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Steady-flow energy-balance calculator

Evaluate a one-inlet, one-outlet steady-flow energy balance from supplied heat, enthalpy, velocity and elevation; inspect shaft-work signs and units.

THERMODYNAMICS · 1.0.0

Steady-flow energy balance

Follow heat, stream energy and shaft work through one inlet and one outlet. Enthalpies are supplied by you, on the same reference basis.

Ẇout = Q̇in + ṁ [(h₁ − h₂) + (C₁² − C₂²)/2000 + g(z₁ − z₂)/1000]

Heat into the control volume is positive. Shaft work out is positive. A negative result is a work-input requirement.

Define the model

Use a decimal point or comma, without thousands separators. Units are fixed and shown beside every input.

Same flow at inlet and outlet; zero or 10⁻⁹ to 10⁶.
Heat in positive, heat loss negative; −10¹² to +10¹².
User-supplied property, same reference as h₂; −10⁷ to +10⁷.
Not temperature. Use the same fluid/property basis; −10⁷ to +10⁷.
Speed magnitude, 0 to 10⁵.
Speed magnitude, 0 to 10⁵.
Common elevation datum; −10⁶ to +10⁶.
Common elevation datum; −10⁶ to +10⁶.

Current calculation

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Method and symbols

Draw a fixed control surface around the chosen device. Steady operation means no stored mass or energy changes with time. Equal inlet and outlet mass flow leaves heat, shaft work and transported stream energy in the balance.

Enthalpy h = u + pv already contains flow work. Do not add a separate inlet/outlet pv term. Use properties from one valid source and reference; this tool does not identify the fluid or calculate properties from temperature.

Write Q̇ − Ẇ = ṁ [Δh + Δ(C²/2)/1000 + gΔz/1000], where Δ means outlet minus inlet. Rearranging gives the displayed shaft-power equation. The result table uses inlet-minus-outlet contributions to shaft power.

Velocity-squared and gz have units J/kg. Divide by 1000 to combine with kJ/kg. Then kg/s × kJ/kg = kJ/s = kW. The standard-gravity value used is g = 9.80665 m/s².

Changing both enthalpies by the same reference offset leaves the result unchanged. A common shift of both elevations also cancels. A first-law balance is necessary; it does not establish second-law feasibility.

Method and symbols
QuantityValueUnit
ṁMass flow ratekg/s
h₁, h₂Specific enthalpy at inlet/outletkJ/kg
C₁, C₂Stream speed magnitudesm/s
z₁, z₂Elevations from one datumm
Q̇inHeat-transfer rate into control volumekW
ẆoutShaft-work rate out of control volumekW
Control-volume energy transfersOne mass stream enters on the left and exits on the right. The upper arrow defines positive heat into the control volume. The lower arrow defines positive shaft work out. Negative values reverse those directions.Control volumedE/dt = 0Inlet · ṁh₁, C₁, z₁Outlet · ṁh₂, C₂, z₂Heat in +Shaft out +
Reference arrows show positive directions, not the sign of the current numerical result.

A maritime teaching example

Synthetic shipboard auxiliary-expander exercise. These prescribed enthalpies are teaching inputs, not a steam table or measured machinery record. Use ṁ = 2 kg/s, Q̇ = −12 kW, h₁ = 600, h₂ = 450 kJ/kg, C₁ = 30, C₂ = 80 m/s, z₁ = 4 and z₂ = 8 m.

  1. Enthalpy: 2 × (600 − 450) = +300 kW.
  2. Kinetic: 2 × (30² − 80²)/2000 = −5.5 kW.
  3. Potential: 2 × 9.80665 × (4 − 8)/1000 = −0.0784532 kW.
  4. Ẇout = −12 + 300 − 5.5 − 0.0784532 = 282.4215468 kW.
  5. q = −12/2 = −6 kJ/kg; w = 282.4215468/2 = 141.2107734 kJ/kg.
A maritime teaching example
QuantityValueUnit
Heat into control volume-12kW
Enthalpy contribution300kW
Kinetic contribution-5.5kW
Potential contribution-0.0784532kW
Net shaft power out282.4215468kW

Assumptions and limits

  • One inlet and one outlet, with equal mass flow and no accumulation; not a filling vessel, start-up or multi-stream heat exchanger.
  • Enthalpies must share a reference and compatible fluid/composition basis. Negative enthalpy values are not automatically invalid; the zero of enthalpy is reference-dependent.
  • The shaft term represents all work other than flow work. Any other work transfer must be explicitly included in that interpretation.
  • Zero mass flow is an algebraic limiting case only. No q or w is reported at zero flow. No inverse flow sizing is offered because zero or tiny energy differences can make it indeterminate or unstable.
  • No efficiencies, turbine performance map, pipe-loss model, property database or equipment approval are inferred. Large velocities require a separately justified flow/property model.

These are numerical scope limits, not permissible equipment operating limits or proof that an ideal-gas approximation is suitable.

Check your understanding

If both enthalpies rise by 500 kJ/kg, does shaft power change?

No. Only h₁ − h₂ enters this balance.

Why does a higher outlet speed reduce shaft output with the other inputs fixed?

More energy leaves as kinetic energy, leaving less for shaft work.

Does Ẇout < 0 make the calculation invalid?

No. It indicates required work input under this sign convention.

Primary references

Curriculum link: bounded teaching applications of first-law and energy-balance topics in İTÜ GMI 213 Thermodynamics. Official course catalogue

Related context

The method explanation and worked example are on this page. The articles below provide additional context.

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