Question

Steam is condensed in a closed system at a constant pressure of 75 kPa from a saturated vapor to a saturated liquid by rejecting heat to a thermal energy reservoir at \(37^{\circ} \mathrm{C}\) Determine the second-law efficiency of this process. Take \\[ T_{0}=25^{\circ} \mathrm{C} \text { and } P_{0}=100 \mathrm{kPa} \\].

Short answer

Question: Determine the second-law efficiency of a process involving steam at 75 kPa saturated vapor being cooled and compressed to saturated liquid at the same pressure. The surrounding temperature is 25°C. Answer: The second-law efficiency of this process is 0.

Step-by-step solution

Find the initial and final temperatures of the steam

Find the initial and final properties of the steam at saturated vapor and saturated liquid state at the given pressure (75 kPa). You can refer to the steam tables for these values.

Calculate the heat transfer during the process

Calculate the heat transfer by using the specific enthalpy difference between the initial and final states: \$Q = m(h_{2} - h_{1})$ where: \(Q\) = Heat transfer during the process \(h_{1}\) = Specific enthalpy of the steam at the initial state (saturated vapor) \(h_{2}\) = Specific enthalpy of the steam at the final state (saturated liquid) \(m\) = mass of the steam We can obtain \(h_{1}\) and \(h_{2}\) from the steam tables using the given pressure (75 kPa).

Find the reversible work

The reversible work is given by the equation: \$W_{rev} = T_{0} \cdot Q(T_{2}) - T_{0} \cdot Q(T_{1})$ where: \(W_{rev}\) = Reversible work \(T_{0}\) = Surrounding temperature (given as \(25^{\circ}C\) or 298.15 K) \(Q(T_{1})\) = Heat transferred to the surrounding at initial temperature (\(T_{1}\) = initial temperature of steam from step 1) \(Q(T_{2})\) = Heat transferred to the surrounding at final temperature (\(T_{2}\) = final temperature of steam from step 1) To find \(Q(T_{1})\) and \(Q(T_{2})\), we will need to perform a separate analysis using the heat engine model.

Determining the heat transfer to the surroundings

Model this process as a heat engine operating between the steam at temperatures \(T_{1}\) and \(T_{2}\), and a thermal energy reservoir at \(37^{\circ}C\) (310.15 K). Considering the Carnot efficiency for the heat engine, we can find the heat transfer to the surroundings: \$Q(T_{2}) = \frac{T_{2}}{T_{1}} \cdot Q$ and \$Q(T_{1}) = Q - Q(T_{2})$

Calculate the second-law efficiency

Use the obtained values of heat transfer and reversible work to calculate the second-law efficiency: \$η_{II} = \frac{W_{act}}{W_{rev}}$ where: \(η_{II}\) = Second-law efficiency \(W_{act}\) = Actual work (0, since there is no work output in the process) In this case, the work output is 0, so the second-law efficiency will be 0: \$η_{II} = \frac{0}{W_{rev}} = 0$ Hence, the second-law efficiency of this process is 0.