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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?

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1 Answer 1

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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

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    $\begingroup$ 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. $\endgroup$
    – John Snow
    Dec 8, 2014 at 5:04
  • $\begingroup$ With standard reference data, you simply look up the data. There is no process, except for the last line of my answer. $\endgroup$
    – user7951
    May 20, 2019 at 17:01

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