Query 9

#$&*

course Phy 122

2/14 1755hrsI believe that this completes module one Sir. My OIC will give me the test next week, when he has some time. I am curious to see how I will do. Thanks for squaring me away earlier.

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:

580 + 273.15 = 853.15K

380 + 273.15 = 653.15K

853.15K - 653.15K = 200K / 853.15K = .2344 * 100 = 23%

This means that the maxe = 23%

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: I guess that K conversion for the purposes of this book is 273, and not what I learned a ways back of 273.15. Either way, the answer turned out he same.

Your Self-Critique Rating:3

@&

273.15 is good, but the 2- or 3-significant figure information given does not justify the last two significant figures of that quantity.

Had the temperatures been given as 380.38 and 579.93, for example, you would have wanted to use the 273.15.

*@

*********************************************

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

(700C + 273)K = 973K

(27C + 273)K = 310K

973K - 310K = 663K/973K = .681(efficiency is always less than 1) * 100 = 68% = Carnot e

Table 15-4 really doesn’t give me any defining information so I conducted the problem similarly to the one above and the example in the book.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution: 1

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%.

Your Self-Critique: my answer was close so I think that I am doing this correctly.

Your Self-Critique Rating:2

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

The heat transfer to the environment came from the system, so it must be put back.

10kj + 8.5kj = 18.5kj

With 10kj of work, the efficiency would be the work / by the input:

e = 10kj/18.5kj = .541*100 = 54%

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique:3

Your Self-Critique Rating: nailed it.

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

It looks the equations above, just done in reverse:

22% = .22e

.22 * 6 * 10^9 = 1.32 * 10E9

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique: I am thinking that the ‘d = the derivative.

Your Self-Critique Rating:2

@&

`d stands for capital Delta.

For a derivative you would use just plain d.

*@

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Before the upgrade:

.36 * 2.5E14 = 9E13

After the upgrade:

36% + 3.32% = 39.32% = .3932

.3932*2.5E14 = 9.83E13 so the difference is 8.3E12 as an increase. Not sure of the heat transfer to the environment but thinking about it, I would guess it would be the same. You are not increasing the energy input, your increasing the energy output through an upgrade making it more efficient, not working harder.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique: forgot to label my energy in J. I think that I am thinking about it correcty.

Your Self-Critique Rating:3

*********************************************

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 .)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

(550C + 273)K = 823K, (20C + 273)K = 293K

823K - 293K = 530K/823K = .644 * 100 = 64%

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique: It takes a while of starring at the word problems for me to figure out what is actually being asked. When I look at the solution section, I can see how I did the math right but labeling it doesn’t come easy for me.

Your Self-Critique Rating:2

"

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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

Query 9

#$&*

course Phy 122

2/14 1755hrsI believe that this completes module one Sir. My OIC will give me the test next week, when he has some time. I am curious to see how I will do. Thanks for squaring me away earlier.

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: 580 + 273.15 = 853.15K 380 + 273.15 = 653.15K 853.15K - 653.15K = 200K / 853.15K = .2344 * 100 = 23% This means that the maxe = 23% 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: I guess that K conversion for the purposes of this book is 273, and not what I learned a ways back of 273.15. Either way, the answer turned out he same. Your Self-Critique Rating:3

@& 273.15 is good, but the 2- or 3-significant figure information given does not justify the last two significant figures of that quantity.

Had the temperatures been given as 380.38 and 579.93, for example, you would have wanted to use the 273.15. *@

********************************************* 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: (700C + 273)K = 973K (27C + 273)K = 310K 973K - 310K = 663K/973K = .681(efficiency is always less than 1) * 100 = 68% = Carnot e Table 15-4 really doesn’t give me any defining information so I conducted the problem similarly to the one above and the example in the book. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution: 1 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%. Your Self-Critique: my answer was close so I think that I am doing this correctly. Your Self-Critique Rating:2

********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The heat transfer to the environment came from the system, so it must be put back. 10kj + 8.5kj = 18.5kj With 10kj of work, the efficiency would be the work / by the input: e = 10kj/18.5kj = .541*100 = 54% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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. Your Self-Critique:3 Your Self-Critique Rating: nailed it.

