Unit 03 · Lesson 06 of 09

How do pumps and losses change the balance?

A pump supplies the head the system requires.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do

Keep machine and loss terms visible.
Two large reservoirs simplify the endpoints.
Convert head to shaft power.

Before you touch the controls

Does a more efficient pump change required hydraulic head at a specified flow?

An experiment you can change

Size a pump for elevation and friction

Predict the direction of change, move one control, then explain the response using the equation.

Volume flow5 L/sElevation rise10 mPump efficiency0.7

h_p=10+0.2(5)^2=15\,\mathrm m,\quad \dot W_{in}=\frac{1000(9.81)(510^{-3})(15)}{0.7\times1000}=1.0511\,\mathrm{kW}

Efficiency changes shaft power, not the hydraulic head required by the chosen flow and elevation. Kinetic correction α matters when endpoint profiles are not negligible.

Static lift10 m

Loss head5 m

Model assumptions and units

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

h_p=\Delta z+\frac{\Delta p}{\rho g}+\frac{\alpha_2V_2^2-\alpha_1V_1^2}{2g}+h_L,\quad \dot W_{shaft,in}=\frac{\rho g\dot Vh_p}{\eta_p}

Keep machine and loss terms visible.

Between sections, pressure, kinetic, and elevation heads are balanced along with pump additions, turbine removals, and irreversible losses. The kinetic correction factor α accounts for a nonuniform velocity profile.

\frac{p_1}{\rho g}+\alpha_1\frac{V_1^2}{2g}+z_1+h_p-h_t=\frac{p_2}{\rho g}+\alpha_2\frac{V_2^2}{2g}+z_2+h_L

Two large reservoirs simplify the endpoints.

Both surfaces can be at atmospheric pressure with negligible speed. Then pump head must cover static elevation rise plus loss head. Losses usually rise with flow and depend on piping and fittings.

h_p=\Delta z+h_L

Convert head to shaft power.

Hydraulic power is density times g times volume flow times pump head. Divide by pump efficiency to get shaft input; for a turbine the efficiency multiplies available hydraulic power instead.

\dot W_{hyd}=\rho g\dot Vh_p,\quad\dot W_{shaft,in}=\frac{\dot W_{hyd}}{\eta_p}

Check kinetic corrections.

Fully developed laminar circular-pipe flow has α=2; turbulent pipe flow often has α closer to 1. At large reservoir surfaces the speed term is negligible, so the precise α there has little influence.

A worked example

Lift water with a lossy pipe.

For $q=5\,\mathrm{L/s}$ and $h_L=0.2q^2$ , loss head is 5 m. A 10 m lift needs $h_p=15\,\mathrm m$ .
Hydraulic power is $1000(9.81)(0.005)(15)=735.75\,\mathrm W$ .
At 70% pump efficiency shaft input is $735.75/0.70=1051.1\,\mathrm W=1.0511\,\mathrm{kW}$ .

Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Supply lift and loss

Pump 5 L/s up 10 m with loss $hL = 0.2$ q² m and efficiency 0.70. Find shaft power in kW.

Hint

First find required head, then hydraulic power, then divide by efficiency.

Explore the visual solution

Volume flow5 L/sElevation rise10 mPump efficiency0.7

h_p=10+0.2(5)^2=15\,\mathrm m,\quad \dot W_{in}=\frac{1000(9.81)(510^{-3})(15)}{0.7\times1000}=1.0511\,\mathrm{kW}

Efficiency changes shaft power, not the hydraulic head required by the chosen flow and elevation. Kinetic correction α matters when endpoint profiles are not negligible.

Static lift10 m

Loss head5 m

Model assumptions and units

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

h_p=\Delta z+\frac{\Delta p}{\rho g}+\frac{\alpha_2V_2^2-\alpha_1V_1^2}{2g}+h_L,\quad \dot W_{shaft,in}=\frac{\rho g\dot Vh_p}{\eta_p}

Controls explore variations. The steps below retain the original question’s values.

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

h_p=\Delta z+\frac{\Delta p}{\rho g}+\frac{\alpha_2V_2^2-\alpha_1V_1^2}{2g}+h_L,\quad \dot W_{shaft,in}=\frac{\rho g\dot Vh_p}{\eta_p}

2 · Compare

Efficiency improves

At the same 5 L/s and 10 m lift, raise efficiency to 1. Find shaft power and inspect whether required head changes.

Hint

Head depends on the system; efficiency converts hydraulic power to shaft input.

Explore the visual solution

Volume flow5 L/sElevation rise10 mPump efficiency1

h_p=10+0.2(5)^2=15\,\mathrm m,\quad \dot W_{in}=\frac{1000(9.81)(510^{-3})(15)}{1\times1000}=0.73575\,\mathrm{kW}

Efficiency changes shaft power, not the hydraulic head required by the chosen flow and elevation. Kinetic correction α matters when endpoint profiles are not negligible.

Static lift10 m

Loss head5 m

Model assumptions and units

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

h_p=\Delta z+\frac{\Delta p}{\rho g}+\frac{\alpha_2V_2^2-\alpha_1V_1^2}{2g}+h_L,\quad \dot W_{shaft,in}=\frac{\rho g\dot Vh_p}{\eta_p}

Controls explore variations. The steps below retain the original question’s values.

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

h_p=\Delta z+\frac{\Delta p}{\rho g}+\frac{\alpha_2V_2^2-\alpha_1V_1^2}{2g}+h_L,\quad \dot W_{shaft,in}=\frac{\rho g\dot Vh_p}{\eta_p}

3 · Challenge the model

Double flow

Pump 10 L/s up 10 m at efficiency 0.70. Find shaft power and compare with the 5 L/s case.

Hint

Loss head quadruples while flow doubles.

Explore the visual solution

Volume flow10 L/sElevation rise10 mPump efficiency0.7

h_p=10+0.2(10)^2=30\,\mathrm m,\quad \dot W_{in}=\frac{1000(9.81)(1010^{-3})(30)}{0.7\times1000}=4.2043\,\mathrm{kW}

Efficiency changes shaft power, not the hydraulic head required by the chosen flow and elevation. Kinetic correction α matters when endpoint profiles are not negligible.

Static lift10 m

Loss head20 m

Model assumptions and units

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

h_p=\Delta z+\frac{\Delta p}{\rho g}+\frac{\alpha_2V_2^2-\alpha_1V_1^2}{2g}+h_L,\quad \dot W_{shaft,in}=\frac{\rho g\dot Vh_p}{\eta_p}

Controls explore variations. The steps below retain the original question’s values.

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

h_p=\Delta z+\frac{\Delta p}{\rho g}+\frac{\alpha_2V_2^2-\alpha_1V_1^2}{2g}+h_L,\quad \dot W_{shaft,in}=\frac{\rho g\dot Vh_p}{\eta_p}

Your learning path

Ready to carry this forward?

Mark this lesson complete when you can explain the main idea and work through its practice. Completion is your choice.

Completion is saved on this browser.