Given:
Temperature of water, =
=273 +(-6) =267 K
Temperature surrounding refrigerator, =
=273 + 21 =294 K
Specific heat given for water, = 4.19 KJ/kg/K
Specific heat given for ice, = 2.1 KJ/kg/K
Latent heat of fusion, = 335KJ/kg
Solution:
Coefficient of Performance (COP) for refrigerator is given by:
Max =
= = 9.89
Coefficient of Performance (COP) for heat pump is given by:
Max =
= 10.89
Answer:
If the heat engine operates for one hour:
a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.
b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.
In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.
Explanation:
The Carnot efficiency is obtained as:
Where is the atmospheric temperature and
is the maximum burn temperature.
For the case (B), the efficiency we will use is:
The work done by the engine can be calculated as:
where Hv is the heat value.
If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.
If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.