query_9

#$&*

course Phy 232

6/15/12 submitted around 1:22 PM

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

I am a university student

Confidence Rating:

.............................................

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:

Your Self-Critique Rating:

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

Your Solution:

I am university student

Confidence Rating:

.............................................

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:

Your Self-Critique Rating:

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?

Your Solution:

We can use the formula of: e_mas=1-T_c/T_H

If we plug in our value then we get:1-279.15K/300.15K=7.0%

Next we can find how much the cold water absorbs by using the formula:Q_H/t=P_out/e=210kW/.070=3000kW

Now we can solve for cold water:Q_c/t=Q_H/t -W/t

We can fill in numbers: 3000kW-210kW=2,790kW

To find the rate we can use the formula: (Q_c/t)/cdeltaT

Filling in the values we get: 2.8E6W*3600s/h /4190J/kgK*4K=6E5L/h

Confidence Rating: 3

.............................................

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: OK

Your Self-Critique Rating: 3

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

query_9

#$&*

course Phy 232

6/15/12 submitted around 1:22 PM

Your solution, attempt at 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

I am a university student

Confidence Rating:

.............................................

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:

Your Self-Critique Rating:

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

Your Solution:

I am university student

Confidence Rating:

.............................................

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:

Your Self-Critique Rating:

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?

Your Solution:

We can use the formula of: e_mas=1-T_c/T_H

If we plug in our value then we get:1-279.15K/300.15K=7.0%

Next we can find how much the cold water absorbs by using the formula:Q_H/t=P_out/e=210kW/.070=3000kW

Now we can solve for cold water:Q_c/t=Q_H/t -W/t

We can fill in numbers: 3000kW-210kW=2,790kW

To find the rate we can use the formula: (Q_c/t)/cdeltaT

Filling in the values we get: 2.8E6W*3600s/h /4190J/kgK*4K=6E5L/h

Confidence Rating: 3

.............................................

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.

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: OK

Your Self-Critique Rating: 3

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

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query_9

#$&*

course Phy 232

6/17/12 submitted around 12:41 PM

Your solution, attempt at 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

I am a university student

Confidence Rating:

.............................................

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:

Your Self-Critique Rating:

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

Your Solution:

I am university student

Confidence Rating:

.............................................

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:

Your Self-Critique Rating:

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?

Your Solution:

We can use the formula of: e_mas=1-T_c/T_H

If we plug in our value then we get:1-279.15K/300.15K=7.0%

Next we can find how much the cold water absorbs by using the formula:Q_H/t=P_out/e=210kW/.070=3000kW

Now we can solve for cold water:Q_c/t=Q_H/t -W/t

We can fill in numbers: 3000kW-210kW=2,790kW

To find the rate we can use the formula: (Q_c/t)/cdeltaT

Filling in the values we get: 2.8E6W*3600s/h /4190J/kgK*4K=6E5L/h

Confidence Rating: 3

.............................................

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.

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: OK

Your Self-Critique Rating: 3

Self-critique (if necessary):

------------------------------------------------

Self-critique rating: