Question

The following heat engines produce power of 95,000 kW. Determine in each case the rates at which heat is absorbed from the hot reservoir and discarded to the cold reservoir.

a. A carnot engine operates between heat reservoir at 750K and 300K.

b. A practical engine operates between the same heat reservoirs but with a thermal efficiency n = 0.35

*Please show are the formulas and steps how to do this problem please.

Answer 1

a) Carnot cycle

Th= 750K

Tc=300 K

W= 95000 kW

W= Qh-Qc=95000

Delta S1= entropy change of isothermal expansion at Th,

              =Qh / Th

Delta S2= entropy change of isothermal compression at Tc

               =Qc/TC

for carnot cycle

Delta S1=Delta S2

Qh/Th = Qc /Tc

Qh= Qc Th/Tc

Qh= Qc 750/300

Qh= 2.5 Qc

now, 95000= Qh-Qc

95000= 2.5Qc- Qc

Qc= 63333.33 kW--- Heat discaded to cold reservoir (ans a)

Qh=2.5 Qc

     =2.5*63333

  Qh =158333.33 kW--- heat absorbed from hot reservoir ( ans a)

b) Practical engine

efficiency=0.35 = W/ Qh

so Qh= W/ 035

        =95000 / 0.35

      = 271428.571 kW--- heat absored from hot reservoir (ans b)

Now, efficiency of cycle also given as = 1- Qc/ Qh

0.35=1-Qc/Qh

Qc/Qh=1-0.35

Qc/Qh =0.65

Qc=0.65 *Qh

Qc= 0.65*271428.571

=176428.571 kW  ------- heat discarded to cold reservoir (ans b)