Unit 03 · Lesson 05 of 09

When can we use Bernoulli?

An equation earns its cancellations.

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

What you’ll be able to do

Audit every assumption.
The ideal reservoir prediction.
Add the missing physics.

Before you touch the controls

What happens to a reservoir discharge prediction when irreversible losses are added?

An experiment you can change

Compare ideal and lossy predictions

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

Loss coefficient K1 Pump head0 m

V_2=\sqrt{\frac{2(9.81)(5+0)}{1+1}}=7.0036\,\mathrm{m/s}

Loss does not destroy total energy; mechanical energy becomes internal energy. Adding pump head or losses requires the extended equation rather than an unchanged Bernoulli constant.

Speed with same pump, no loss9.905 m/s

Speed with selected loss7.004 m/s

Model assumptions and units

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K = 0 and no pump; extended energy accounting handles K and pump head.

z_1+h_p=z_2+\frac{V_2^2}{2g}+K\frac{V_2^2}{2g},\quad V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

Audit every assumption.

Check steadiness, density change, machine work, viscous losses, and whether the compared points lie on the relevant streamline. A large reservoir makes its surface speed small; an atmospheric free jet fixes its outlet static pressure.

The ideal reservoir prediction.

For a reservoir discharging 5 m lower into the same atmospheric pressure, ideal Bernoulli predicts that elevation head becomes exit kinetic head. It ignores all losses and gives an upper-speed estimate for this passive setup.

V_{ideal}=\sqrt{2g\Delta z}

Add the missing physics.

A loss coefficient K produces loss head $KV^2/(2g)$ . Pump head supplies additional mechanical energy. Rearranging the extended balance gives a different prediction that contains both effects.

V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

Mechanical loss is not energy destruction.

Viscous dissipation converts mechanical energy into internal energy. Total energy remains conserved even though the useful pressure/speed/elevation head decreases. A heat-transfer term belongs in a full thermal balance when temperature change matters.

A worked example

A reservoir outlet with losses.

Let $\Delta z=5\,\mathrm m$ , no pump, and $K=1$ .
Ideal speed is $\sqrt{2(9.81)(5)}=9.9045\,\mathrm{m/s}$ .
With loss, speed is $\sqrt{98.1/2}=7.0036\,\mathrm{m/s}$ . Both kinetic and loss heads are 2.5 m.

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

Add an outlet loss

An open reservoir discharges 5 m lower with $\mathrm{K} = 1$ and no pump. Find exit speed.

Hint

Both exit kinetic head and loss head use the same speed.

Explore the visual solution

Loss coefficient K1 Pump head0 m

V_2=\sqrt{\frac{2(9.81)(5+0)}{1+1}}=7.0036\,\mathrm{m/s}

Loss does not destroy total energy; mechanical energy becomes internal energy. Adding pump head or losses requires the extended equation rather than an unchanged Bernoulli constant.

Speed with same pump, no loss9.905 m/s

Speed with selected loss7.004 m/s

Model assumptions and units

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K = 0 and no pump; extended energy accounting handles K and pump head.

z_1+h_p=z_2+\frac{V_2^2}{2g}+K\frac{V_2^2}{2g},\quad V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

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

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K = 0 and no pump; extended energy accounting handles K and pump head.

z_1+h_p=z_2+\frac{V_2^2}{2g}+K\frac{V_2^2}{2g},\quad V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

2 · Compare

Include a pump explicitly

With $\mathrm{K} = 1,$ add 5 m pump head to the same reservoir system. Find exit speed.

Hint

Pump head adds to the available elevation head before losses.

Explore the visual solution

Loss coefficient K1 Pump head5 m

V_2=\sqrt{\frac{2(9.81)(5+5)}{1+1}}=9.9045\,\mathrm{m/s}

Loss does not destroy total energy; mechanical energy becomes internal energy. Adding pump head or losses requires the extended equation rather than an unchanged Bernoulli constant.

Speed with same pump, no loss14.007 m/s

Speed with selected loss9.905 m/s

Model assumptions and units

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K = 0 and no pump; extended energy accounting handles K and pump head.

z_1+h_p=z_2+\frac{V_2^2}{2g}+K\frac{V_2^2}{2g},\quad V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

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

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K = 0 and no pump; extended energy accounting handles K and pump head.

z_1+h_p=z_2+\frac{V_2^2}{2g}+K\frac{V_2^2}{2g},\quad V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

3 · Challenge the model

Large losses

Set $\mathrm{K} = 5$ with no pump. Find exit speed and determine the ratio of loss head to exit kinetic head.

Hint

The ratio hL/(V²/2g) is K.

Explore the visual solution

Loss coefficient K5 Pump head0 m

V_2=\sqrt{\frac{2(9.81)(5+0)}{1+5}}=4.0435\,\mathrm{m/s}

Loss does not destroy total energy; mechanical energy becomes internal energy. Adding pump head or losses requires the extended equation rather than an unchanged Bernoulli constant.

Speed with same pump, no loss9.905 m/s

Speed with selected loss4.044 m/s

Model assumptions and units

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K = 0 and no pump; extended energy accounting handles K and pump head.

z_1+h_p=z_2+\frac{V_2^2}{2g}+K\frac{V_2^2}{2g},\quad V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

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

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K = 0 and no pump; extended energy accounting handles K and pump head.

z_1+h_p=z_2+\frac{V_2^2}{2g}+K\frac{V_2^2}{2g},\quad V_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

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