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The steady energy balance relates heat, shaft work, and enthalpy, kinetic and potential energy carried by flowing streams.
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Formula
Q̇,Ẇ in W; ṁ in kg/s; h,v2/2,gz in J/kg
| Symbol | Meaning | Unit |
|---|---|---|
| Q̇ | Heat-transfer rate into the control volume | W |
| Ẇ | Technical work rate out of the control volume | W |
| Specific enthalpy including flow work | J/kg | |
| Mean section velocity | m/s | |
| Elevation above a reference | m |
\dot Q-\dot W+\sum_{in}\dot m(h+v^2/2+gz)-\sum_{out}\dot m(h+v^2/2+gz)=0Assume steady operation and negligible stored-energy change; nonuniform velocity profiles may require a kinetic-energy correction factor. Q̇ is positive inward and Ẇ positive outward. Enthalpy h=u+p/ρ includes flow work. Sum every inlet and outlet; convert kJ/kg and watts consistently before calculation.
Assume steady operation and negligible stored-energy change; nonuniform velocity profiles may require a kinetic-energy correction factor. Q̇ is positive inward and Ẇ positive outward. Enthalpy h=u+p/ρ includes flow work. Sum every inlet and outlet; convert kJ/kg and watts consistently before calculation.
The steady energy balance relates heat, shaft work, and enthalpy, kinetic and potential energy carried by flowing streams.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Steady-flow energy balance for a control volume” and check its unit and magnitude.
A turbine has ṁ=2 kg/s, h1=300 kJ/kg, v1=20 m/s, h2=400 kJ/kg, v2=5 m/s, equal elevations and is adiabatic.
Ẇ=ṁ[(h1−h2)+(v12−v22)/2]=2[−100000+187.5] W=−199.625 kW.
Negative means work input under the stated outward-positive convention.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: Q̇,Ẇ in W; ṁ in kg/s; h,v2/2,gz in J/kg.
Formula cluster
The steady energy balance relates heat, shaft work, and enthalpy, kinetic and potential energy carried by flowing streams.
A turbine has ṁ=2 kg/s, h1=300 kJ/kg, v1=20 m/s, h2=400 kJ/kg, v2=5 m/s, equal elevations and is adiabatic. Ẇ=ṁ[(h1−h2)+(v12−v22)/2]=2[−100000+187.5] W=−199.625 kW. Negative means work input under the stated outward-positive convention.
Balance steady streams in turbines, compressors, nozzles and heat exchangers.
Here is how to work through a typical Steady-flow energy balance for a control volume (Q̇ − Ẇ + Energiezufluss − Energieabfluss = 0) task step by step:
Task
Work through the numerical example for “Steady-flow energy balance for a control volume” and check its unit and magnitude.
Solution path
A turbine has ṁ=2 kg/s, h1=300 kJ/kg, v1=20 m/s, h2=400 kJ/kg, v2=5 m/s, equal elevations and is adiabatic.
Ẇ=ṁ[(h1−h2)+(v12−v22)/2]=2[−100000+187.5] W=−199.625 kW.
Negative means work input under the stated outward-positive convention.
Mass entering or leaving must be pushed across the control surface. The associated flow work p/ρ combines with internal energy u to form enthalpy h=u+p/ρ. That is why stream energy is h+v2/2+gz. Adding p/ρ again would double-count flow work. Signs are fixed here: heat inward is positive, technical work outward is positive; enthalpy includes p/ρ. Application: Balance steady streams in turbines, compressors, nozzles and heat exchangers. The numerical example A turbine has ṁ=2 kg/s, h1=300 kJ/kg, v1=20 m/s, h2=400 kJ/kg, v2=5 m/s, equal elevations and is adiabatic. Ẇ=ṁ[(h1−h2)+(v12−v22)/2]=2[−100000+187.5] W=−199.625 kW. Negative means work input under the stated outward-positive convention anchors the idea in a concrete measurement case.
Under the stated convention, Ẇ is positive when work leaves the control volume. For an adiabatic turbine with no elevation change, Ẇ=ṁ[(h1−h2)+(v12−v22)/2]. If total stream energy decreases from inlet to outlet, delivered work is positive. The balance sign depends on the declared convention. Signs are fixed here: heat inward is positive, technical work outward is positive; enthalpy includes p/ρ. The corresponding units are Q̇,Ẇ in W; ṁ in kg/s; h,v2/2,gz in J/kg. A turbine has ṁ=2 kg/s, h1=300 kJ/kg, v1=20 m/s, h2=400 kJ/kg, v2=5 m/s, equal elevations and is adiabatic. Ẇ=ṁ[(h1−h2)+(v12−v22)/2]=2[−100000+187.5] W=−199.625 kW. Negative means work input under the stated outward-positive convention. This checks that the reference quantities are compatible.
If energy storage changes appreciably during startup, tank cooling or cyclic operation, the balance needs an accumulation term. Electrical work, chemical energy or multiple heat boundaries must also be included when relevant. The steady shortcut is not a universal replacement for every open process. Signs are fixed here: heat inward is positive, technical work outward is positive; enthalpy includes p/ρ. Scope: Assume steady operation and negligible stored-energy change; nonuniform velocity profiles may require a kinetic-energy correction factor. Q̇ is positive inward and Ẇ positive outward. Enthalpy h=u+p/ρ includes flow work. Sum every inlet and outlet; convert kJ/kg and watts consistently before calculation. A practical application is Balance steady streams in turbines, compressors, nozzles and heat exchangers.
Write the quantity and unit from the variable list next to every value: Q̇,Ẇ in W; ṁ in kg/s; h,v2/2,gz in J/kg. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example A turbine has ṁ=2 kg/s, h1=300 kJ/kg, v1=20 m/s, h2=400 kJ/kg, v2=5 m/s, equal elevations and is adiabatic. Ẇ=ṁ[(h1−h2)+(v12−v22)/2]=2[−100000+187.5] W=−199.625 kW. Negative means work input under the stated outward-positive convention. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. Signs are fixed here: heat inward is positive, technical work outward is positive; enthalpy includes p/ρ.
The background states this starting condition: Assume steady operation and negligible stored-energy change; nonuniform velocity profiles may require a kinetic-energy correction factor. Q̇ is. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Signs are fixed here: heat inward is positive, technical work outward is positive; enthalpy includes p/ρ.
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Curated exam set for "Steady-flow energy balance for a control volume":
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Which formula describes Steady-flow energy balance for a control volume?
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How do you rearrange Steady-flow energy balance for a control volume for Base equation for the target quantity?
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Which common mistake happens with Steady-flow energy balance for a control volume?
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