Cyclic processes and their effeciency, Carnot cycle Homework Mark Scheme

Homework Mark Scheme

1.

a.   Power is the rate of useful work form an engine so W (which here represents the rate in MW) = 120 MW and e = W/Q_H = 0.40 = 120 MW/Q_H giving Q_H = 300 MW

b.   The rate of heat input from the combustion of oil is 300 Joules per second.  Since oil provides 4.4 × 10⁷ joules per kilogram burned we can divide to find the number of kg per second that must be combusted:
Dm/Dt = (300 × 10⁶ J/s) ÷ (4.4 × 10⁷ J/kg) = 6.82 kg/s

c.   Q_C = Q_H – W = 180 MW

1.      .

a.       P = Fv (from mechanics) = mgv = (10 kg)(10 m/s²)(4 m/s) = 400 W

b.   = 0.4 or 40%

c.   i.          With an efficiency of 0.4 and useful work done at the rate of 400

W we have e = (W/t)/(Q_H/t) or (Q_H/t) = 1000 W

      ii.         (Q_C/t) = Q_H/t) – (W/t) = 600 W

3.

a.   Since P_a = P_b, V_a/T_a = V_b/T_b giving T_b = 750 K

b/c. DU_ab = 3/2 nRDT = (1.5)(1 mole)(8.32 J/mol-K)(750 K – 250 K) = 6240 J
W_ab = –PDV = –(1.2 × 10⁵ Pa)(51 × 10^–3 m³ – 17 × 10^–3 m³) = –4080 J
Q = DU – W = 10,320 J

d.   W = –PDV = 0 (no area under the line)

e.   In a cycle DU = 0 so W = –Q = –1800 J

f.    = 0.66