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
