Thermodynamics Problems- Solutions:

(1) Find the maximum theoretical efficiency of an engine for which steam is supplied at a temperature of 572F and condensed at a temperature of 95F.

Soln. :

572F = 5/9 (F – 32)C = 5/9(572 – 32)C = 5/9(540C) = 300 C

95F = 5/9(95 – 32)C = 5/9(63C) = 35 C

T2 = 300 C = (300 + 273)K = 573 K

T1 = (35 + 273) = 308K

Then: Eff = (T2 – T1)/ T2 = (573 K – 308 K)/ 572 K = 265K/ 572 K = 0.46

(2)Suppose that a nuclear plant generates heat in the amount of 2 x 10^7 J from its fuel rods at a temperature of 1200C and delivers it to a specialized condenser at 140C, to drive a turbine. Assuming the basis of a Carnot engine and cycle, obtain the efficiency for this exchange. Estimate the amount of heat lost from the plant if the actual coefficient of performance is 1.01

Soln: Q_h = 2.0 x 10^7 J

T_h= 1200 C = (1200 + 273)K = 1473K

T_c = 140 C = (140 + 273)K = 413K

Eff.= (T_h – T_c)/ T_h = (1473K – 413K)/ (1473K) = 0.72

Then the theoretical coefficient of performance, or COP = 1/ 0.72 = 1.38

The actual COP = 1.01, so the fraction of theoretical work done is:

COP(act.)/ COP (th.) = 1.01/1.38 = 0.73

COP (th)= (Q_h/ W(th)) = 1.38 so W (th) = (2.0 x 10^7 J)/ 1.38 = 1.44 x 10^7 J

0.73(1.4 x 10^7 J) = 1.04 x 10^7 J

So, heat lost = (1.44 x 10^7 J – 1.04 x 10^7 J) = 0.36 x 10^7 J = 3.6 x 10^6 J

(3) The surface temperature on the planet Mars varies from 183 K (night time) to 268 K (day) near one of its poles. Obtain these temperatures in Celsius and Fahrenheit temperatures. (Note for the latter, take: F = 9C/5 + 32).

Soln:

Night time: In Celsius:

183K = (183 – 273)C = -90 C

In Fahrenheit: F = 9(-90C)/5 + 32 = -130F

Day time: In Celsius:

268K = (268 – 273)C = -5C

In Fahrenheit: F = 9(-5C)/5 + 32 = (-9 + 32)F = 23F

(4)A heat engine operates in a Carnot cycle between 80C and 350 C. It absorbs some 2 x 10^4 J of heat per cycle from the hot reservoir. The duration of each cycle is 1 second. Find: (a) the maximum power output of this engine, and (b) the amount of heat it expels at each cycle.

Soln:

a) Q_h = 2 x 10^4 J

T_c = 80 C = (80 + 273)K = 353 K

T_h = 350 C = (350 + 273)K = 623 K

Carnot efficiency = e(c) = (623K – 350K)/ 623K = 0.43

Net work = W(net) = (Q_h – Q_c) = e(c)W (net)

So: W(net) = 2 x 10^4 J(0.43) = 8.67 x 10^3 J

Power P(net)= W(net)/( 1 s) = (8.67 x 10^3J/ 1s)= 8.67 kW

b) Heat expelled in each cycle = Q_c

But: Q_c = Q_h – W(net) = 2 x 10^4J – 8.67 x 10^3 K = 11.33 kJ

(5) A heat pump powered by an electric motor absorbs heat from outside at 5 C and exhausts heat inside in the form of hot air at 40 C. Find: (a) the maximum coefficient of performance of the heat pump, and (b) the fraction of the theoretical work actually done if the COP = 3.2

Soln:

(a) The max. COP is the theoretical value or COP (th)= Q_h/W

Q_h/W = T_h/ (T_h – T_c) = 313K/ (313K – 278K) = 313K/ 35K = 8.94

(b) The actual COP = 3.2 so the fraction of theoretical work actually done is:

W(net) = COP(ac)/ COP (th) = 3.2/ 8.94 = 0.358 or 35.8%

(6) Technically speaking, the 2nd law of thermodynamics applies only to closed systems. Solar radiation injects on average 1360 watts per square meter onto the Earth's surface, or 1360 J each second per sq. meter. Given this fact, and that plants absorb a good deal of this for the process of photosynthesis, show why the creationist argument that "evolution violates the 2nd law of thermodynamics" doesn't hold up. (Hint: NO quantitative work is needed!)

Ans. Evolving animals are dependent on plants for their nutrition and hence further evolution. (Nutrition provides the ‘fuel’ for evolution). Since plants are open to receiving radiant energy from the Sun, and animals feed on them, then animals are also open to the indirect reception of solar energy. Since neither plants nor animals are therefore closed bio-systems (in terms of energy) then evolution cannot violate the 2nd law of thermodynamics.

(6) In an Otto cycle, during the power stroke, the gas expands from volume V2 -> V1 while the efficiency attained overall is 0.64. To what volume (V1) did the gas expand during the power stroke. (Take the ratio of specific heats as 7/5)

The efficiency of an Otto cycle:

Eff(Otto) = 1 – 1/(V1/V2)^(y-1)

y = 7/5 = 1 2/5 = 1.4 and (y- 1) = (1.4 -1) = 0.4

We know: Eff(Otto) = 0.64

0.64 = = 1 – 1/(V1/V2)^0.4

So: 1/(V1/V2)^0.4 = 0.36 or:

(V2/V1)^0.4 = 0.36

Raising each side to the 5/2 power:

(V2/V1) = (0.36)^5/2 = 0.078

Or V2 = 0.078V1

So that V1 expands to roughly 12.8x the volume V2 during the power stroke (This can be checked algebraically by using the original Otto cycle equation and substituting 12.8 for the ratio (V1/V2)!)

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