Question

Plot the head loss in a 500 m long 200 mm cast iron pipe for the following flow rates; 0.001, 0.01, 0.1, and 1 m3/s. Use (a) the Moody Diagram and (b) the Swamee-Jain equation to find the friction factor and (c) the Hazen-Williams Eq. Discuss why the losses are similar or different for parts a and c

Answer 1

Ans) Case 1 - When Q = 0.001 m³/s

A = (/4) d²

= 0.785 x 0.2 x0.2

=0.0314 m²

We know,

Q = A x V

0.001 = 0.0314 x V

V= 0.0318 m/s

Hence Reynold Number = VD/

=1000 x 0.0318 x 0.2 / 0.001

= 6360

For cast iron pipe , roughness = 0.26 mm

e/d = 026 /2000

= 0.0013

a) According to Moody diagram , for Reynold number 6360 and e/d = 0.0013

f = 0.036

Therefore , headloss (H_L) = f L V² /( 2gd)

= 0.036 x 500 x 0.0318²  / (2 x 9.81 x .2)

= 0.00463 m

b) Swamee jain equation

f = 0.25 / [Log ( e/3.7D + 5.74/ R^0.9 )]²

= 0.25 / Log[ (0.0013/3.7 + 5.74 / 6360^0.9)]²

= 0.25/ 6.75

= 0.037

Therefore, head loss = 0.037 x 500 x 0.0318² / (2 x 9.81 x 0.2)

= 0.00478 m

c) By Hazen William Equation

H_L = L (V / 0.849CR^0.63)^1.85

where C = 130 for cast iron pipe

R = hydraulic radius equal to A/P

R = (/4)d² / d

= d/4

= 0.2 /4 or 0.05

= 500(0.0318 /0.849x 130 x 0.05^0.63)^1.85

= 500 x 0.00000926

= 0.00463 m

Hence we can see that result obtained from (a) and (c) are approximately equal

Case 2) When Q = 0.01 m³

A = 0.0314 m²

V = Q/A

= 0.318 m/s

Reynold Number = 63600

e/d = 0.0013

a) Hence from Moody diagram

f = 0.024

H_L = 0.024 x 500 x 0.318² / (2x9.81 x 0.2)

= 0.309 m

b) From Swamee Jain equation

f = 0.25 / [ Log (0.0013/3.7 + 5.74/ 63600^0.9)]²

=0.024

Hence , again H_L = 0.309 m

c) From Hazen William Equation

H_L = 500[ 0.318 /0.849 x130 x 0.05^0.63]^1.81

= 0.316 m

Case 3) When Q = 0.10 m³

A = 0.0318 m²

V =3.18 m/s

R = 636000

e/d = 0.0013

a) From Moody diagram

f = 0.02

H_L = 0.02x500x3.18²/ (2x9.81x.2)

= 25.7 m

b) From Swamee Jain equation

f = 0.25/ [Log(0.0013/3.7 + 5.74/636000^0.9]²

= 0.02

Again H_L = 25.7 m

c) Hazen William Equation

H_L = 500[3.18 / (0.849 x 130 x 0.05^0.63)]^1.81

= 24.8 m

Case 4 - When Q = 1m³

A =0.0318 m²

V = Q/A

= 31.44 m/s

R = 6360000

e/d =0.0013

a) From Moody diagram

f = 0.021

H_L = 0.02 x500x31.44² / (2x9.81x0.2)

= 2519.04 m

b) From Swamee Jain Equation

f = 0.25 / Log [0.0013/3.7 + 5.74/6360000^0.9]²

= 0.021

Again , H_L = 2519.04 m

c) From Hazen William Equation

H_L = 500[31.44 /(0.849 x130x 0.05^0.63)]^1.81

= 500 x 4.97

= 2485 m

Hence, we can see that, for small value of discharge(Q), Head loss from (a) and(c) is almost equal but as the discharge increases the variation of head loss with both equation also increases.