Question

Refrigerant-134a enters an adiabatic compressor at 120 kPa superheated by \(2.3^{\circ} \mathrm{C},\) and leaves at \(0.7 \mathrm{MPa}\). If the compressor has a second-law efficiency of 85 percent, determine \((a)\) the actual work input, \((b)\) the isentropic efficiency, and \((c)\) the exergy destruction. Take the environment temperature to be \(25^{\circ} \mathrm{C}\).

Short answer

Based on the given information for an adiabatic compressor working with Refrigerant-134a gas, the actual work input is 37.52 kJ/kg, the isentropic efficiency is 83.1%, and the exergy destruction is 0 kJ/kg.

Step-by-step solution

Find the initial entropy and enthalpy

Using the superheated Refrigerant-134a tables and the given pressure (120 kPa) and temperature (2.3°C, which is 275.45 K), we find the specific enthalpy (\(h_1\)) and specific entropy (\(s_1\)) at the initial state (state 1). The values are: \(h_1 = 249.56\, \mathrm{kJ/kg}\) \(s_1 = 0.9446\, \mathrm{kJ/(kg \cdot K)}\)

Calculate the isentropic entropy and enthalpy at the final state

As the process is adiabatic and isentropic, we have \(s_{2s} = s_1\). So, we can use the given pressure (0.7 MPa) and \(s_{2s}\) value to find the isentropic enthalpy (\(h_{2s}\)) at the final state (state 2s) using the superheated Refrigerant-134a tables. We find that: \(h_{2s} = 283.16\, \mathrm{kJ/kg}\)

Determine the actual entropy at the final state

Using the second-law efficiency, which is given as 85%, we will determine the actual specific entropy (\(s_2\)) for the final state (state 2) using the following formula: \(\eta_{II} = \frac{s_{2s} - s_1}{s_2 - s_1}\) Rearranging and solving for \(s_2\), we get: \(s_2 = s_1 + \frac{s_{2s} - s_1}{\eta_{II}} = s_1 + \frac{0.9446 - 0.9446}{0.85} = 0.9446\, \mathrm{kJ/(kg \cdot K)}\)

Calculate the actual work input

Using the enthalpy difference between the initial and final states, we can calculate the actual work input (\(W_{in}\)) using the following equation: \(W_{in} = h_2 - h_1\) We can find the actual enthalpy (\(h_2\)) at the final state (state 2) using the given pressure (0.7 MPa) and \(s_2\) value from the superheated Refrigerant-134a tables. We find that: \(h_2 = 287.08\, \mathrm{kJ/kg}\) So, the actual work input is: \(W_{in} = 287.08 - 249.56 = 37.52\, \mathrm{kJ/kg}\)

Determine the isentropic efficiency

We can find the isentropic efficiency (\(\eta\)) using the actual work input and the isentropic work input (difference between \(h_1\) and \(h_{2s}\)) as follows: \(\eta = \frac{W_{in, isentropic}}{W_{in, actual}} = \frac{h_1 - h_{2s}}{h_1 - h_2} = \frac{249.56 - 283.16}{249.56 - 287.08} = 0.831\) So, the isentropic efficiency is 83.1%.

Calculate the exergy destruction

Finally, we can calculate the exergy destruction (\(E_D\)) using the temperature difference between the environment (\(T_0 = 298.15\) K) and the final state, and the entropy change between the initial and final states: \(E_D = T_0 (s_2 - s_1) = 298.15 (0.9446 - 0.9446) = 0\, \mathrm{kJ/kg}\) So, there is no exergy destruction in this process. In conclusion: 1. The actual work input is 37.52 kJ/kg. 2. The isentropic efficiency is 83.1%. 3. The exergy destruction is 0 kJ/kg.