# How to calculate initial and final temperatures during supercritical steam expansion?

Ten $\mathrm{kg}$ of steam at $500~\mathrm{bar}$ is expanded at constant pressure until its volume increases to seven times its initial value of $0.01~\mathrm{m^3}$.
(a) Calculate the initial and final temperature of the steam
(b) Calculate the heat that must be supplied to carry out the process

An attempt at solving this problem:
$$\Delta H + \Delta E_k + \Delta E_p = Q - W_S\ \ \text{(open system)}$$

Taking out unnecessary terms leaves us with
\begin{aligned} \Delta H &= Q\\ 500~\mathrm{bar} &= 5\cdot10^7~\mathrm{Pa}\\ \Delta H &= H_f - H_i\\ H_i &= (5\cdot10^7~\mathrm{Pa})\cdot(0.01~\mathrm{m^3}) = 5\cdot10^5~\mathrm{J}\\ H_f &= (5\cdot10^7~\mathrm{Pa})\cdot(0.07~\mathrm{m^3}) = 3.5\cdot10^6~\mathrm{J}\\ \end{aligned}

Converting to kJ/kg (10 kg of steam) \begin{aligned} \hat{H}_i &= 50~\mathrm{kJ/kg}\\ \hat{H}_f &= 350~\mathrm{kJ/kg}\\ \end{aligned}

And that's where I stopped. I tried looking up the values of $\hat{H}_i$ and $\hat{H}_f$ in steam tables to interpolate the temperatures, but the values are too low. Since the pressure is too high, I also know the ideal gas law $(PV=nRT)$ won't apply to this scenario. Can someone point me in the right direction?

Since supercritical steam is not an ideal gas, you might want to use standard reference data (e.g. provided by NIST).

Initial state
mass m0 = 10 kg
pressure p0 = 500 bar = 50.0 MPa
volume V0 = 0.01 m3
density ρ0 = 1000 kg/m3
temperature T0 = 340.53 K
specific enthalpy h0 = 323.06 kJ/kg

Final state
mass m1 = 10 kg
pressure p1 = 500 bar = 50.0 MPa
volume V1 = 0.07 m3
density ρ1 = 142.86 kg/m3
temperature T1 = 925.86 K
specific enthalpy h1 = 3453.11 kJ/kg

Heat transfer
heat Q = ΔH = (h1h0) • m = 31.3 MJ

• Hi @Loong. Generally with answers to questions such as this, it's preferred that you show the processes of how to arrive at the answer. – John Snow Dec 8 '14 at 5:04
• With standard reference data, you simply look up the data. There is no process, except for the last line of my answer. – Loong May 20 at 17:01