# How can I calculate the power to drive a water pump based on discharge from two hoses?

I am trying to solve the following problem:

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.

Here are my workings for part (a) of the question.

Head lost at nozzle $H_p = H - 4fLv^2/ D2g$
or Hp = Head at inlet to hoses - frictional head loss in each hose.
$C_v = 0.97$ and Jet Velocity $V = C_v \sqrt {2g * Hp}$, so then I worked out $H_p$ from this to be 31.208 m.
For continuity of flow, flow from hoses = flow from nozzle, so $v = V (d^2/ D^2)$ and $v = 0.03375/ D^2$.
I have then put this in the frictional head loss formula $4fLV^2/ D2g$ and this gives $1.43521875 * 10^-3/ D^5 * 19.62$.
Note $2g = 19.62$ and $H = 5 * \text{frictional head loss}$.

So lastly I put this in the formula $H_p = H - 4fLv^2/ D2g$. I solved for $D$ and I got 98.7 mm for the hose diameter.

Is this on the correct path for Part (a) of the question?

How can I work out the power for Part (b)?

• Hint: $Power = \frac{Energy}{Time}$ . Find the kinetic energy per unit volume of water; you know the flow rate so you know the total energy per unit time. Multiply by $\frac{1}{70percent}$ to get power to the pump. (oops, and the power lost in the hoses themselves, too) – Carl Witthoft Sep 14 '16 at 14:18

This is the methodology for calculating the power. I do not agree with Carl Witthofts comment because it does not take into account the major and minor losses or the potential energy (from height difference), and I think he is acknowledging that by saying oops.

1. Find your flow rate (velocity times area x 2 for each hose!)
3. Find your minor loss head(from the nozzle and whatever other components there are, if any).
6. Fluid Power is equal to your total flow times your total pressure drop from (5)
7. Your motor input power is equal to Fluid Power/0.7 (efficiency)

To account for the 20% maximum hose power loss portion, 8. Isolate your major losses (hose losses) only and calculate the power to overcome those losses only. 9. Take that power and compare it to the power calculated in 7. 10. If it's less than 20% of the power, you're good! If not, then increase the power of the motor until your power to overcome the hose losses is equal to 20% of the motor power.

• (1) Flow rate = jet velocity (24m/s) x area of jet from nozzle x 2 (2) Major loss of head is frictional head loss from length of both hoses and $Hf = 4fLv^2/ D2g$. (3) I am not sure about minor loss head as I thought that losses from nozzle are negligible and nozzle actually creates kinetic energy $V^2/ D2g$ (4) When you say Find your head from height difference of 3 m, do you mean calculate Head from pump using Bernoulli equation. If this is the case wouldn't this give same answer as step 5 where heads from 2, 3, and 4 are added together to give total pressure . – Rob Wilkinson Jan 30 '17 at 12:45
• (1) Flow rate = jet velocity (24m/s) x area of jet from nozzle x 2 (2) Major loss of head is frictional head loss from length of both hoses and Hf=4fLv2/D2g . (3) I am not sure about minor loss head as I thought that losses from nozzle are negligible and nozzle actually creates kinetic energy V2/D2g (4) When you say Find your head from height difference of 3 m, do you mean calculate Head from pump using Bernoulli equation. If this is the case wouldn't this give same answer as step 5 where heads from 2, 3, and 4 are added together to give total pressure . – Rob Wilkinson Mar 22 '18 at 14:57
• Hi Prevost May you please get back to me on this . Lift of elevation (z2 - z1) = 3 m Head Loss is from pipe = 4 * f * L * v1^2 / d * 2g. Isn't exit velocity head from jet velocity of the nozzle V2^2 / 2g. The problem I am having though, is that I think jet velocity will be different than 24 m/s in part (a) because in part (a) there was no change in elevation. However in part (b) there is. How can you calculate power when you don't know jet velocity? I believe that i am correct for part (a) – Rob Wilkinson Sep 27 at 3:55
• Hi Prevost Can you please show me in step by step calculation how you would get 31.1 kW for part (b)? – Rob Wilkinson Sep 27 at 3:57

(a) Head lost in friction $H_f = 4fLv^2/ D2g$, 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 $H_p = H - 4fLv^2/ D2g$
or $H_p = Head at inlet to hoses - frictional head loss in each hose. If jet velocity = V and$C_v = 0.97$Jet Velocity$V = C_v \sqrt {2g * Hp}$, where$V = 24m/s

$H_p = V^2/ {2g * C_v^2}$ = (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 (d^2/ D^2)$= 24 (0.0375)^2 / D^2 =$v = 0.03375/ D^2$. I have then put this in the frictional head loss formula$4fLV^2/ D2g$= 4*0.007*45*(0.03375D^-2)^2 / (D * 19.62) =$1.43521875*10^-3*D^-4/ D * 19.62$=$1.43521875 * 10^-3/ (D^5 * 19.62)$. 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 * 4fLv^2/ D2g so$H_p = 5 * (4fLv^2/ D2g)$- (4fLv^2/ D2g) = 4 * (4fLv^2/ D2g) = 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) Power from pump P = wQHp, where w = unit weight of water, Q = flow through pipes and Hp = Head from pump Now we have to use Bernoulli equation with pump head term in it as well as head loss Z1 + P1/pg + V1^2/2g + Hp = Z2 + P2/pg + V2^2/2g + Hf In this formula we assume that P1 ands P2 are atmospheric pressures at entry to the pump and exit of the nozzle, so P1 = P2 = 0 Z1 = 0 and Z2 = 3m V1 = velocity in hoses and V2 = jet velocity from nozzle = 24m/s We have to calculate velocity in hoses and then frictional head loss From before$v = V (d^2/ D^2)$so v = 24 * (0.0375^2)/ (0.09872^2) = 3.463m/s so V1 = 3.463m/s Now we have to put V1 into frictional head loss formula$H_f = 4fLv^2/ D2g\$

so Hf = 4*0.007*45*(3.463)^2 / (0.09872 * 19.62) = 7.80m

Now we have to put values in Bernoulli equation to find head from pump

so using Z1 + P1/pg + V1^2/2g + Hp = Z2 + P2/pg + V2^2/2g + Hf

0 + 0 + (3.463^2)/2g + Hp = 3 + 0 + (24^2)/2g + 7.8 Hp = 3 + (24^2)/2g + 7.8 - (3.463^2)/2g = 3 + 29.3578 + 7.8 - 0.611 Hp = 39.5468m

Flow from one hose Q = 1/4 Pi*D^2*v = 1/4 Pi*0.09872^2* 3.463 = 0.026507m^3/s

Power from one hose P = wQHp = 9810 * 0.026507 * 39.5468 = 10284W = 10.284kW as there are 2 hoses, total Power from hoses P = 10.284 * 2 = 20.567kW

as the pump has an efficiency of 70 per cent then Power from pump = 20.567/ 0.7 = 29.382kW

IN MY TEXT BOOK THE ANSWER WAS GIVEN AS 31.1kW, IS THERE SOMETHING ELSE I SHOULD TAKE INTO CONSIDERATION?