A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 10 V2/2g (V is velocity in the pipe), how much power will have to be supplied to the flow by the pump to produce this discharge? Assume α = 1.0 at all locations.

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss (Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4. Q/π. v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D = [ 4 * 9m³/s / 3.142 * (√0.98m) ] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 * (11.69m/s) ² / 2 * 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π. r², where r = D/2

Therefore, A = π. (D/2) ²

A = 3.142 * [0.98m/2]² = 0.75m²

Therefore, Force = Pressure * Area

Force F = 683.25N/m² * 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force * distance d moved by the water

W = F. d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v * t

Distance d = v * t = 11.69m/s * 1.298s = 15.17m.

Hence,

Work Done W = Force * distance

W = 512.44N * 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done / time

P = 7775.56joules/1.298s

P = 5990.4joules/s.