Question

You are part of the water supply engineering team of your council. Your team is working on a certain part of cast iron piping of a water distribution system involving a parallel section. Both parallel pipes have a diameter of 30 cm, and the flow is fully turbulent. One of the branches (pipe A) is 1000 m long while the other branch (pipe B) is 3000 m long. If the flow rate through pipe A is 0.4 m³/s, determine the flow rate through pipe B and the head loss in pipe B. Disregard minor losses. Consider the water temperature to be 15°C. Assume that the flow is fully turbulent (i.e. the friction factor is independent of Reynolds number).

Answer 1

Solution:- the values given in the question are as follows:

length of pipe A(L1)=1000 m

flow rate through pipe A(Q₁)=0.4 m^3/s

diameter of both pipe(d)=30 cm

length of pipe B(L2)=3000 m

flow rate through pipe B(Q₂)=?

head loss in pipe B(H_L1)=?

both pipes are in parallel

temperature of water is 15^oC

Parallel combination:-

let the friction factor for both cast iron pipe is 0.26 mm

both pipe are in parallel so the head loss is same in both pipes

H_L1=H_L2

(f₁L1*v₁^2)/2gd₁=(f₂L2*v₂^2)/2gd₂

(4*f₁*L1*Q₁^2)/(2gd₁^5)=(4*f₂*L2*Q₂^2)/(2gd₂^5) , [Eq-1]

where,. f₁=f₂=0.26 mm

f₁=f₂=0.00026

d₁=d₂=30 cm , or 0.30 m

L1=1000 m , L2=3000 m

Q₁=0.4 m^3/s

values put in above equation-(1) and calculate the value of Q₂

(16*0.00026*1000*0.4^2)/(2g*0.30^5)=(16*0.00026*3000*Q₂^2)/(2g*0.30^5)

all values put in m in above equation

1000*0.4^2=3000*Q₂^2

Q₂^2=(0.4^2)/3

Q₂=0.23094 m^3/s

flow rate in pipe B(Q₂)=0.2309 m^3/s

Calculating head in pipe B(H_L2)-

head loss in pipe B(H_L2)=16*f₂*L2*Q₂^2/(2gd₂^5)

head loss in pipe B(H_L2)=(16*0.00026*3000*0.2309^2)/(2*9.81*3.14*0.30^5)

head loss in pipe B(H_L2)=0.665368/0.4705

head loss in pipe B(H_L2)=1.414174 m

head loss in pipe B(H_L2)=1.4142 m

[Ans]