Question

For this homework question do not calculate anything for the individual devices (turbine, boiler, etc.). Treat the entire cycle as a “black box” and use the results for thermal efficiency of a Carnot cycle as presented in class or found in Equation 5.1 and 5.6. This cycle operates between 200 deg. C (boiler) and 10 kPa (condenser). Determine the thermal efficiency of the cycle if it is a Carnot cycle. (Be sure to use the absolute temperature in equation 5.6!) If the Carnot cycle produces 2500 kW of power (work output), determine the required rate of high-temperature heat input and required rate of low-temperature heat output.

Answer 1

Efficiency for a carnot cycle = 1- (Tc/Th) where Tc & Th are cold and hot temperatures in K

now, Th = bolier temeprature = 200DegC = 200+273 = 473 K;

since 10kPa is the condenser pressure, we can calcultae the Tc (saturation temperature of water at 10 kPa) = 45.8 DegC = 45.8+273 = 318.8 K

efficiency = 1-318.8/473 = 0.32 or 32%

net work = 2500 kW  

we know that for carnot cycle W = Qh-Qc where Qh = high temperature heat input & Qc = low temperature heat input

for carnot cycle we know that Qh/Th = Qc/Tc

thus Qh/473 = Qc/318.8 or Qh = 1.48Qc

thus Qh-Qc = W = 1.48Qc-Qc = 2500 or Qc = 5208.3 kW

Qh = 1.48Qc = 1.48*5208.3 = 7708.3