Question

A frictionless piston is used to provide a constant pressure of 400 kPa in a cylinder containing steam originally at 200 degree C with a volume of 2 m3. If 3500 kJ of heat is added.Calculate the final temperature

Answer 1

According to the first law of thermodynamics, $Q-W = \Delta U$. The work done during the isobaric process (the pressure is const, P =400 kPa) is

$W = \int PdV = P(V_2 - V_1) =3500 \cdot 10^3$ J.

We cannot consider this steam as ideal gas, because for such temperature (which definitely becomes much higher after the work 3500 kJ is done) the molecules of gas begin unavoidably interact. The is no really good model that could describe real gases, instead there are a lot of experimental data.

The mass before and after remains unchanged. Using the steam tables, this is expressed as

$m = \frac{V_1}{v_1}= \frac{2}{0.5342} = 3.744 \; kg.$

The volume $V_2$ is written as $V_2=mv_2 = 3.744 v_2$ . The first law of thermodynamics is then, finding $u_1$, internal energy per mass unit, from the steam tables,

$3500 - (400) (3.744v_2-2)=(u_2 - 2647) \cdot(3.744)$

This requires a trial-and-error process. One plan for obtaining a solution is to guess a value for $v_2$ and calculate $u_2$ from the equation above. If this value checks with the $u_2$ from the steam tables at the same temperature, then the guess is the correct one. For example, guess $v_2$ = 1.0m 3 /kg. Then the equation gives $u_2$ = 3395 kJ/kg. From the steam tables, with P = 0.4 MPa, the $u_2$ value allows us to interpolate $T_2$ = 654°C and the $v_2$ gives $T_2$ = 600°C. Therefore, the guess must be revised. Try $v_2$ = 1.06 m3/kg. The equation gives $v_2$ = 3372 kJ/kg. The tables are interpolated to give $T_2$ = 640°C; for $v_2$ , $T_2$ = 647°C. The actual $v_2$ is a little less than 1.06 m3/kg, with the final temperature being approximately $T_2$ =644°C.

Answer: $T_2$ =640°C