ES 211:Thermodynamics Tutorial 10

Q1 An insulated rigid tank contains 0.9 kg of air at 150 kPa and 20 oC. A paddle wheel inside the tank is now rotated by an external power source until the temperature in the tank rises to 55 oC. If the surrounding air is at T0=20 oC, determine (a) the exergy destroyed and (b) the reversible work for this process.

(a) The exergy destroyed during a process can be determined from an exergy balance, or directly from Xdestroyed = T0Sgen. We will use the second approach. But first determining the entropy generated from an entropy balance, Taking cv=0.718 kJ/kg oC and substituting, the exergy destroyed becomes: (b) The reversible work, which represents the minimum work input Wrev,in in this case, can be determined from the exergy balance by setting the exergy destruction equal to zero.

Since ∆KE= ∆PE=0 and V2=V1. Noting that T0(S2-S1)=T0 ∆Ssystem=21 Since ∆KE= ∆PE=0 and V2=V1. Noting that T0(S2-S1)=T0 ∆Ssystem=21.4 kJ, the reversible work becomes Therefore, a work input of just 1.2 kJ would be sufficient to accomplish the process if all the irreversibilities were eliminated.

Q2 A 5-kg block initially at 350 oC is quenched in an insulated tank that contains 100 kg of water at 30 oC. Assuming the water that vaporizes during the process condenses back in the tank and the surroundings are at 20 oC and 100 kPa, determine (a) the final equilibrium temperature, (b) the exergy of the combined system at the initial and the final states, and (c) the wasted work potential during this process.

(a) Noting that no energy enters or leaves the system during the process, the application of the energy balance gives: By using specific heat values for water and iron at room temperature, the final equilibrium temperature Tf becomes: (b) For an incompressible substance exergy X can be calculated by:

Where T is the temperature of the specified state and T0 is the temperature of the surroundings. At the initial state: Similarly, the total exergy at the final state is (c) Xdestroyed can be calculated as follows:

Q3 An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17 oC, and 800 kJ/kg of heat is transferred to air during the constant-volume heat addition process. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure that occur during the cycle, (b) the net work output, (c) the thermal efficiency and (d) the mean effective pressure for the cycle.

(a) Determining the temperature and pressure of air at the end of the isentropic compression process: Process 1-2 (isentropic compression of an ideal gas): Process 2-3 (constant-volume heat addition)

(b) Process 3-4 (isentropic expansion of an ideal gas): Process 4-1 (constant volume heat rejection): (c) Thermal efficiency of the cycle is determined from its definition:

(d) The mean effective pressure is determined from its definition:

Q4 An ideal Diesel cycle with air as the working fluid has a compression ration of 18 and a cut off ration of 2. At the beginning of the compression process, the working fluid is at 100 kPa, 27 oC and 1917 cm3. Utilizing the cold-air-standard assumptions, determine (a) the temperature and pressure of air at the end of each process, (b) the net work output and the thermal efficiency and (c) the mean effective pressure.

(a) Determining the volume at the end of each process: Process 1-2 (isentropic compression of an ideal gas) Process 2-3 (constant pressure heat addition to an ideal gas): Process 3-4 (isentropic expansion of an ideal gas)

(b) The net work for a cycle is equivalent to the net heat transfer (b) The net work for a cycle is equivalent to the net heat transfer. But first we find the mass of air: Process 2-3 is a constant pressure heat addition process, for which the boundary work and ∆u terms can be combined into ∆h: Process 4-1 is a constant volume heat rejection process and the amount if heat rejected is:

Then the thermal efficiency becomes (c) The MEP is determined by: