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First Law of Thermodynamics (Open System, Steady Flow)

Quantifies the energy balance for an open system operating under steady-flow conditions.

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\dot{Q} - \dot{W} = o u t \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z) - in \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z)

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Core idea

Overview

The First Law of Thermodynamics for open systems, also known as the steady-flow energy equation, is a fundamental principle stating that energy is conserved. For a steady-flow system, the rate of energy entering the system must equal the rate of energy leaving the system plus the rate of energy accumulation within the system (which is zero for steady state). This equation accounts for heat transfer, work transfer, and the energy carried by mass flow, including enthalpy, kinetic, and potential energy components. For the purpose of this calculator, a single inlet and single outlet are assumed.

When to use: Apply this equation to analyze devices like turbines, compressors, nozzles, diffusers, heat exchangers, and pumps where mass flows in and out of a control volume. It is crucial for calculating energy transfer rates, determining unknown fluid properties at inlets or outlets, or sizing components in power plants and refrigeration cycles. Ensure the system is at steady state and identify all energy interactions.

Why it matters: This law is the bedrock of thermal system design and analysis in engineering. It enables engineers to predict performance, optimize efficiency, and troubleshoot energy-related issues in a vast array of applications, from power generation to HVAC systems and chemical processes. Its mastery is essential for developing sustainable and efficient energy solutions.

Symbols

Variables

$\dot{Q}$ = Heat Transfer Rate, $\dot{W}$ = Work Transfer Rate, $\overset{m}{˙}$ = Mass Flow Rate, $h_{in}$ = Specific Enthalpy (Inlet), $h_{o u t}$ = Specific Enthalpy (Outlet)

\dot{Q}

Heat Transfer Rate

\dot{W}

Work Transfer Rate

\overset{m}{˙}

Mass Flow Rate

kg/s

h_{in}

Specific Enthalpy (Inlet)

kJ/kg

h_{o u t}

Specific Enthalpy (Outlet)

kJ/kg

V_{in}

Velocity (Inlet)

m/s

V_{o u t}

Velocity (Outlet)

m/s

g

Gravitational Acceleration

m/s²

z_{in}

Elevation (Inlet)

m

z_{o u t}

Elevation (Outlet)

m

Walkthrough

Derivation

Formula: First Law of Thermodynamics (Open System, Steady Flow)

The First Law of Thermodynamics for open systems states that the rate of energy entering a control volume equals the rate of energy leaving it, plus any accumulation, under steady-flow conditions.

The system operates under steady-flow conditions (properties at any point do not change with time).
The control volume is fixed in space.
Only one inlet and one outlet are considered for simplification, but the principle extends to multiple streams.
Energy transfer occurs via heat, work, and mass flow.

Start with the General Energy Balance:

The rate of change of energy within the control volume ( $E_{C V}$ ) equals the net rate of heat transfer in, minus the net rate of work done by, plus the net rate of energy carried by mass flow.

\frac{d E _{C V}}{d t} = \dot{Q} - \dot{W} + in \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z) - o u t \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z)

Apply Steady-Flow Condition:

For steady-flow, the properties within the control volume do not change with time, so the rate of energy accumulation is zero.

\frac{d E _{C V}}{d t} = 0

Rearrange for Steady-Flow Energy Equation:

Substitute the steady-flow condition into the general energy balance equation.

0 = \dot{Q} - \dot{W} + in \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z) - o u t \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z)

Final Form (as presented):

Rearrange the equation to isolate the net heat and work transfer terms on one side, showing they balance the net energy carried by mass flow. This form is particularly useful for analyzing engineering devices with inlets and outlets.

\dot{Q} - \dot{W} = o u t \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z) - in \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z)

Result

\dot{Q} - \dot{W} = o u t \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z) - in \sum \overset{m}{˙} (h + \frac{V ^{2}}{2} + g z)

Source: Cengel, Y. A., & Boles, M. A. (2015). Thermodynamics: An Engineering Approach (8th ed.). McGraw-Hill Education.

