Question

Consider a 100 mm internal diameter plastic pipe with a length of 1000 m and an absolute roughness of 0.01 mm. Assume the kinematic viscosity of water to be 1.14 mm2/s.
a) What should the secondary loss coefficient k be to have a secondary loss one tenth of the friction loss at Re = 4000?
[41]
b) Calculate the friction, secondary and total head losses for the pipe at a Re = 105. Use the secondary loss coefficient found in a).
[15 m]
c) Did the proportional contribution of the secondary loss to the total head loss increase or decrease between (a) and (b)? Why did it change?

Answer 1

Ans a) Roughness of pipe , e = 0.01 mm

Diamter of pipe ,D = 100 mm or 0.10 m

=> Relative roughness,e/D = 0.01mm / 100mm = 0.0001

Given, Reynold number, Re = 4000

Since, Re = 4000, flow is turbulent so according to Moody digram for Re = 4000 and e/D = 0.0001 , Darcy friction factor, f = 0.040

Also, Re = V D /

where, = Kinematic viscosity of water = 1.14 mm2/s or 1.14 x 10-6 m2/s

=> 4000 = V (0.1) / 1.14 x 10-6

=> V = 0.045 m/s

We know, friction head loss, Hf = f L V2 / 2 g D

=> Hf = 0.04(1000)(0.045)2 / (2 x 9.81 x 0.10)

=> Hf = 0.0413 m

According to question, secondary loss, Hs = Hf / 10 = 0.00413 m

Also, Hs = k V2 / 2 g

where, k = loss coeffcient

=> k (0.045)2 / (2 x 9.81) = 0.00413

=> k = 40.02 41

Hence, secondary loss coeffcient is 41

Ans b) Given, Re = 105 > 4000

Since, Reynold number is more than 4000, flow is turbulent

   According to Moody digram for Re = 105 and e/D = 0.0001 , Darcy friction factor, f = 0.0185

Also, Re = V D /

=> 105 = V (0.1) / 1.14 x 10-6

=> V = 1.14 m/s

Friction head loss, Hf = f L V2 / 2 g D

=> Hf = 0.0185(1000)(1.14)2 / (2 x 9.81 x 0.10)

=> Hf = 12.25 m

Also,secondary loss, Hs = k V2 / 2 g

where, k = loss coeffcient = 41 from part (a)

=> Hs = 41 (1.14)2 / (2 x 9.81)

=> Hs = 2.71 m

=> Total head loss = Friction loss + Secondary loss = 12.25 + 2.71 = 14.96 m 15 m

Ans c) In part (a), secondary loss is one-tenth of frictional loss  

=> Hs = 0.10 Hf

In part (b), secondary loss is 2.71 m which is about 0.22 times the frictional loss

So, we can clearly see tht proportional contribution of secondary loss increses between part (a) to (b) which is obvious because secondary loss is proportional to square of velocity. As velocity in part(b) is more than as compared to part (a) , contribution of secondary loss is also more in part (b) then part (a) .