If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

 

009.  `query 9

 

 

 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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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.

 

 

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Question:   

Openstax:  Prin only: A gas-cooled nuclear reactor operates between hot and cold reservoir temperatures of 700ºC and 27.0ºC . (a) What is the maximum efficiency of a heat engine operating between these temperatures? (b) Find the ratio of this efficiency to the Carnot efficiency of a standard nuclear reactor (found in Example 15.4).

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Given Solution: 

The maximum possible Carnot efficiency at two temperatures is based on the absolute temperatures of the hot and cold reservoirs.  For this problem we get

e_max = (T_H - T_C) / T_H = (973 K - 310 K) / (973 K), or around 65%.

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Question:   

Openstax: A certain heat engine does 10.0 kJ of work and 8.50 kJ of heat transfer occurs to the environment in a cyclical process. (a) What was the heat transfer into this engine? (b) What was the engine’s efficiency?

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Given Solution: 

The energy to do the work and the energy transferred to the environment must be put into the engine.  So this engine requires an input of 10 kJ + 8.5 kJ = 18.5 kJ of heat.

It does 10 kJ of work, so its efficiency, which is the ratio of work done to energy input, is

e = (work done) / (energy input) = 10 kJ / 18.5 kJ = .53, or 53%, roughly.

 

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Question:   

Openstax:  (a) What is the work output of a cyclical heat engine having a 22.0% efficiency and 6.00×10^9 J of heat transfer into the engine? (b) How much heat transfer occurs to the environment?

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Given Solution:   

The work done is 22% of the energy input, or

`dW = .22 * 6 * 10^9 J = 1.3 * 10^9 J.

The rest of the energy input goes to the environment.

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Question:   

Openstax:  Assume that the turbines at a coal-powered power plant were upgraded, resulting in an improvement in efficiency of 3.32%. Assume that prior to the upgrade the power station had an efficiency of 36% and that the heat transfer into the engine in one day is still the same at 2.50×10^14 J . (a) How much more electrical energy is produced due to the upgrade? (b) How much less heat transfer occurs to the environment due to the upgrade?

 

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Given Solution: 

Previously the energy produced was

.36 * 2.5 * 10^14 J = 9 * 10^13 J.

After the improvement the efficiency is about 39.3% so the energy produces is

.393 * 2.5 * 10^14 J = 9.8 * 10^13 J, roughly.

The difference is about 8 * 10^12 J.  This heat now goes into the electrical energy produced by the plant, and the heat transferred to the environment decreases by the same amount.

 

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Question:   

Openstax:  A coal-fired electrical power station has an efficiency of 38%. The temperature of the steam leaving the boiler is 550ºC . What percentage of the maximum efficiency does this station obtain? (Assume the temperature of the environment is 20ºC .)

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Given Solution:   

The maximum efficiency of an engine running between 550 C and 20 C is

e_max = (T_h - T_c) / (T_h) = (823 K - 293 K) / (823 K) = 63%, roughly.

Operating at 38%, the station is at

38% / (63%) = 60%, roughly, of the maximum possible efficiency.

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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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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. **

 

 

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Question:    univ phy problem 20.45 11th edition 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?

 

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Given Solution: 

** work done / thermal energy required = .07 so thermal energy required = work done / .07.

 

Translating directly to power, thermal energy must be extracted at rate 210 kW / .07 = 3,000 kW.  The cold water absorbs what's left after the 210 kW go into work, or 2,790 kW.

 

Each liter supplies 4186 J for every degree, or about 17 kJ for the 4 degree net temp change of the water entering and exiting the system.   Needing 3,000 kJ/sec this requires about 180 liters / sec, or about 600 000 liters / hour (also expressible as about 600 cubic meters per hour).

 

Comment from student:  To be honest, I was surprised the efficiency was so low.

 

 Efficiency is low but the energy is cheap and environmental impact in the deep ocean can be negligible so the process can be economical. **

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