Free formulas

Rearrangements

Solve for $\dot{Q}$

First Law of Thermodynamics (Open System): Make $\dot{Q}$ the subject

\dot{Q} = \dot{W} + \overset{m}{˙} [(h_{o u t} - h_{in}) + \frac{V _{o u t}^{2} - V _{in}^{2}}{2} + g (z_{o u t} - z_{in})]

To make $\dot{Q}$ (rate of heat transfer) the subject, move the work transfer term to the right side of the equation.

Difficulty: 3/5

Solve for $\dot{W}$

First Law of Thermodynamics (Open System): Make $\dot{W}$ the subject

\dot{W} = \dot{Q} - \overset{m}{˙} [(h_{o u t} - h_{in}) + \frac{V _{o u t}^{2} - V _{in}^{2}}{2} + g (z_{o u t} - z_{in})]

To make $\dot{W}$ (rate of work done) the subject, rearrange the equation to isolate the work term.

Difficulty: 3/5

Solve for $\overset{m}{˙}$

First Law of Thermodynamics (Open System): Make $\overset{m}{˙}$ the subject

\overset{m}{˙} = \frac{Q ˙ - W ˙}{( h _{o u t} - h _{in} ) + \frac{V _{o u t}^{2} - V _{in}^{2}}{2} + g ( z _{o u t} - z _{in} )}

To make $\overset{m}{˙}$ (mass flow rate) the subject, divide the net energy transfer by the specific energy change per unit mass.

Difficulty: 4/5

Solve for $h_{in}$

First Law of Thermodynamics (Open System): Make $h_{in}$ the subject

h_{in} = h_{o u t} - \frac{Q ˙ - W ˙}{m ˙} + \frac{V _{o u t}^{2} - V _{in}^{2}}{2} + g (z_{o u t} - z_{in})

To make $h_{in}$ (specific enthalpy at inlet) the subject, isolate the enthalpy difference term and then solve for $h_{in}$ .

Difficulty: 4/5

Solve for $h_{o u t}$

First Law of Thermodynamics (Open System): Make $h_{o u t}$ the subject

h_{o u t} = h_{in} + \frac{Q ˙ - W ˙}{m ˙} - \frac{V _{o u t}^{2} - V _{in}^{2}}{2} - g (z_{o u t} - z_{in})

To make $h_{o u t}$ (specific enthalpy at outlet) the subject, isolate the enthalpy difference term and then solve for $h_{o u t}$ .

Difficulty: 4/5

Solve for $V_{in}$

First Law of Thermodynamics (Open System): Make $V_{in}$ the subject

V_{in} = V_{o u t}^{2} - 2 (\frac{Q ˙ - W ˙}{m ˙} - (h_{o u t} - h_{in}) - g (z_{o u t} - z_{in}))

To make $V_{in}$ (velocity at inlet) the subject, isolate the kinetic energy term, then solve for $V_{in}$ .

Difficulty: 4/5

Solve for $V_{o u t}$

First Law of Thermodynamics (Open System): Make $V_{o u t}$ the subject

V_{o u t} = V_{in}^{2} + 2 (\frac{Q ˙ - W ˙}{m ˙} - (h_{o u t} - h_{in}) - g (z_{o u t} - z_{in}))

To make $V_{o u t}$ (velocity at outlet) the subject, isolate the kinetic energy term, then solve for $V_{o u t}$ .

Difficulty: 4/5

Solve for $g$

First Law of Thermodynamics (Open System): Make $g$ the subject

g = \frac{1}{z _{o u t} - z _{in}} (\frac{Q ˙ - W ˙}{m ˙} - (h_{o u t} - h_{in}) - \frac{V _{o u t}^{2} - V _{in}^{2}}{2})

To make $g$ (gravitational acceleration) the subject, isolate the potential energy term and then solve for $g$ .

Difficulty: 4/5

Solve for $z_{in}$

First Law of Thermodynamics (Open System): Make $z_{in}$ the subject

z_{in} = z_{o u t} - \frac{1}{g} (\frac{Q ˙ - W ˙}{m ˙} - (h_{o u t} - h_{in}) - \frac{V _{o u t}^{2} - V _{in}^{2}}{2})

To make $z_{in}$ (elevation at inlet) the subject, isolate the potential energy term and then solve for $z_{in}$ .

