Question

A smooth concrete pipe (1.5-ft diameter) carries water from a reservoir to an industrial treatment plant 1 mile away and discharges it into the air over a holding tank. The pipe leaving the reservoir is 3 ft below the water surface and runs downhill on a 1:100 slope. Determine the flow rate (in cfs, ft3/s) if the water temperature is 40°F (4°C) and minor losses are negligible.

Answer 1

ANSWER:

Q = 0.17ft3/s

EXPLANATION: since the water runs downhill on a 1:100 slope, that means the flow is laminar.

Using poiseuille equation:

Q = (π × D^4 × ∆P) ÷ (128 × U × ∆X)

Q is the volume flow rate.

π is pie constant value at 3.142

D is the diameter of the pipe

∆P is the pressure drop

U is the viscosity

∆X is the length of the pipe or distance of flow.

Form the question, we are to determine U then Find Q

Therefore;

D = 1.5ft

∆P = 1pa since the minor losses are negligible.

∆X = 1mile = 5280ft.

STEP1: FIND U

Viscosity is a function of the temperature of the liquid. An increase in temperature increases the viscosity of the liquid.

We know that at room temperature, which is 25°C the viscosity of water is 8.9×10^-4pa.s . We can find the viscosity of water at 4°C by cross multiplying.

Therefore;

25°C = 8.9×10^-4pa.s

4°C = U

Cross multiply

U25°C = 4°C × 8.9×10^-4pa.s

U25°C = 0.00356°C.pa.s

Therefore;

U = 0.00356°C.pa.s ÷ 25°C

U = 1.424×10^-4pa.s

Therefore at 4°C the viscosity of water in the pipe is 1.424×10^-4pa.s

STEP2: FIND Q

Imputing the values into poiseuille equation above.

Q = (3.142 × (1.5ft)^4 × 1pa) ÷ (128 × 1.424×10^-4pa.s × 5280ft)

Q = 15.906375pa.ft4 ÷ 96.239616pa.s.ft

Therefore;

Q = 0.16547887ft3/s

Approximately;

Q = 0.17ft3/s