Suggested Response - Q2

*** You have a primary refrigeration process using ammonia (R717). The vaporisation temperature is -10°C, and the condensation temperature is 30°C. Saturated vapour leaves the evaporator. The cooling requirement for the brewery is 500 kW.

a)    Draw the key points on the Mollier diagram provided.

b)    For each stage of the OVC, describe the process the ammonia is undergoing, and the equipment involved.

c)    Calculate the COP for the system described.

d)    What would the COP be if the ammonia was superheated by 10°C leaving the evaporator, and subcooled by 5°C leaving the condenser?

e)    Describe the ways in which the COP can be improved.

f)     If the condensation temperature changes to 20°C and the vaporisation temperature changes to -5°C, what will the new COP be?

g)    Comment on the COP values you have calculated.

h)    Calculate the flow rate of refrigerant required for the initial conditions.

Figure: Mollier diagram for ammonia

Figure: Mollier diagram for ammonia

Response:

a)    The process is shown by the points ABCD.

 

b)    AB is the vaporiser. This process happens at constant temperature (isothermic) and constant pressure (isobaric).

BC is the compressor. Here, we have an increase in pressure and temperature. We assume this to be isentropic (constant entropy).

CD is the condenser. This happens at constant pressure (isobaric). The temperature decreases until all the superheat is removed. Thereafter, it occurs at constant temperature (isothermic).

DE is the expansion valve. The pressure and the temperature both drop. This happens at constant enthalpy (isenthalpic).

 

c)    To calculate the cooling load (Qe):

Q subscript e space equals space H subscript B space minus space H subscript A space equals space 1 comma 433 space minus space 323 space equals space 1 comma 110 space k J space k g to the power of negative 1 end exponent

To calculate the compressor work (Qw):

Q subscript w space equals space H subscript C space minus space H subscript B space equals space 1 comma 650 space minus space 1 comma 433 space equals space 2 comma 17 space k J space k g to the power of negative 1 end exponent

To then calculate the COP:

C O P space equals space Q subscript e over Q subscript w space equals space fraction numerator 1 comma 110 over denominator 217 end fraction space equals space 5.11

 

d)    The new cycle is shown by the points A1B1C1D1.

The cooling load (Qe) would be:

Q subscript e space equals space H subscript B 1 end subscript space minus space H subscript A 1 end subscript space equals space 1 comma 470 space minus space 300 space equals space 1 comma 160 space k J space k g to the power of negative 1 end exponent

The compressor work (Qw) would be:

Q subscript w space equals space H subscript C 1 end subscript space minus space H subscript B 1 end subscript space equals space 1 comma 687 space minus space 1 comma 470 space equals space 217 space k J space k g to the power of negative 1 end exponent

The COP would then be:

C O P space equals space Q subscript e over Q subscript w space equals space fraction numerator 1 comma 160 over denominator 217 end fraction space equals space 5.34

 

e)    COP can be improved by increasing the temperature of evaporation and decreasing the temperature of condensation.

 

f)    The new cycle is shown by the points A2B2C2D2.

The cooling load (Qe) would be:

Q subscript e space equals space H subscript B 2 end subscript space minus space H subscript A 2 end subscript space equals space 1 comma 420 space minus space 300 space equals space 1 comma 120 space k J space k g to the power of negative 1 end exponent

The compressor work (Qw) would be:

Q subscript w space equals space H subscript C 2 end subscript space minus space H subscript B 2 end subscript space equals space 1 comma 580 space minus space 1 comma 420 space equals space 160 space k J space k g to the power of negative 1 end exponent

The COP would then be:

C O P space equals space Q subscript e over Q subscript w space equals space fraction numerator 1 comma 120 over denominator 160 end fraction space equals space 7

 

g)    The second option is  more effective as it reduces the work done by the compressor. The first option only provides more cooling, but the compressor duty remains virtually unchanged.

 

h)    If the user requires 500 kW of cooling, which is in fact 500 kJ s-1.

The cooling available is:

capital delta h space equals space H subscript B space minus space H subscript A space equals space 1 comma 433 space minus space 323 space equals space 1 comma 110 space k J space k g to the power of negative 1 end exponent

Thus, to calculate the flow rate of refrigerant:

q space equals space m increment h

m space equals space fraction numerator Q over denominator increment h end fraction space equals space fraction numerator 500 over denominator 1 comma 110 end fraction space equals space 0.45 space k g space s to the power of negative 1 end exponent

m space equals space 0.45 space k g space s to the power of negative 1 end exponent space x space fraction numerator 3 comma 600 over denominator 1 comma 000 end fraction space equals space 1.62 space t o n n e s space p e r space h o u r

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