Difficulty: 4/5

Solve for $z_{o u t}$

First Law of Thermodynamics (Open System): Make $z_{o u t}$ the subject

z_{o u t} = z_{in} + \frac{1}{g} (\frac{Q ˙ - W ˙}{m ˙} - (h_{o u t} - h_{in}) - \frac{V _{o u t}^{2} - V _{in}^{2}}{2})

To make $z_{o u t}$ (elevation at outlet) the subject, isolate the potential energy term and then solve for $z_{o u t}$ .

Difficulty: 4/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph displays a straight line where the heat transfer rate scales proportionally with the mass flow rate, with the slope determined by the combined differences in enthalpy, kinetic energy, and potential energy. For an engineering student, this linear relationship means that increasing the mass flow rate requires a proportional increase in heat transfer to maintain the energy balance, where small values represent low-throughput systems and large values represent high-capacity industrial processes. The most important feature of this curve is that the linear relationship means doubling the mass flow rate will exactly double the heat transfer rate, provided the work transfer and energy differences remain constant.

Graph type: linear

Why it behaves this way

Intuition

Visualize a fixed, imaginary box (control volume) through which fluid continuously flows, while heat and work simultaneously cross its boundaries, all in a steady, unchanging manner.

\dot{Q}

Rate of heat transfer into the control volume

Heat added to the system increases its total energy; heat removed decreases it.

\dot{W}

Rate of work done by the control volume

Work done *by* the system (e.g., a turbine) removes energy; work done *on* the system (e.g., a compressor) adds it.

\overset{m}{˙}

Mass flow rate

Represents how much mass, and thus how much associated energy, is crossing the boundary per unit time.

h

Specific enthalpy of the fluid

Combines the internal energy of the fluid with the 'flow work' (energy needed to push fluid across the boundary), representing the total energy content per unit mass of the flowing fluid.

\frac{V ^{2}}{2}

Specific kinetic energy of the fluid

Energy per unit mass due to the bulk motion of the fluid; faster fluid carries more kinetic energy.

Specific potential energy of the fluid

Energy per unit mass due to the fluid's elevation in a gravitational field; higher fluid carries more potential energy.

\sum_{o u t}

Summation over all outlets

Accounts for the total energy carried out of the system by all exiting mass streams.

\sum_{in}

Summation over all inlets

Accounts for the total energy carried into the system by all entering mass streams.

Signs and relationships

-\dot{W}: The negative sign indicates that work done *by* the system (e.g., a turbine producing power) removes energy from the control volume. If work were done *on* the system (e.g., a compressor), $\dot{W}$ would be negative
-\sum_{in} \dot{m} (h + \frac{V^2}{2} + gz): This term represents the rate of energy *entering* the control volume via mass flow. Since the right side of the equation represents the net energy *leaving* the system via mass flow (energy out minus energy in), the.

Free study cues

Insight

Canonical usage

The equation balances energy transfer rates (power) with the net change in energy carried by mass flow, requiring consistent units for power and mass-specific energy.

Common confusion

The most frequent error is a magnitude mismatch between enthalpy (usually kJ/kg) and kinetic/potential energy (usually J/kg or m2/s2), leading to a factor of 1000 error in SI calculations.

Dimension note

This equation is not dimensionless; it is a balance of power (Energy/Time).

Unit systems

$Q_{d} o t, W_{d} o t$ W or BTU/h - Represents the rate of heat and work transfer; must be consistent with the units of mass flow rate times specific energy.

$m_{d} o t$ kg/s or lb/h - Mass flow rate; in steady state, the sum of mass flow in equals the sum of mass flow out.

$h$ J/kg or BTU/lb - Specific enthalpy, representing internal energy plus flow work (u + Pv).

$V^{2} /2$ m^2/s^2 or ft^2/s^2 - Specific kinetic energy; 1 m^2/s^2 is equivalent to 1 J/kg.

gzm^2/s^2 or ft-lbf/lb - Specific potential energy; must be converted to match the units of enthalpy (e.g., kJ/kg or BTU/lb).

