In reference to this example problem (only partially shown), I need to understand why for $\Delta h$ the change in enthalpy, which is $\Delta P/\rho + \Delta u$, the change in pressure is ignored in the solution even though it is considerably large. Could somebody please explain this to me?


2 Answers 2


Starting Point

The starting point of the difference form of the steady state energy balance equation is below.

$$ \dot{q} - \dot{w} = \dot{m}\Delta\left(\tilde{U} + \frac{p}{\rho} + \frac{v^2}{2} + gz\right) $$

In this, $\dot{q}$ is the heat flow (J/s - negative leaving the system), $\dot{w}$ is the Clausius form of work (J/s - positive when done by the system), $\dot{m}$ is the mass flow (kg/s), $\tilde{U}$ is the specific internal energy (J/kg), $p$ is pressure (Pa), $\rho$ is density (kg/m$^3$), $v$ is velocity (m/s), $g$ is gravity (m$^2$/s), and $z$ is height (m).

For any system, the definition of specific enthalpy is $\tilde{H} \equiv \tilde{U} + p\tilde{V} = \tilde{U} + \frac{p}{\rho}$. The expression becomes as below.

$$ \dot{q} - \dot{w} = \dot{m}\Delta\left(\tilde{H} + \frac{v^2}{2} + gz\right) $$


Take the control volume as the fluid inside the compression flow. Allow adiabatic so that $\dot{q} = 0$. Redefine $w$ as purely shaft work because the system (fluid) does not expand or contract as a whole (see below about incompressible fluids or ideal gases).

The system is horizontal in $z$. This is identical to stating categorically that $\Delta z \equiv 0$. The somewhat questionable approach to propose to neglect change in potential energy misses the clarity that zero change in height is a fact of the choice of control volume and not a (randomly applied) assumption.

The velocity of the fluid at any point is found from volumetric flow $\dot{V}$ and area $A$.

$$\dot{m} = \dot{V}\rho = vA\rho$$

Incompressible Fluid

For an incompressible fluid AND ONLY FOR THIS CASE, we can use the conservation of energy expression to obtain this expression at any point in the flow.

$$ v = v_i \frac{A_i}{A} $$

The statement that we "neglect" changes in kinetic energy can therefore only mean that we correspondingly assume that the area changes are negligible. Otherwise, the full representation of the energy balance is as below.

$$ - \dot{w}_s = \dot{m}\Delta\tilde{H} +\frac{\dot{m}v_i^2}{2} \left(\frac{A_i^2}{A_f^2} - 1 \right) $$

For a system with constant specific heat capacity $\tilde{C}_p$ (J/kg K), this can be written as below using change in temperature $\Delta T = T_f - T_i$ ($^o$K or $^o$C).

$$ - \dot{w}_s = \dot{m}\tilde{C}_p\Delta T + \frac{\dot{m}v_i^2}{2} \left(\frac{A_i^2}{A_f^2} - 1 \right) $$

Ideal Gas

The opposing case to an incompressible fluid is the ideal gas. In this case, we obtain the expression below using molar mass $M$ (kg/mol) and gas law constant $R$ (J/mol K).

$$ \dot{m} = \frac{\dot{n}}{M} = \frac{p}{MRT}vA $$

This leads to a modification of the velocity equation as

$$ v = v_i \frac{\beta_i A_i}{\beta A} $$

with $\beta = p/MRT$. For a system with negligible pressure drop from inlet to outlet, the energy balance becomes as below for an ideal gas.

$$ - \dot{w}_s = \dot{m}\tilde{C}_p\Delta T + \frac{\dot{m}v_i^2}{2} \left(\frac{\beta_i^2 A_i^2}{\beta_f^2 A_f^2} - 1 \right) $$


The solution states that potential energy change is negligible. In fact, $\Delta z = 0$ is an absolute truth, not a required assumption.

The solution states that kinetic energy change is negligible. This can be a gross mistake depending on the type of fluid (incompressible or ideal gas), the extent of the throttling ratio $A_f/A_i$, and for an ideal gas the ratio of temperature and pressure drops through the system.


When calculating flow energy, you should always take pressure or temperature changes into account as they are linked by ideal gas law ($PV = nRT$). In this case, calculating change in enthalpy due to temperature change is equivalent to calculating it using pressure change.

Table of thermodynamic quantities for specific cases from https://en.wikipedia.org/wiki/Table_of_thermodynamic_equations

  • $\begingroup$ Hi Alexandre. Thank you for your reply. Do you know a source that could support what you've said (that dU and dP represent the same value for a gas)? $\endgroup$
    – Lespiegle
    May 5, 2019 at 20:18
  • $\begingroup$ I have linked a table from Wikipedia. I am not saying dU and dP represent the same value. Don't get confused with the $C_v$ and $C_p$ which are linked with the following relation : $C_p = C_v + R$ From the general equation $H=U+pV$, using $Q=0$ and $C_p$-$C_v$ relation you should be able to show that $\Delta h = C_p*\Delta T$. $\endgroup$
    – Rusoiba
    May 7, 2019 at 8:59

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