Page 265 - Schaum's Outline of Theory and Problems of Applied Physics

250 THERMODYNAMICS [CHAP. 21

with an ideal gas that has a movable piston at one end. Figure 21-6 shows the four stages in the operation of a
Carnot engine. The engine does work during the two expansions, and work is done on the engine during the two
compressions; the net work done per cycle is the area enclosed by the curve.
A Carnot engine is the most efficient engine that can operate between the temperatures T 1 and T 2 at which
heat is absorbed and exhausted. If heat Q 1 is absorbed at the absolute temperature T 1 and heat Q 2 is given off at
the absolute temperature T 2 , then Q 1 /Q 2 = T 1 /T 2 in such an engine. Its efficiency is therefore

work output W Q 1 − Q 2 Q 2 T 2
Efficiency (ideal) = = = = 1 − = 1 −
heat input Q 1 Q 1 Q 1 T 1
The smaller the ratio between T 2 and T 1 , the more efficient the engine. Because no reservoir can exist at a
◦
temperature of0Kor0 R, which is absolute zero, no heat engine can be 100 percent efficient.

SOLVED PROBLEM 21.7
The Carnot engine uses only isothermal and adiabatic processes in its operating cycle. Why?
An engine has maximum efficiency when all the processes that occur in its operation are reversible without the
performance of work, since any other processes must necessarily involve the waste of energy. Heat flow from a hot
reservoir to a cooler one is not reversible in this sense because the natural direction of heat flow is from hot to cold
(in fact, this is an alternate statement of the second law of thermodynamics). However, in an isothermal process, the
heat flow occurs at a constant temperature, so the process can be reversed without any work being lost. An adiabatic
process is also reversible in the same sense because no heat enters or leaves a system during such a process. Hence
an engine that uses only isothermal and adiabatic processes is the most efficient possible.

SOLVED PROBLEM 21.8
6
A 1-MW (10 -W) generating plant has an overall efficiency of 40 percent. How much fuel oil whose heat
of combustion is 45 MJ/kg does the plant burn each day?
In 1 day the plant produces

10
6
W = Pt = (10 W)(3600 s/h)(24 h/day) = 8.64 × 10 J
of electric energy. Since its efficiency is 0.4 and Eff = work output/heat input, we have
10
work output 8.64 × 10 J 11
Heat input = = = 2.16 × 10 J
Eff 0.4
The mass of fuel required to supply this amount of heat is

11
2.16 × 10 J
3
m = = 4.8 × 10 kg
6
45 × 10 J/kg
SOLVED PROBLEM 21.9
A Carnot engine absorbs 1 MJ of heat from a reservoir at 300 C and exhausts heat to a reservoir at 150 C.
◦
◦
Find the work it does.
◦
◦
The intake and exhaust temperatures are, respectively, T 1 = 300 C + 273 = 573 K and T 2 = 150 C + 273 =
423 K. The engine efficiency is
T 2 423 K
Eff = 1 − = 1 − = 1 − 0.74 = 0.26
T 1 573 K
and so the work done is
6
5
W = (Eff)(heat intake) = (0.26)(10 J) = 2.6 × 10 J