One free problem

Practice Problem

A steam turbine operates under steady-flow conditions. Steam enters at an enthalpy of 2800 kJ/kg and velocity of 50 m/s at an elevation of 10 m. It exits at an enthalpy of 2600 kJ/kg and velocity of 150 m/s at an elevation of 5 m. The mass flow rate is 2 kg/s, and the turbine produces 50 kW of work. Calculate the rate of heat transfer to or from the turbine.

Work Transfer Rate50 kW

Mass Flow Rate2 kg/s

Specific Enthalpy (Inlet)2800 kJ/kg

Specific Enthalpy (Outlet)2600 kJ/kg

Velocity (Inlet)50 m/s

Velocity (Outlet)150 m/s

Gravitational Acceleration9.81 m/s²

Elevation (Inlet)10 m

Elevation (Outlet)5 m

Solve for: $Q_{d} o t$

Hint: Remember to convert kinetic and potential energy terms to kJ/kg by dividing by 1000.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Analyzing the power output of a steam turbine in a power plant or the cooling capacity of a refrigeration cycle compressor.

Study smarter

Tips

Always ensure consistent units (e.g., kJ/s for power, kJ/kg for specific enthalpy, m/s for velocity).
Carefully define your control volume and identify all inlets and outlets.
Pay attention to the sign convention for heat and work (heat added to system is positive, work done by system is positive).
Simplify terms if kinetic or potential energy changes are negligible (e.g., for heat exchangers or slow-moving fluids).

Avoid these traps

Common Mistakes

Incorrectly applying sign conventions for heat and work.
Forgetting to include all energy forms (enthalpy, kinetic, potential) or assuming they are negligible when they are not.
Mixing units (e.g., using kJ for enthalpy and J for kinetic energy without conversion).
Applying the equation to unsteady-flow systems without modification.

Common questions

Frequently Asked Questions

The First Law of Thermodynamics for open systems states that the rate of energy entering a control volume equals the rate of energy leaving it, plus any accumulation, under steady-flow conditions.

Apply this equation to analyze devices like turbines, compressors, nozzles, diffusers, heat exchangers, and pumps where mass flows in and out of a control volume. It is crucial for calculating energy transfer rates, determining unknown fluid properties at inlets or outlets, or sizing components in power plants and refrigeration cycles. Ensure the system is at steady state and identify all energy interactions.

This law is the bedrock of thermal system design and analysis in engineering. It enables engineers to predict performance, optimize efficiency, and troubleshoot energy-related issues in a vast array of applications, from power generation to HVAC systems and chemical processes. Its mastery is essential for developing sustainable and efficient energy solutions.

Incorrectly applying sign conventions for heat and work. Forgetting to include all energy forms (enthalpy, kinetic, potential) or assuming they are negligible when they are not. Mixing units (e.g., using kJ for enthalpy and J for kinetic energy without conversion). Applying the equation to unsteady-flow systems without modification.

Analyzing the power output of a steam turbine in a power plant or the cooling capacity of a refrigeration cycle compressor.

Always ensure consistent units (e.g., kJ/s for power, kJ/kg for specific enthalpy, m/s for velocity). Carefully define your control volume and identify all inlets and outlets. Pay attention to the sign convention for heat and work (heat added to system is positive, work done by system is positive). Simplify terms if kinetic or potential energy changes are negligible (e.g., for heat exchangers or slow-moving fluids).

References

Sources

Fundamentals of Heat and Mass Transfer by Incropera, DeWitt, Bergman, Lavine, 7th Edition
Thermodynamics: An Engineering Approach by Yunus A. Cengel and Michael A. Boles, 8th Edition
Transport Phenomena by R. Byron Bird, Warren E. Stewart, and Edwin N. Lightfoot, 2nd Edition
Wikipedia: First law of thermodynamics
Moran & Shapiro, Fundamentals of Engineering Thermodynamics
Cengel & Boles, Thermodynamics: An Engineering Approach
NIST CODATA
Cengel and Boles Thermodynamics: An Engineering Approach

First Law of Thermodynamics (Open System, Steady Flow)

Overview

Variables

Derivation

Start with the General Energy Balance:

Apply Steady-Flow Condition:

Rearrange for Steady-Flow Energy Equation:

Final Form (as presented):

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

First Law of Thermodynamics (Closed System)

Bernoulli's Equation

Continuity Equation

Frequently Asked Questions

Sources