Unit 02 · Lesson 09 of 09

How do we balance energy in a flowing system?

Mass carries energy through the boundary.

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

What you’ll be able to do

Write the complete control-volume balance.
Use the device to choose simplifications.
Nozzle speed comes from an enthalpy drop.

Before you touch the controls

Where does the speed increase in an adiabatic nozzle come from?

An experiment you can change

An adiabatic nozzle turns enthalpy into speed

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

Enthalpy drop50 kJ/kgInlet speed40 m/s

V_2=\sqrt{40^2+2000(50)}=318.75\,\mathrm{m/s}

Temperature and enthalpy can fall while total energy is conserved. This energy balance alone does not determine nozzle area or validate every proposed outlet state.

Enthalpy converted50 kJ/kg

Kinetic-energy gain50 kJ/kg

Model assumptions and units

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

\dot Q-\dot W_s=\dot m\left[h_2-h_1+\frac{V_2^2-V_1^2}{2}+g(z_2-z_1)\right],\quad V_2=\sqrt{V_1^2+2000(h_1-h_2)}

Write the complete control-volume balance.

Energy accumulation equals heat input minus shaft work output plus energy transported by entering mass minus energy transported by leaving mass. At steady state the accumulation term vanishes.

\frac{dE_{CV}}{dt}=\dot Q-\dot W_s+\sum_{in}\dot m\left(h+\frac{V^2}{2}+gz\right)-\sum_{out}\dot m\left(h+\frac{V^2}{2}+gz\right)

Use the device to choose simplifications.

A nozzle usually has no shaft work; if insulated it has negligible heat transfer. A turbine produces shaft work; a compressor requires it. Heat exchangers may have negligible work while transferring energy between separate streams. Name the boundary before canceling heat terms.

Nozzle speed comes from an enthalpy drop.

For a steady adiabatic nozzle with negligible elevation change, inlet enthalpy plus kinetic energy equals the outlet total. Rearranging and converting kJ/kg to J/kg gives the live equation.

V_2=\sqrt{V_1^2+2000(h_1-h_2)}

Turbines and throttles differ.

An adiabatic turbine with small speed and height changes delivers work approximately equal to mass flow times enthalpy drop. An insulated throttle with no shaft work and small kinetic change instead has approximately equal inlet and outlet enthalpy; temperature need not stay constant.

\dot W_{turbine}\approx\dot m(h_1-h_2),\quad h_1\approx h_2\;\text{(throttle)}

A worked example

An accelerating nozzle.

Take $V_1=40\,\mathrm{m/s}$ and $h_1-h_2=50\,\mathrm{kJ/kg}$ .
The balance gives $V_2^2=40^2+2000(50)=101600\,\mathrm{m^2/s^2}$ .
Thus $V_2=318.75\,\mathrm{m/s}$ . No heat input was needed; the stream converted enthalpy into kinetic energy.

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

Accelerate an adiabatic nozzle

An adiabatic nozzle has inlet speed 40 m/s and enthalpy drop 50 kJ/kg. Find outlet speed.

Hint

Convert the enthalpy drop to J/kg inside the squared-speed equation.

Explore the visual solution

Enthalpy drop50 kJ/kgInlet speed40 m/s

V_2=\sqrt{40^2+2000(50)}=318.75\,\mathrm{m/s}

Temperature and enthalpy can fall while total energy is conserved. This energy balance alone does not determine nozzle area or validate every proposed outlet state.

Enthalpy converted50 kJ/kg

Kinetic-energy gain50 kJ/kg

Model assumptions and units

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

\dot Q-\dot W_s=\dot m\left[h_2-h_1+\frac{V_2^2-V_1^2}{2}+g(z_2-z_1)\right],\quad V_2=\sqrt{V_1^2+2000(h_1-h_2)}

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

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

\dot Q-\dot W_s=\dot m\left[h_2-h_1+\frac{V_2^2-V_1^2}{2}+g(z_2-z_1)\right],\quad V_2=\sqrt{V_1^2+2000(h_1-h_2)}

2 · Compare

A smaller enthalpy drop

A nozzle enters at 100 m/s and drops enthalpy by 10 kJ/kg. Find exit speed.

Hint

Add V1² to 2000 times the enthalpy drop before taking the square root.

Explore the visual solution

Enthalpy drop10 kJ/kgInlet speed100 m/s

V_2=\sqrt{100^2+2000(10)}=173.21\,\mathrm{m/s}

Temperature and enthalpy can fall while total energy is conserved. This energy balance alone does not determine nozzle area or validate every proposed outlet state.

Enthalpy converted10 kJ/kg

Kinetic-energy gain10 kJ/kg

Model assumptions and units

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

\dot Q-\dot W_s=\dot m\left[h_2-h_1+\frac{V_2^2-V_1^2}{2}+g(z_2-z_1)\right],\quad V_2=\sqrt{V_1^2+2000(h_1-h_2)}

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

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

\dot Q-\dot W_s=\dot m\left[h_2-h_1+\frac{V_2^2-V_1^2}{2}+g(z_2-z_1)\right],\quad V_2=\sqrt{V_1^2+2000(h_1-h_2)}

3 · Challenge the model

No conversion available

With zero enthalpy drop and inlet speed 100 m/s, find exit speed under the nozzle assumptions. Could the nozzle reach 200 m/s without another energy change?

Hint

Compare the required kinetic-energy increase with the balance.

Explore the visual solution

Enthalpy drop0 kJ/kgInlet speed100 m/s

V_2=\sqrt{100^2+2000(0)}=100\,\mathrm{m/s}

Temperature and enthalpy can fall while total energy is conserved. This energy balance alone does not determine nozzle area or validate every proposed outlet state.

Enthalpy converted0 kJ/kg

Kinetic-energy gain0 kJ/kg

Model assumptions and units

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

\dot Q-\dot W_s=\dot m\left[h_2-h_1+\frac{V_2^2-V_1^2}{2}+g(z_2-z_1)\right],\quad V_2=\sqrt{V_1^2+2000(h_1-h_2)}

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

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

\dot Q-\dot W_s=\dot m\left[h_2-h_1+\frac{V_2^2-V_1^2}{2}+g(z_2-z_1)\right],\quad V_2=\sqrt{V_1^2+2000(h_1-h_2)}

Your learning path

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