Open Query 9

course Phys 202

009. `query 8

Question: prin phy and gen phy problem 15.19 What is the maximum efficiency of a heat engine operating between temperatures of 380 C and 580 C?

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Your Solution:

e = 1 – TL/TH

e = 1 – 653 K/ 853 K

e = .23

Confidence Rating: 3

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Given Solution:

The maximum possible efficiency is (T_h - T_c) / T_h, where T_h and T_c are the absolute max and min operating temperatures.

T_h is (580 + 273)K = 853 K and T_c is (380 + 273) K = 653 K, so the maximum theoretical efficiency is

max efficiency = (T_h - T_c) / T_h = (853 K - 653 K) / (853 K) = .23, approx.

This means that the work done by this engine will be not greater than about 23% of the thermal energy that goes into it.

Your Self-Critique: OK

Your Self-Critique Rating: OK

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Question: query gen phy problem 15.26 source 550 C -> Carnot eff. 28%; source temp for Carnot eff. 35%?

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Your Solution:

e = 1 – TL/TH

.28 = 1 – TL/823 K

TL = 593 K

e = 1 – TL/TH

.35 = 1 – 593 K / TH

TH = 912 K = 639 C

Confidence Rating: 3

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Given Solution:

** Carnot efficiency is eff = (Th - Tc) / Th.

Solving this for Tc we multiply both sides by Th to get

eff * Th = Th - Tc so that

Tc = Th - eff * Th = Th ( 1 - eff).

We note that all temperatures must be absolute so we need to work with the Kelvin scale (adding 273 C to the Celsius temperature to get the Kelvin temperature)

If Th = 550 C = 823 K and efficiency is 30% then we have

Tc =823 K * ( 1 - .28) = 592 K.

Now we want Carnot efficiency to be 35% for this Tc. We solve eff = (Th - Tc) / Th for Th:

Tc we multiply both sides by Th to get

eff * Th = Th - Tc so that

eff * Th - Th = -Tc and

Tc = Th - eff * Th or

Tc = Th ( 1 - eff) and

Th = Tc / (1 - eff).

If Tc = 576 K and eff = .35 we get

Th = 592 K / ( 1 - .35 ) = 592 C / .6 = 912 K, approx.

This is (912 - 273) C = 639 C. **

Your Self-Critique: OK

Your Self-Critique Rating: OK

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Question: univ phy problem 20.44 (18.40 10th edition) ocean thermal energy conversion 6 C to 27 C

At 210 kW, what is the rate of extraction of thermal energy from the warm water and the rate of absorption by the cold water?

&#Your work looks very good. Let me know if you have any questions. &#