Assingment 09

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course PHY 232

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://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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Your Solution:

We are given two temperatures so assume

T2=580C+273=853k

T1=380C+273=653k

Then the max efficiency=T2-T1/T2=853-653/853=0.23

confidence rating #$&*:

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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:

The given solution has a similar approach to the problem as my solution.

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:

Since we know eff=T2-T1/T2 we can solve for T1=T2-(eff*T2)

Where T2=550C+273=823K

T1=823K*(1-0.28)=593K

Therefore we can do the same process and solve for T2=T1/(1-eff)=593/1-0.35=593/0.35=593/0.65=912k

Hence converting the temp back to C=912-273=639C

confidence rating #$&*:

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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:

The given solution has the same idea and approach i took to solve the problem.

Your Self-Critique Rating:Ok

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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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Your Solution:

Since Work/thermal energy=0.07

then if we rearrange the equation above and solve for thermal energy =Work/0.07

Therefore the rate of the thermal energy extracted = 210KW/0.07=3000KW

while for the cold water absorbed will be 3000-210=2790KW

confidence rating #$&*:

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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. **

Your Self-Critique:

The given solution is well detailed although i still don't understand this statement in quotes ""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).""

Your Self-Critique Rating:OK"

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In abbreviated form:

4186 J per degree

4 degrees implies 4 * 4186 J or about 17 kJ.

3000 kJ / sec, with 17 kJ per liter, requires about 180 liters every second.

etc.

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Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#