## Pump power for hoses feeding to nozzles

## Pump power for hoses feeding to nozzles

(OP)

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 well, as some things that 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

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

This is my logic for part (b) as well, as some things that 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

## RE: Pump power for hoses feeding to nozzles

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

## RE: Pump power for hoses feeding to nozzles

I suspect that you have a rounding error. You should be using significant digits in your entire calculation.