Question

4. An ideal Carnot power generation cycle using air is represented by the diagram below. The properties of the state points are given in the table below. Take the properties of air to be cp 0040.717. and RBhthe ns: and G, have been 1.004 chosen as the average values between the high temperature and low temperature processes, in order for you to not have to worry about the temperature dependence of cp and c, for this problem.) Recall that y kg K kg K Separately calculate the work done by or on the system for each process. Things to remember as you try to accomplish this: a. Boundary work (on a per mass basis) is givn by w Pdv . The polytropic and ideal gas equations can be used to solve for pressure, P, as a function of specific volume, v, for the purposes of solving the integral. . The ideal gas equation is Pv-RT A (reversible) adiabatic process can be represented by the polytropic equation, P-constant, with n-y The constant for a polytropic equation can be solved for using the starting or ending properties given in the table below, subbed into the polytropic equation (l found the state point values for you, so that they would have the same constant at the start and end of polytropic/adiabatic processes). . Positive values for boundary work mean that the specific volume of the system is increasing and that the system is therefore doing work. Negative values mean the specific volume is decreasing and therefore work is being done on the system. b. Calculate the net work done by the cycle (on a per mass basis). Hint: the boundary work integral above will yield positive numbers for work done by the system and negative numbers for work done on the system if used properly Use the first law to determine the heat transferred TO the system (on a per mass basis) during the relevant isothermal process. Hint: Using the first law on each isothermal process will tell you whether the high temperature or low temperature isothermal process is the one receiving the heat (rather than transferring it to the surroundings). Calculate the thermal efficiency of this cycle using your answers to parts b and c above in the equation nth- c. d. Wnet,out Now use the equation Mch-1 (if you're still confused about temperature conversions, try the following, too: use the above equation, but use the equivalent Celsius temperatures TL23°C and TH 27°C and see if the η th is the same. You could also convert the original Kelvin to Rankine, and calculate the thermal efficiency with the Rankine temperatures. Absolute, thermodynamic temperature scales will all yield the same thermal efficiency, but other scales will yield different, untrue values for efficiency). e. and compare to the answer for part d Pressure (kPa) Temperature (K) Specific volume 250 189.3 100 132.1 0.3444 0.4549 0.7175 0.5433 300 State point 1 State point 2 State point 3 State point 4 300 250 250 -1 to 2 (isothermal expansion) 220 3 to 4 (isothermal compression) 4 to 1 (adiabatic compression) 200 180 160 140 120 4 100 0.3 0.35 0.4 0.450.5 0.55 0.6 0.65 70.75 v (mA3/kg)

Answer 1

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