[Rob Wilk]

I did a Civil Engineering course some years ago and from my textbook I had a question on a pump with hoses. I think I have the correct answer to part (a) as can be seen below, but I am struggling with part (b) and need some help.

This is my logic for part (b) as you will there are some things I am trying to understand :
To calculate power I believe that we need to find the pump term Hp in bernoulli equation

Z1 + P1 / (Rho * g) + V1^2 / 2g + Hp = Z2 + P2 / (Rho * g) + V2^2 / 2g + Hf (frictional Head losses)

Z1 = 0 m (Because this is the level of the pump) and Z2 = 3 m because that is where jet of water comes out nozzles

In part (a) the head at the nozzle was calculated while in the process of finding the diameter of hoses.
Is this head used in part (b) of this question ?

In part (a) they say that jet velocity is 24 m/s when jet diameter is 37.5 mm and nozzle is at the same level as the pump.
What I want to understand is how does the jet velocity and diameter differ when the nozzle is 3 m above the pump ?

I assume that the pressures P1 and P2 = 0 because they are atmospheric.

V2 is jet velocity from nozzle and is V1 = 0 or is V1 = hose velocity ?

Here is the question below
A pump feeds two hoses, each of which is 45 m long and is fitted with a nozzle. Each nozzle has a coefficient of velocity of 0.97 and discharges a 37.5 mm diameter jet of water at 24 m/s when the nozzle is at the same level as the pump. If the power lost in overcoming friction in the hoses is not to exceed 20 per cent of the hydraulic power available at the inlet end of the hoses, calculate (a) the diameter of the hoses, taking f = 0.007, and (b) the power required to drive the pump if its efficiency is 70 percent and it draws its water supply from a level 3 m below the nozzle ?

These are the calculations I have so far

a) Head lost in friction Hf = 4 * f * L * v^2/ D * 2g, where L = Length of hose, D = Hose diameter, v = hose velocity and f = friction coefficient

If H = Head at inlet to pump and hoses

Head at nozzle Hp = H − 4 * f * L * v^2/ D * 2g or Hp = Head at inlet to hoses - frictional head loss in each hose.

If jet velocity = V and Cv = 0.97 Jet Velocity V=Cv √(2g ∗ Hp), where V = 24m/s

Hp = V^2/(2g ∗ Cv^2) = (24^2) / 2g * (0.97)^2, so Hp = 31.208 m

For continuity of flow, flow from hoses = flow from nozzle, so 1/4 Pi * D^2 * v = 1/4 * Pi * d^2 * V, where d = jet diameter

v = V * (d2/D2) = 24 *(0.0375)^2 / D^2 = v = 0.03375 / D2.

I have then put this in the frictional head loss formula 4 * f * L * V2/D * 2g = 4 * 0.007 * 45 * (0.03375 D^-2)^2 / (D * 19.62) = 1.43521875 ∗ 10−3 ∗ D−4/ D ∗ 19.62 = 1.43521875 ∗ 10−3/ (D5 ∗ 19.62).

Note 2g = 19.62

If power lost in overcoming friction in hoses is not to exceed 20 per cent of hydraulic power available at inlet end of hoses, then H = 5 * 4 * f * L * v^2/ D2g so Hp = 5 ∗ (4 * f * L * v2/ D * 2g) - (4 * f * L * v^2/ D2g) = 4 * (4 * f * L * v^2/ D * 2g) = 4 * 1.43521875 * 10^-3/ (D^5 * 19.62) Hp = 5.740875*10^-3/ (D^5*2g)

So 2g * Hp = 5.740875 * 10^-3/ D^5 = 19.62 * 31.2018 = 612.1798m

D^5 = 5.740875 * 10^-3 / 612.179, so D = (5.740875 * 10^-3/ 612.1798)^0.2 = 0.09872m

so Hose Diameter = 98.72 mm IN MY TEXT BOOK ANSWER WAS GIVEN AS 98.5mm, WOULD THIS BE DUE TO ROUNDING?

(b) Need to get an understanding on what I mentioned above

I have given it some thought on how to get the answer of 31.1 kW and I am very close, I got 31.2 kW by doing it this way below.

Is this meant to be the way this is calculated, this is what I have been trying to figure out.


From part (a) the power head at the nozzle was calculated as 31.208 m, when the pump is at the same elevation or level as the nozzle. This is the kinetic energy from the water jet.

I think you know that Power Head = Total Head - Frictional Head Loss for transmission of power by pipeline or hoses in this case.

The question ignores small loss of head from nozzle (probably because it is not much in this case)


So now in part (b) we have to find pump head term Hp in order to find power.


Bernoulli equation for this is now :


Z1 + P1 / (Rho g) + V1^2 /2g + Hp = Z2 + P2 / (Rho g) + V2^2 /2g + Hf


For part (b) the 31.2018 m head available at nozzle = V2^2 / 2g term in bernoulli equation


Z1 = level of pump = 0 m

Z2 = 3 m (outlet of nozzle)


P1 = P2 = 0 (because inlet of pump and outlet of nozzle are at atmospheric pressure)

V1 = 0 (because it is assumed water is taken from where the velocity is 0 at pump inlet )


Note: V2 = Jet Velocity / Cv = 24 / 0.97 = 24.742 m/s and this is velocity leaving nozzle where it reaches atmosphere. This is where the water area is slightly smaller than 37.5 mm diameter at nozzle outlet and this velocity is slightly larger than when it initially leaves the nozzle at 24 m/s.

I don't think we need to know this for this question because I believe that we are using the 31.2018 m head at nozzle calculated in part (a)


and Hf = Frictional Head losses from hoses


Now power lost in overcoming friction in hoses is not to exceed 20 per cent of the hydraulic power available at the inlet end of the hoses.

So this means 80 % of head is available at nozzle and this is 4 times the 20 % lost in the hoses.


So Hf = 31.2018 / 4 = 7.8 m

or H = 31.2018 * (5/4) = 39.002 m Total Head


Rearranging Bernoulli Equation to find Hp

Hp = (Z2 - Z1) + (P2 - P1) / (Rho * g) + (V2^2 - V1^2) / 2g + Hf


with P1, P2 and V1 = 0


we have Hp = (Z2 - Z1) + V2^2 / 2g + Hf


Hp = 3 + 31.2018 + 7.8

Pump Head term Hp = 42.002 m


Power from one Hose = P = w *Q * Hp


Now we want to find the flow = Q for the power


flow Q is either

Flow through the pipe = ( 24 (0.0375^2) / (0.09872^2) ) * ( ( ( pi 0.09872^2)/4) = 0.026507 m^3/s

or Flow leaving nozzle = 24 * ( ( pi 0.0375^2) / 4 ) = 0.026507 m^3/s


so Flow = Q = 0.026507 m^3/s or 26.507 litres /sec


w = = Density per unit Weight = 9.81 * 1000 = 9,810 N/m^3


P = 9,810 * 0.026507 * 42.002 = 10,922.085 Watts


For 2 Hoses, P = 22,556.88 Watts (This is power output, which is less than input, due to efficiency)


As the pump has an efficiency of 70 per cent, then

Total Input Power = 21,844.17 / 0.70 = 31,205.96 Watts

so Power required to drive pump = 31.206 kW

this is just over 100 Watts more than the 31.1 kW answer