********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: It looks the equations above, just done in reverse: 22% = .22e .22 * 6 * 10^9 = 1.32 * 10E9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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. Your Self-Critique: I am thinking that the ‘d = the derivative. Your Self-Critique Rating:2

@& `d stands for capital Delta.

For a derivative you would use just plain d. *@

********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Before the upgrade: .36 * 2.5E14 = 9E13 After the upgrade: 36% + 3.32% = 39.32% = .3932 .3932*2.5E14 = 9.83E13 so the difference is 8.3E12 as an increase. Not sure of the heat transfer to the environment but thinking about it, I would guess it would be the same. You are not increasing the energy input, your increasing the energy output through an upgrade making it more efficient, not working harder. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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. Your Self-Critique: forgot to label my energy in J. I think that I am thinking about it correcty. Your Self-Critique Rating:3

********************************************* 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 .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: (550C + 273)K = 823K, (20C + 273)K = 293K 823K - 293K = 530K/823K = .644 * 100 = 64% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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. Your Self-Critique: It takes a while of starring at the word problems for me to figure out what is actually being asked. When I look at the solution section, I can see how I did the math right but labeling it doesn’t come easy for me. Your Self-Critique Rating:2

"

Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

"

Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!

`gr51

Query 9

#$&*

course Phy 122

2/14 1755hrsI believe that this completes module one Sir. My OIC will give me the test next week, when he has some time. I am curious to see how I will do. Thanks for squaring me away earlier.

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:

580 + 273.15 = 853.15K

380 + 273.15 = 653.15K

853.15K - 653.15K = 200K / 853.15K = .2344 * 100 = 23%

This means that the maxe = 23%

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: I guess that K conversion for the purposes of this book is 273, and not what I learned a ways back of 273.15. Either way, the answer turned out he same.

Your Self-Critique Rating:3

@&

273.15 is good, but the 2- or 3-significant figure information given does not justify the last two significant figures of that quantity.

Had the temperatures been given as 380.38 and 579.93, for example, you would have wanted to use the 273.15.

*@

*********************************************

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

(700C + 273)K = 973K

(27C + 273)K = 310K

973K - 310K = 663K/973K = .681(efficiency is always less than 1) * 100 = 68% = Carnot e

Table 15-4 really doesn’t give me any defining information so I conducted the problem similarly to the one above and the example in the book.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution: 1

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%.

Your Self-Critique: my answer was close so I think that I am doing this correctly.

Your Self-Critique Rating:2

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

The heat transfer to the environment came from the system, so it must be put back.

10kj + 8.5kj = 18.5kj

With 10kj of work, the efficiency would be the work / by the input:

e = 10kj/18.5kj = .541*100 = 54%

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique:3

Your Self-Critique Rating: nailed it.

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

It looks the equations above, just done in reverse:

22% = .22e

.22 * 6 * 10^9 = 1.32 * 10E9

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique: I am thinking that the ‘d = the derivative.

Your Self-Critique Rating:2

@&

`d stands for capital Delta.

For a derivative you would use just plain d.

*@

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Before the upgrade:

.36 * 2.5E14 = 9E13

After the upgrade:

36% + 3.32% = 39.32% = .3932

.3932*2.5E14 = 9.83E13 so the difference is 8.3E12 as an increase. Not sure of the heat transfer to the environment but thinking about it, I would guess it would be the same. You are not increasing the energy input, your increasing the energy output through an upgrade making it more efficient, not working harder.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique: forgot to label my energy in J. I think that I am thinking about it correcty.

Your Self-Critique Rating:3

*********************************************

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 .)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

(550C + 273)K = 823K, (20C + 273)K = 293K

823K - 293K = 530K/823K = .644 * 100 = 64%

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

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.

Your Self-Critique: It takes a while of starring at the word problems for me to figure out what is actually being asked. When I look at the solution section, I can see how I did the math right but labeling it doesn’t come easy for me.

Your Self-Critique Rating:2

"

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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