Q1,Q2,Q3 are flow rates to be delivered at the tanks on the top of the respectiv

Question

Q1,Q2,Q3 are flow rates to be delivered at the tanks on the top of the respective buildings. V, V1,V2,V3 are valves.

Design the pipe diameters d1, d2, d3 and the pump power so that Q1=Q2=Q3=4.5 m^3/hr when all V,V1,V2,V3 are open. It is given that d1,d2,d3<= .12 m (the diameter of the feed pipe).

Q1,Q2,Q3 are flow rates to be delivered at the tanks on the top of the respective buildings. V, V1,V2,V3 are valves. Design the pipe diameters d1, d2, d3 and the pump power so that Q1=Q2=Q3=4.5 m^3/hr when all V,V1,V2,V3 are open. It is given that d1,d2,d3

Explanation / Answer

Kinematic viscosity of water v = 0.8*10^-6 m^2/s

Q = 4.5 m^3 / hr = 4.5 / (60*60) m^3 /s = 0.00125 m^3/s

In the pump Q_pump = Q1 + Q2 + Q3 = 0.00375 m^3/s

V_pump = Q_pump / A_pump = 0.00375 / (3.14*0.12^2 /4) = 0.3317 m/s

V1 = Q1 / A1 = Q / (pi*d1^2 /4) = 4Q / (pi*d1^2) = 4*0.00125 / (pi*d1^2) = 0.005 / (pi*d1^2)

V2 = Q2 / A2 = Q / (pi*d2^2 /4) = 4Q / (pi*d2^2) = 4*0.00125 / (pi*d2^2) = 0.005 / (pi*d2^2)

V3 = Q3 / A3 = Q / (pi*d3^2 /4) = 4Q / (pi*d3^2) = 4*0.00125 / (pi*d3^2) = 0.005 / (pi*d3^2)

Reynolds number in pipe attached to pump = V_pump*d / v = 0.3317*0.12 / (0.8*10^-6) = 49761

From Moody chart, for smooth pipe and Re = 49761 we get, f = 0.021

Flowrate between d1 and d2 is = Q1 + Q2 = 0.0025 m^3 /s

Velocity between d1 and d2 is 0.0025 / (3.14*0.12^2 /4) = 0.2211 m/s

Re = 0.2211*0.12 / (0.8*10^-6) = 33174

From Moody chart, for smooth pipe and Re = 33174 we get, f = 0.022

Head loss = (fL/d) *V^2 /(2g) = (0.022*65 / 0.12)*0.2211^2 / (2*9.81) = 0.0296 m

Flowrate between d2 and d3 is = Q3 = 0.00125 m^3 /s

Velocity between d2 and d3 is 0.00125 / (3.14*0.12^2 /4) = 0.1106 m/s

Re = 0.1106*0.12 / (0.8*10^-6) = 16587

From Moody chart, for smooth pipe and Re = 16587 we get, f = 0.028

Head loss = (fL/d) *V^2 /(2g) = (0.028*65 / 0.12)*0.1106^2 / (2*9.81) = 0.00946 m

Head loss in pipe d1 = (f1*L1/d1)*V1^2 / (2g)

Let pressure at V be P.

So,

P/(rho*g) - (f1*L1/d1)*V1^2 / (2g) = 30

P/(rho*g) - 0.0296 - (f2*L2/d2)*V2^2 / (2g) = 20

P/(rho*g) - 0.0296 - 0.00946 - (f3*L3/d3)*V3^2 / (2g) = 20

Or,

P/(rho*g) - (f1*(70+30)/d1)*[0.005 / (pi*d1^2)]^2 / (2g) = 30

P/(rho*g) - 0.0296 - (f2*(30+20)/d2)*[0.005 / (pi*d2^2)]^2 / (2g) = 20

P/(rho*g) - 0.0296 - 0.00946 - (f3*(30+20)/d3)*[0.005 / (pi*d3^2)]^2 / (2g) = 20

Or,

P/(rho*g) - 0.00258*f1 / d1^5 = 30

P/(rho*g) - 0.0296 - 0.00129*f2 / d2^5 = 20

P/(rho*g) - 0.0296 - 0.00946 - 0.00129*f3 / d3^5 = 20

Therefore,

0.0296 + 0.00129*f2 / d2^5 - 0.00258*f1 / d1^5 = 10

0.00946 + 0.00129*f3 / d3^5 - 0.00129*f2 / d2^5 = 0

We estimate f1, f2 and f3 from Moody chart.

f1 = 0.02, f2 = 0.025, f3 = 0.03

0.0296 + 3.23*10^-5 / d2^5 - 5.16*10^-5 / d1^5 = 10

0.00946 + 3.87*10^-5 / d3^5 - 3.23*10^-5 / d2^5 = 0

Assuming d3 = 0.1 m, we get d2 = 0.096 m and d1 = 0.099 m

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