Cyclic processes and their effeciency, Carnot cycle Mark Scheme

1.

C

2.

Q_H = 100 J and Q_C = 60 J;

A

3.

(use absolute temperatures)

D

4.

C

5.

where T_H µ p_BV_B (the highest temperature) and T_C µ p_DV_D (the lowest temperature) gives e_C = (6p₀V₀ – p₀V₀)/p₀V₀

E

6.

The heat input for this engine occurs during process DðAðB and the heat exhaust is BðCðD

If P₀V₀ corresponds to temperature T₀, the temperatures at points A, B, C and D respectively are 3T₀, 6T₀, 2T₀ and T₀.  The change in temperature for each process is then AB = +3T₀, BC = –4T₀, CD = –T₀ and DA = +2T₀.

We also have P₀V₀ = nRT₀

For the isochoric process, where W = 0, Q = DU = 3/2 nRDT

          DA: Q = 3/2 nR(2T₀) = 3nRT₀

          BC: Q = 3/2 nR(–4T₀) = –6nRT₀

For the isobaric processes, where W = –pDV = –nRDT,
          Q = DU – W = 3/2 nRDT + nRDT = 5/2 nRDT

          AB: Q = 5/2 nR(3T₀) = 7.5nRT₀

          CD: Q = 5/2 nR(–T₀) = –2.5nRT₀

Putting it all together gives us Q_input = Q_DA + Q_AB = 10.5nRT₀ and Q_exhaust­ = –8.5nRT₀

B

7.

E

8.

(use absolute temperature) gives e_c = 0.074

b.   e = W/Q_H, or Q_H = W/e = (100 MW)/(0.074) = 1350 MW and Q_C = Q_H – W = 1250 MW (note Q may represent heat in Joules or rate in Watts)

c.   AB is isothermal so DT = 0.  It is an expansion so W is – and Q = –W
BC is adiabatic so Q = 0.  Temperature drops so DT is negative.
CD is isothermal so DT = 0.  It is a compression so W is + and Q = –W
BC is adiabatic so Q = 0.  Temperature rises so DT is positive.