An ideal gas at 2000 kPa is throttled adiabatically to 150 kPa at the rate of 20

Question

An ideal gas at 2000 kPa is throttled adiabatically to 150 kPa at the rate of 20 mol/sec. Determine the rate of entropy generation and the rate of lost work if T ?=300K.

Explanation / Answer

follow this this will help you 1- The surface temperature of the compressor remains constant. Therefore, it can be treated as a thermal reservoir. 2- The compressor operates at steady-state with negligible changes in kinetic and potential energies. 3- Air behaves as an ideal gas. Diagram : Equations / Data / Solve : Part a.) We can use the following equations to evaluate the internal and total entropy generation rates. Eqn 2 If we place the system boundary far from the surface of the compressor, then ALL of the irreversibilities are inside the system because the temperature at the system boundary is the same as the temperature of the surroundings. So, heat exchange between this system and the surroundings is reversible. This explains why using Tsurr in Eqn 2 tells us that we are computing the TOTAL entropy generation rate. Because with this system, there are no irreversibilities OUTSIDE the system boundary. Notice that heat transfer from this sytem does occur, but it occurs with zero temperature driving force, so it is reversible ! Eqn 3 Using our usual system boundary, right at the surface of the compressor, heat exchange with the surroundings is irreversible. So, by using Eqn 3, with Tsurf instead of Tsurr, we have excluded the external irreversibility due to heat transfer through a finite temperature difference. So, this entropy generation equation gives us the INTERNAL entropy generation only ! The external entropy generation rate is equal to the rate at which the entropy of the universe (compressor and surroundings) changes due only to the heat transfer from the compressor to the surroundings. Eqn 4 The minus sign appears in Eqn 4 because the sign of Q from them the perspective of the system is negative, but from the perspective of the surroundings, Q > 0 because heat is entering the surroundings ! Eqn 4 can be rearranged to give us : Eqn 5 Let's begin by determing DS using the 2nd Gibbs Equation and the Ideal Gas Property Table for air. Eqn 6 So1 0.34649 kJ/kg-K So2 0.0061681 kJ/kg-K R 8.314 J/mole-K MW 28.97 g/mole DS -0.32049 kJ/kg-K Next, we need to evaluate Q, the rate of heat loss from the compressor. We can do this using the 1st Law for open systems operating at steady-state with negligible changes in kinetic and potential energies. Eqn 7 We can solve Eqn 7 for Q : Eqn 8 Now, we can use the Ideal Gas Property Table for air to evaluate H1 and H2. Ho1 87.41 kJ/kg Ho2 208.88 kJ/kg Plug values back into Eqn 7 to evaluate Q : Q -94.27 kW Now, we are able to plug values back into Eqns 2, 3 & 5 to evaluate the entropy generation rates. Sgen,int 0.075418 kW/K Sgen,ext 0.078554 kW/K Sgen,tot 0.153972 kW/K Double check your calculations using : Eqn 9 Part b.) Once we have completed part (a), part (b) is a straightforward application of the relationship between lost work and entropy generation. Eqn 10 Eqn 10 applies for internal, external and total entropy generation and lost work. Wlost,int 22.63 kW Wlost,ext 23.57 kW Wlost,tot 46.19 kW Part c.) Here we must verify that our answer from part (b) is consistent with the defintion of lost work. Eqn 11 In order to do this, we must evaluate WS,rev. The key to doing this is to understand that the reversible process still operates between states 1 and 2. The values of Q and WS are different from those for the real compressor. The fact that ties this part of the problem together and allows us to determine Q and WS for the reversible compressor is that : Eqn 12 We can use this fact and solve Eqn 3 for Qrev : Eqn 13 Qrev -48.07 kW Next, we can apply the 1st Law to the reversible compressor to evaluate WS,rev. We can solve Eqn 7 for WS,rev. The result is : Eqn 14 Plugging values into Eqn 14 yields : WS,rev -108.81 kW Finally, put values into the right-hand side of Eqn 11 : 46.19 kW This matches our result for lost work from part (b). So, we have confirmed the relationship between reversible work, actual work and lost work. Verify : None of the assumptions made in this problem solution can be verified. Answers : a.) Sgen,int 0.075418 kW/K Sgen,ext 0.078554 kW/K Sgen,tot 0.153972 kW/K b.) Wlost,int 22.63 kW Wlost,ext 23.57 kW Wlost,tot 46.19 kW c.) WS,rev - WS,act = 46.19 kW

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