P110/120: Energy/Energy & Technology Fall 2015
Homework 03
P110/120 Homework assignment HW03
1. Can you cool a kitchen by leaving the refrigerator door open? Why or why not?
No. Leaving the door open would cause the refrigerator to run more. When a refrigerator runs, more heat will be put out of the metal coils. The coolness from the inside will be temporary. More heat will be put off as it runs.
2. Why is a condenser needed in a steam electric plant? Why isn’t it better to just recycle the low-pressure steam rather than condense it, releasing heat to the environment?
The condenser kind of closes the loop. It is better to condense than recycle the low pressure steam because more would get lost in the air.
3. If 100 Btu of heat energy …show more content…
If only conductive heat loss was significant for a house, then by what percentage would you lower the heat loss if the temperature were reduced inside from 70F to 66F when the outside temperature is a chilly 20F?
Q= UA change T
70 degrees F-20 degrees F = 50 degrees F
66 degrees F- 20degrees F= 46 degrees F
Difference/initial x 100
50-46/ 50= 0.08 x 100= 8% less heat loss
8. A simple heat engine might make use of the warm air around New York city. Energy could be taken as heat from the atmosphere (assume 30C) and rejected as heat to the Hudson River (10C). What is the maximum efficiency of such an engine for the conversion of thermal into mechanical energy?
Efficiency= 1-(Tc /Th)
1- (273 +10 degrees C)/ 273 +30 degrees C)
1- 0.933 = 0.066 x 100
= 66% efficiency
9. What is the efficiency of an engine that has 3000 J of heat added during combustion and loses 2200 J of heat in the exhaust?
Efficiency= work out put/ total energy input x 100
3000J – 2200 J = 800 J
800J/3000 J= 0.26 x 100= 26.6 % efficiency
10. Would you be willing to financially back an inventor who is marketing a device that they claim takes in 25,000 J of heat at 600 K, expels heat at 300 K, and does 12,000 J of work?
Homework 03
P110/120 Homework assignment HW03
1. Can you cool a kitchen by leaving the refrigerator door open? Why or why not?
No. Leaving the door open would cause the refrigerator to run more. When a refrigerator runs, more heat will be put out of the metal coils. The coolness from the inside will be temporary. More heat will be put off as it runs.
2. Why is a condenser needed in a steam electric plant? Why isn’t it better to just recycle the low-pressure steam rather than condense it, releasing heat to the environment?
The condenser kind of closes the loop. It is better to condense than recycle the low pressure steam because more would get lost in the air.
3. If 100 Btu of heat energy …show more content…
If only conductive heat loss was significant for a house, then by what percentage would you lower the heat loss if the temperature were reduced inside from 70F to 66F when the outside temperature is a chilly 20F?
Q= UA change T
70 degrees F-20 degrees F = 50 degrees F
66 degrees F- 20degrees F= 46 degrees F
Difference/initial x 100
50-46/ 50= 0.08 x 100= 8% less heat loss
8. A simple heat engine might make use of the warm air around New York city. Energy could be taken as heat from the atmosphere (assume 30C) and rejected as heat to the Hudson River (10C). What is the maximum efficiency of such an engine for the conversion of thermal into mechanical energy?
Efficiency= 1-(Tc /Th)
1- (273 +10 degrees C)/ 273 +30 degrees C)
1- 0.933 = 0.066 x 100
= 66% efficiency
9. What is the efficiency of an engine that has 3000 J of heat added during combustion and loses 2200 J of heat in the exhaust?
Efficiency= work out put/ total energy input x 100
3000J – 2200 J = 800 J
800J/3000 J= 0.26 x 100= 26.6 % efficiency
10. Would you be willing to financially back an inventor who is marketing a device that they claim takes in 25,000 J of heat at 600 K, expels heat at 300 K, and does 12,000 J of work?