Heat:
Heat is the amount of energy flowing from one body to another due to the difference in their temperatures.
Energy:
Energy is the capacity to do work. It is an extensive property. It includes the kinetic energy and potential energies. For a closed system, the energy transfer takes place across system boundary by means of only work and heat.
Refrigerator:
It is a device used to transfer the heat from low temperature medium to high temperature medium.
First law of thermodynamics:
It states that energy cannot be created nor destroyed but can be changed from one form to another. Thus, energy is conserved. For an energy imbalance in a system, the energy change inside the system is difference between energy entering into the system and the energy leaving the system.
Energy balance neglecting the kinetic and potential energies is given as
The schematic diagram of a refrigeration system is shown in figure (1).
Coefficient of performance for a refrigerator (COP):
The efficiency of a refrigerator is indicated as the coefficient of performance . It is given by
Here, heat removed from refrigerated space is and work input to the system is .
From the first law of thermodynamics, the energy conversion in refrigeration system is expressed as
…… (1)
Here, the heat rejected to warm environment is .
The relation between temperatures and heat transfers in a refrigeration system can be expressed as
…… (2)
Here, the temperature at warm environment is and the temperature at the refrigerated space is .
From the relations (1) and (2), the COP of refrigerator can be expressed in different forms as:
Calculate the maximum possible coefficient of performance between given temperatures.
Substitute for and for .
(a)
Calculate the new COP for given conditions.
Substitute for and for .
Compare the and .
(b)
Calculate the new COP for given conditions.
Substitute for and for .
Compare the and .
(c)
Calculate the new COP for given conditions.
Substitute equation (1) for .
Substitute for and for .
Compare the and .
(d)
Compare the and .
Substitute 6 for and 6.875 for .
Ans: Part aThe cycle is impossible.
5.43 A refrigeration cycle operating between two reservoirs receives energy QC from a cold reservoir TC = 275 K and re...
A power cycle operating between two thermal reservoirs receives energy QH by heat transfer from a hot reservoir at TH = 2000 K and rejects energy QC by heat transfer to a cold reservoir at TC = 400 K. For each of the following cases determine whether the cycle operates reversibly, operates irreversibly, or is impossible. (a) QH = 1000 kJ, ƞ = 60% (b) QH = 1000 kJ, Wcycle = 850 kJ (c) QH = 1000 kJ, QC =...
As shown in the figure, a reversible power cycle receives energy QH by heat transfer from a hot reservoir at TH and rejects energy QC by heat transfer to a cold reservoir at TC. a) If TH = 1600 K, TC = 400 K, what is the thermal efficiency? b) If TH = 500oC, TC = 20oC, and Wcycle = 1000 kJ, what are QH and QC, each in kJ? c) If ? = 60% and TC = 40oF, what...
A power cycle receives energy QH from a high temperature energy source at TH and rejects energy QL by heat transfer to a low temperature energy sink at TL = 400 K. For each of the following cases determine whether the cycle operate reversibly, irreversibly, or is impossible. (a) QH = 1200 kJ, Wcycle = 1020 kJ. (b) QH = 1200 kJ, QL = 240 kJ. (c) Wcycle = 1400 kJ, QL = 600 kJ. (d) ?=40%.
Problem 1: Two reversible refrigeration cycles are arranged in series. The first cycle receives energy by heat transfer from a cold reservoir at temperature Tc and rejects energy by heat transfer to a reservoir at an intermediate temperature T greater than Te. The second cycle receives energy by heat transfer from the reservoir at temperature T and rejects energy by heat transfer to a higher-temperature reservoir at TH. Obtain an expression for the coefficient of performance of a single reversible...
At steady-state, a refrigeration cycle operating between hot and cold reservoirs at 300 K and 275 K, respectively, removes energy by heat transfer from the cold reservoir at a rate of 600 kW. (a) If the coefficient of performance is 4, determine the power input required, in kW (b) Determine the minimum theoretical power required for any such cycle.
At steady-state, a refrigeration cycle operating between hot and cold reservoirs at 300 K and 275 K, respectively, removes energy by heat transfer from the cold reservoir at a rate of 600 kW. (a) If the coefficient of performance is 4, determine the power input required, in kW (b) Determine the minimum theoretical power required for any such cycle.
4.A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. If the coefficient of performance is 3.3 and TH = 75°F, determine TC, in °F. 5.A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. If TC = -26°C and TH = 40°C, determine the coefficient of performance.
Problem 5.040 SI At steady state, a refrigeration cycle operating between hot and cold reservoirs at 300 K and 275 K, respectively, removes energy by heat transfer from the cold reservoir at a rate of 100 kw. (a) If the cycle's coefficient of performance is 4, determine the power input required, in kW. (b) Determine the minimum theoretical power required, in kW, for any such cycle.
A power cycle operates between hot and cold reservoirs at 500 K and 310 K, respectively. At steady state, the cycle rejects energy by heat transfer to the cold reservoir at a rate of 16 MW. Determine the maxi- mum theoretical power that might be developed by such a cycle, in MW.
Data are provided for two reversible refrigeration cycles. One cycle operates between hot and cold reservoirs at 27°C and 15°C, respectively. The other cycle operates between the same hot reservoir at 27°C and a cold reservoir at -20°C Weycle, If each refrigerator removes the same amount of energy by heat transfer from its cold reservoir, determineeyclc the ratio of the net work input values of the two cycles. Weycle,2 Weycle,1