1. The advanced gas cooled reactors use steel cans to hold the nuclear fuel which is in ceramic form. They can use a maximum temp of 625 degrees Celsius. What is the maximum theoretical efficiency of such a reactor?

2. a car engine raises the temp of the fuel to about 600 degrees Celsius and expels the combustion products at about 80 degrees Celsius. Calculate the efficiency of the engine if it achieves 70% of the maximum theoretical efficiency of a 'perfect' heat engine working between these temperatures.

3. many different energy units are used in industry. Two are the kilowatt hour (k W h) and the tonne-of-coal-equivalent (tce), where 1 tce = 3 x 10^8 J.
explain how many joules are equivalent to 1 k W h.
give the name of 2 other energy units.
why is it important to standardise on a common unit?

4. in the 19th century, coal was the principle fuel used to provide domestic hot water. Sugest how, in a sunny climate, solar power might be used to provide hot water for a small isolated community of about 20 people. Answer should include simple diagrams, calculations for which you may have to make estimates as well as make use of the following data:

solar flux at earth's surface at midday = 1 k W m-2
specific heat capacity of water = 4kJ kg-1 K-1
average volume of hot water needed per day per person = 10 litres

state any assumtions made in arriving at your estimates.

5. the following figures were gven by a central heating consultant to a customer who required a central heating system capable of supplying 25 kW.

supplied cost/kWh efficiency cost of heat used/kWh cost/h

gas 1.5p 77%

Night storage 3p 100%
electricity

Normal price 7p 100%
electricity

oil 2.4p 68%


a) suggest the meaning of effieciency in this case.

b) find the cost of each kWh of heat and the cost of runnig each system for an hour.

1. the advanced gas cooled reactors use steel cans to hold the nuclear fuel which is in ceramic form. They can use a maximum temp of 625 degrees Celsius. What is the maximum theoretical efficiency of such a reactor?


625 C = 898 K

1-\frac {T_{low}}{T_{high}} = 1- \frac {?}{898}

the efficiency depends on the T(low) you pick. If you pick absolute 0, then the efficiency is 1 (100%). If you pick any other temperature, the efficiency will be less than 1. The reactors are "gas cooled", so that implies something around room temperature (273 Kelvins = 0 Celsius). Try that, or you can pick room temperature.



2. a car engine raises the temp of the fuel to about 600 degrees Celsius and expels the combustion products at about 80 degrees Celsius. Calculate the efficiency of the engine if it achieves 70% of the maximum theoretical efficiency of a 'perfect' heat engine working between these temperatures.


600C = 873K. 80C = 353K

Efficiency = 1-\frac {T_{low}}{T_{high}} = 1- \frac {353}{873}



3. many different energy units are used in industry. Two are the kilowatt hour (kWh) and the tonne-of-coal-equivalent (tce), where 1 tce = 3 x 10^8 J.
explain how many joules are equivalent to 1 kWh.
give the name of 2 other energy units.
why is it important to standardize on a common unit?


You can look this up or calculate it if you have enough information. This is trivial.



4. in the 19th century, coal was the principle fuel used to provide domestic hot water. Sugest how, in a sunny climate, solar power might be used to provide hot water for a small isolated community of about 20 people. Answer should include simple diagrams, calculations for which you may have to make estimates as well as make use of the following data:

solar flux at earth's surface at midday = 1 k W m-2
specific heat capacity of water = 4kJ kg-1 K-1
average volume of hot water needed per day per person = 10 litres

state any assumptions made in arriving at your estimates.

[QUOTE]
4. in the 19th century, coal was the principle fuel used to provide domestic hot water. Suggest how, in a sunny climate, solar power might be used to provide hot water for a small isolated community of about 20 people. Answer should include simple diagrams, calculations for which you may have to make estimates as well as make use of the following data:

solar flux at earth's surface at midday = 1 k W m-2
specific heat capacity of water = 4kJ kg-1 K-1
average volume of hot water needed per day per person = 10 litres

state any assumptions made in arriving at your estimates.


Looks like a project.



5. the following figures were given by a central heating consultant to a customer who required a central heating system capable of supplying 25 kW.

supplied cost/kWh efficiency cost of heat used/kWh cost/h

gas 1.5p 77%

Night storage 3p 100%
electricity

Normal price 7p 100%
electricity

oil 2.4p 68%


a) suggest the meaning of efficiency in this case.

b) find the cost of each kWh of heat and the cost of running each system for an hour.


a. efficiency, in this case, probably means "most economical".
b. I think this is something you can figure out. It's easy.

625 C = 898 K

1-\frac {T_{low}}{T_{high}} = 1- \frac {?}{898}

the efficiency depends on the T(low) you pick. If you pick absolute 0, then the efficiency is 1 (100%). If you pick any other temperature, the efficiency will be less than 1. The reactors are "gas cooled", so that implies something around room temperature (273 Kelvins = 0 Celsius). Try that, or you can pick room temperature.



600C = 873K. 80C = 353K

Efficiency = 1-\frac {T_{low}}{T_{high}} = 1- \frac {353}{873}



You can look this up or calculate it if you have enough information. This is trivial.





Looks like a project.



a. efficiency, in this case, probably means "most economical".
b. I think this is something you can figure out. It's easy.

i am not sure how to find out the cost of each kW h of heat used or the cost of runnig each system for an hour. Can you tell me please.

Basically, you look up the cost of heat in the table you gave me:



supplied cost/kWh efficiency cost of heat used/kWh cost/h

gas 1.5p 77%

Night storage 3p 100%
electricity

Normal price 7p 100%
electricity

oil 2.4p 68%


You know the system requires 25 KW.

25\,KW \,\times \, \frac {cost}{1\,KWH} \,\times\, number\,of\,hours = \, cost

Of course, that assumes 100% efficiency. You can divide the cost by the efficiency (as a fraction less than 1) to get the actual cost. You can also do this by dividing 25 KW by the efficiency to get the actual number of KW (input) to provide the 25 KW (output).

Basically, you look up the cost of heat in the table you gave me:



You know the system requires 25 KW.

25\,KW \,\times \, \frac {cost}{1\,KWH} \,\times\, number\,of\,hours = \, cost

Of course, that assumes 100% efficiency. You can divide the cost by the efficiency (as a fraction less than 1) to get the actual cost. You can also do this by dividing 25 KW by the efficiency to get the actual number of KW (input) to provide the 25 KW (output).

basically the supplied cost is given in pences... im not sure if i have to use this and the efficiency in order to work it out... is this right

Yes. That's right.