Library / Thermodynamics / THE-02

Thermodynamics

Steady-flow energy balance

Shaft work from a device with one inlet and one outlet.

Ref: THE-02
Rev: v1
Field: Thermodynamics
Result: w_s [kJ/kg]

q - w_{s} = h_{2} - h_{1} + \frac{V_{2} ² - V_{1} ²}{2} + g Δ z

One inlet and one outlet. Heat in and shaft work out are positive. Use J/kg for every term, m/s for speed V and metres for elevation z. Here kinetic energy is in kJ/kg after dividing by 1000.

Energy going into a flowing stream as heat, minus what comes out as work, equals the rise in its enthalpy, speed energy and height energy.

Calculator

Solve for

w_sShaft work outComputed

kJ/kg

qHeat added

kJ/kg

−500500

h₁Inlet enthalpy

kJ/kg

1004,000

h₂Outlet enthalpy

kJ/kg

1004,000

V₁Inlet speed

m/s

0300

V₂Outlet speed

m/s

0300

ΔzHeight rise

−100100

Enthalpy change Δh−510 kJ/kg
Kinetic-energy change10 kJ/kg

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Shaft work out (kJ/kg) current value 480 kJ/kg

h₁Inlet enthalpyKey driver

−2,6201,280

h₂Outlet enthalpy

−8303,070

qHeat added

01,000

V₁Inlet speed

479524

V₂Outlet speed

446491

ΔzHeight rise

479481

For an adiabatic turbine with negligible speed and height changes, work is simply h₁ − h₂.
Enthalpy changes in gas and steam devices dominate; kinetic terms matter in nozzles and jet engines.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (kJ/kg)	Local step	Change in result
h₁ Inlet enthalpy	100 to 4,000 kJ/kg	−2,620	1,280	3,900	+10%	+66.7 %
h₂ Outlet enthalpy	100 to 4,000 kJ/kg	−830	3,070	3,900	+10%	−56 %
q Heat added	−500 to 500 kJ/kg	0	1,000	1,000	+10%	−0.417 %
V₁ Inlet speed	0 to 300 m/s	478.8	523.8	45	+10%	+0.0547 %
V₂ Outlet speed	0 to 300 m/s	446.3	491.3	45	+10%	−0.492 %
Δz Height rise	−100 to 100 m	479	481	1.961	+10%	0 %

Sweep of Heat added

Heat added (kJ/kg)	Shaft work out (kJ/kg)
−500	0
−375	125
−250	250
−125	375
0	500
125	625
250	750
375	875
500	1,000

Sweep one variable

Vary

Shaft work out (kJ/kg)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

w_s

Shaft work out

Result · kJ/kg

What the equation gives you. Everything else feeds it.

Heat added

kJ/kg · typical −500 to 500

Heat added to the stream (negative if lost). Raises the shaft work one-for-one.

h₁

Inlet enthalpy

kJ/kg · typical 100 to 4,000

Specific enthalpy at inlet. Raises the shaft work one-for-one.

h₂

Outlet enthalpy

kJ/kg · typical 100 to 4,000

Specific enthalpy at outlet. Lowers the shaft work one-for-one.

V₁

Inlet speed

m/s · typical 0 to 300

Flow speed at the inlet. Adds kinetic energy at the inlet.

V₂

Outlet speed

m/s · typical 0 to 300

Flow speed at the outlet. Takes energy into kinetic energy.

Δz

Height rise

m · typical −100 to 100

Elevation change between outlet and inlet. Small for gas turbines; large for hydro.

Worked example with the values above

Δh = 2,690 − 3,200 = −510 kJ/kg; ΔKE = (150² − 50²)/2000 = 10 kJ/kg.
Potential: gΔz = 0 kJ/kg.
w_s = q − Δh − ΔKE − gΔz = 480 kJ/kg.

Assumptions and limits

Steady flow, one inlet and one outlet.
Uniform properties across each port.

Common mistakes

Mixing J/kg and kJ/kg.
Using the wrong sign for heat or work.

Statistics and uncertainty

Steam-table enthalpies have errors well below 1%; flow speeds measured by probes are the weaker input.

Where it is used

Turbine, compressor and nozzle analysis.
Jet-engine and gas-turbine cycle calculations.

Maths you need

Enthalpy: h = u + pv
First law: energy conservation

Related formulas

Practise this formula