Unit 02 · Lesson 08 of 09

Why does flowing fluid carry enthalpy?

Enthalpy packages the energy needed to move mass.

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

What you’ll be able to do

A port needs pressure work.
Combine two terms once.
Add motion and elevation.

Before you touch the controls

Why does a flowing energy balance use enthalpy instead of adding only internal energy?

An experiment you can change

Push a fluid parcel into a control volume

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

Pressure200 kPaSpecific volume0.5 m³/kg

h=250+(200)(0.5)=350\,\mathrm{kJ/kg}

The upstream fluid does pressure work to push mass across the inlet. Do not add pv a second time after using h.

Internal energy u250 kJ/kg

Flow work pv100 kJ/kg

Model assumptions and units

Illustrative parcel with u = 250 kJ/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

w_{flow}=\frac{pA\ell}{m}=pv,\qquad h=u+pv

A port needs pressure work.

Imagine a fluid slug of length ℓ and area A entering. The upstream fluid pushes with force $pA$ through distance ℓ. Divide the work $pA$ ℓ by slug mass; its volume per mass is v.

w_{flow}=\frac{pA\ell}{m}=pv

Combine two terms once.

A moving mass carries internal energy and requires pressure work to pass a port. Their sum is enthalpy. This grouping simplifies the open-system balance; it does not introduce a new physical energy source.

h=u+pv

Add motion and elevation.

Each kilogram crossing the boundary transports enthalpy plus kinetic and gravitational potential energy. Use consistent units: $V^2/2$ is J/kg for speed in m/s, while many tables give h in kJ/kg.

e_{flow}=h+\frac{V^2}{2}+gz

Distinguish shaft work.

Shaft and electrical work are separate control-volume transfers. Flow work is absorbed into the inlet and outlet enthalpy terms. When a pump raises liquid pressure, its shaft input supplies the required energy including irreversibilities.

A worked example

A parcel with a flow-work contribution.

For $u=250\,\mathrm{kJ/kg}$ , $p=200\,\mathrm{kPa}$ , and $v=0.5\,\mathrm{m^3/kg}$ , $pv=100\,\mathrm{kJ/kg}$ .
Then $h=250+100=350\,\mathrm{kJ/kg}$ .
At mass rate $0.2\,\mathrm{kg/s}$ , the enthalpy transport rate is $0.2(350)=70\,\mathrm{kW}$ before adding bulk kinetic and potential terms.

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

Package pressure work

For an illustrative parcel with $u = 250 \mathrm{kJ}$ /kg, $p = 200 \mathrm{kPa}$ , and $v = 0.50$ m³/kg, find h.

Hint

kPa m³/kg has units of kJ/kg.

Explore the visual solution

Pressure200 kPaSpecific volume0.5 m³/kg

h=250+(200)(0.5)=350\,\mathrm{kJ/kg}

The upstream fluid does pressure work to push mass across the inlet. Do not add pv a second time after using h.

Internal energy u250 kJ/kg

Flow work pv100 kJ/kg

Model assumptions and units

Illustrative parcel with u = 250 kJ/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

w_{flow}=\frac{pA\ell}{m}=pv,\qquad h=u+pv

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

Illustrative parcel with u = 250 kJ/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

w_{flow}=\frac{pA\ell}{m}=pv,\qquad h=u+pv

2 · Compare

A larger pressure contribution

For $u = 250 \mathrm{kJ}$ /kg, $p = 500 \mathrm{kPa}$ , and $v = 0.20$ m³/kg, find h. Compare with the preceding parcel.

Hint

Pressure alone is not the flow work; use the product pv.

Explore the visual solution

Pressure500 kPaSpecific volume0.2 m³/kg

h=250+(500)(0.2)=350\,\mathrm{kJ/kg}

The upstream fluid does pressure work to push mass across the inlet. Do not add pv a second time after using h.

Internal energy u250 kJ/kg

Flow work pv100 kJ/kg

Model assumptions and units

Illustrative parcel with u = 250 kJ/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

w_{flow}=\frac{pA\ell}{m}=pv,\qquad h=u+pv

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

Illustrative parcel with u = 250 kJ/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

w_{flow}=\frac{pA\ell}{m}=pv,\qquad h=u+pv

3 · Challenge the model

Separate h from motion

Find h for $u = 250 \mathrm{kJ}$ /kg, $p = 100 \mathrm{kPa}$ , and $v = 1$ m³/kg. Then determine the extra kinetic contribution at 100 m/s.

Hint

Kinetic energy is V²/2 = 5000 J/kg = 5 kJ/kg.

Explore the visual solution

Pressure100 kPaSpecific volume1 m³/kg

h=250+(100)(1)=350\,\mathrm{kJ/kg}

The upstream fluid does pressure work to push mass across the inlet. Do not add pv a second time after using h.

Internal energy u250 kJ/kg

Flow work pv100 kJ/kg

Model assumptions and units

Illustrative parcel with u = 250 kJ/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

w_{flow}=\frac{pA\ell}{m}=pv,\qquad h=u+pv

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

Illustrative parcel with u = 250 kJ/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

w_{flow}=\frac{pA\ell}{m}=pv,\qquad h=u+pv

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.