# Why the constant-pressure specific heat can not be used to calculate the enthalpy change of steam?

I'm trying to calculate the enthalpy change of the saturated steam by using the constant-pressure specific heat. The corresponding parameters are as follows.

• $$T_1=110 ^\circ C$$
• $$C_{pg,1}=2.12 kJ/kg/K$$
• $$h_{g,1}=2691.07 kJ/kg$$

and

• $$T_2=140 ^\circ C$$
• $$C_{pg,2}=2.31 kJ/kg/K$$
• $$h_{g,2}=2733.44 kJ/kg$$

The above steam properties come from an online steam properties calculator (http://www.peacesoftware.de/einigewerte/wasser_dampf_e.html).

So the enthalpy change should be $$\Delta h_{g} = h_{g,2}-h_{g,1}= 42.37 kJ/kg$$.

While when I calculate the enthalpy change using the quation $$\Delta h=C_p \Delta T$$, I get a different result.

• $$\Delta h=C_p \Delta T=\overline{C_{pg}}(T_2-T_1)=66.45 kJ/kg$$
• $$\overline{C_{pg}}=\frac{C_{pg,1}+C_{pg,2}}{2}$$

The latter value is much greater than the former. Even I use the more precise equation:

$$\Delta h = \int_{T_1}^{T_2} {C_{pg}} dT=66.31 kJ/kg$$

What am I doing wrong?

• Why are you using $C_p$ in the first place? Pressure seems to change a lot in this process: pressure at 110 degrees is 1.4 bar, pressure at 140 degrees is 3.6 bar. Not an isobaric process. Mar 3, 2023 at 9:51
• @JánLalinský in the section 3.11 P115 of the book Fundamentals of Thermodynamics, 8th Edition link, it goes " enthalpy of an ideal gas is a function of the temperature only and is independent of the pressure......" and "From state 1 the high temperature can be reached by a variety of paths, and in each case the final state is different. However, regardless of the path, the change in internal energy is the same, as is the change in enthalpy......". Mar 3, 2023 at 10:11
• @JánLalinský so I think, regardless of the thermodynamic process, the enthalpy change between to states is equal to that of an Isobaric process with the same temperature range. Mar 3, 2023 at 10:23
• It is true that $dh = c_p dT$ holds for ideal gas even if $c_p$ changes with temperature, like we have here. However, water vapor need not be ideal gas under certain circumstances. For non-ideal gases, the relation $dh = c_pdT$ need not hold, because change in molar volume $dv$ matters as well. So I think your calculation with these numbers shows that water vapor does not obey laws of ideal gas well enough. Try to repeat the calculations with some different source of the numbers, it may be due to erroneous numbers. Mar 3, 2023 at 12:15
• It has got to be a non-ideal gas effect. Calculate the difference in residual enthalpies between the two states using Eqn. 6.58 of Smith and Van Ness. Mar 3, 2023 at 12:19

## 1 Answer

The heat capacity values that you cite from your reference seem too high. From the superheated steam tables in this temperature range, at a pressure of 0.06 bars (simulating the ideal gas limit), the heat capacity value is only 1.9 kJ/kg.C. For saturation pressures at these same temperatures, because of non-ideal gas effects (residual enthalpy), the apparent heat capacity will be even lower than this, as you have shown.

So the problem is that the heat capacity (from the literature reference) is too high for either the ideal gas limit or for the actual saturations conditions (which also include the real gas residual enthalpy).

You can use your steam tables to qualify the residual enthalpy effect by subtracting the enthalpy at temperature T and 0.06 bars (in the superheated steam tables) from the enthalpy at temperature T in the saturated steam tables.