Question

One kilogram of water contained in a piston–cylinder assembly, initially saturated vapor at 460 kPa, is condensed at constant pressure to saturated liquid. Consider an enlarged system consisting of the water and enough of the nearby surroundings that heat transfer occurs only at the ambient temperature of 25 C. Assume the state of the nearby surroundings does not change during the process, and ignore kinetic and potential energy effects. For the enlarged system, determine the heat transfer, in kJ, and the entropy production, in kJ/K.

Answer 1

Answer:

$Q = 2118.075\,kJ$, $S_{gen} = 2.0836\,\frac{kJ}{K}$

Explanation:

The process is modelled after the First and Second Law of Thermodynamic:

$-Q + m\cdot P\cdot (\nu_{1} - \nu_{2}) = m\cdot (u_{2}-u_{1})$

$-\frac{Q}{T_{surr}} + S_{gen} = m\cdot (s_{2}-s_{1})$

The properties of the fluid are obtained from steam tables:

Inlet

$\nu = 0.40610\,\frac{m^{3}}{kg}$

$u = 2557.8\,\frac{kJ}{kg}$

$s = 6.8490\,\frac{kJ}{kg\cdot K}$

Outlet

$\nu = 0.001089\,\frac{m^{3}}{kg}$

$u = 626.03\,\frac{kJ}{kg}$

$s = 1.8285\,\frac{kJ}{kg\cdot K}$

The heat transfer is:

$Q = m\cdot [P\cdot (\nu_{1}-\nu_{2})+(u_{1}-u_{2})]$

$Q = (1\,kg)\cdot \left[(460\,kPa)\cdot \left(0.40610\,\frac{m^{3}}{kg} - 0.001089\,\frac{m^{3}}{kg} \right) + \left(2557.8\,\frac{kJ}{kg} - 626.03\,\frac{kJ}{kg} \right)\right]$

$Q = 2118.075\,kJ$

Lastly, the entropy production is:

$S_{gen} = \frac{Q}{T_{surr}} + m\cdot (s_{2}-s_{1})$

$S_{gen} = \frac{2118.075\,kJ}{298.15\,K} + (1\,kg)\cdot \left(1.8285\,\frac{kJ}{kg\cdot K}-6.8490\,\frac{kJ}{kg\cdot K} \right)$

$S_{gen} = 2.0836\,\frac{kJ}{K}$