to calculate the head loss of the straight pipes - C_h= 110 for steel and 120 for cast iron. Also list the minor loss coefficient

Answer 1

Ans) We know,as per Hazen William equation,

Major head loss = 4.52 LQ^1.852 / (C^1.852 x D^4.87)

where L = length of pipe (ft)

Q = discharge (in gpm)

C = pipe roughness coefficient

D = diameter of pipe ( inch)

For 8" steel pipe,

Q = 1.63 ft³/s or 731.6 gpm

friction loss (H1) = 4.52(2000)(731.6)^1.852 / (110^1.852 x 8^4.87)

=12.08 ft

For 6" cast iron pipe ,

H2 = 4.52(1500)(731.6)^1.852 / (120^1.852 x 6^4.87)

= 31.30 m

Total head loss due to friction(H_f) = H1 + H2

= 12.08 + 31.30

H_f = 43.38 ft  

Now, minor losses in above problems are:

1) For 8" pipe

a) Loss due to entry

b) Loss due to valve

c) Loss due to contraction

Velocity in pipe = discharge / area

Area = (/4)D²

= 0.785 x (8/12)²

= 0.348 ft²

Velocity(V1) = 1.63/0.348

= 4.67 ft/s

Similarly Velocity in second pipe (V2)= Q/A2

A2 = (/4)(D₂)²

= 0.785 x 0.5 x 0.5

= 0.196 ft²

Velocity (V2) = 1.63 /0.196

= 8.30 ft/s

(a) Loss due to entry:

H_e = 0.5 V² /2g

= 0.5(4.67)²/ (2 x 32.2)

H_e = 0.17 ft

(b) Loss due to valve:

H_v = K V² / 2g

where, K = (L_e/d) x f

For swing check valve, equivalent ratio (L_e /d) = 100

For 8" pipe , f = 0.014

Hence,K = 100 x 0.014

= 1.4

H_v = 1.4 (4.67)²/ (2x32.2)

H_v= 0.47 ft

For fully open gate valve, (L_e/d )= 8 and f = 0.014

K = 8 x 0.014

= 0.112

H_g = 0.112 x (4.67)²/(2x32.2)

H_g = 0.038 ft

(c) Loss due to contraction

H_c = K(V₂)² /(2g)

For 20 deg contraction. K = 0.14

H_c = 0.14 x (8.3)²/ (2x32.2)

H_c= 0.15 ft

Now,

2) For 6" pipe

Minor loss in this case are:

a) Loss due to bend

Bend loss = KV² /(2g)

For 45^o valve, K = 0.24

H_b = 0.24 x 8.3² /(2x32.2)

H_b= 0.26 ft

Since there are six valves ,Head loss = 6 x 0.26

H_b= 1.56 ft

For 90^o bend , K = 0.45

H_b = 0.45 x 8.3² /(2x32.2)

H_b = 0.48 ft

   Since , there are two valves, total head loss H = 2 x 0.48

H_b = 0.96 ft