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in Physics by (20 points)

Let’s assume that during isolation at home due to COVID-19, you are going to the kitchen to do an experiment and measure level of our standing of thermodynamic by considering a real invention of human mind. Our fridge is empty, and we need it to cool down a bottle of water which we take it from tap at 20 oC. It is enjoyable we will drink our water at 5 oC. We are keen to know how the air temperature in the kitchen is under the influence of quantity of water that we are to put in our fridge. Our kitchen is a complete cubic with the dimensions 5 m×6 m×4 m. The density of air in the kitchen is constant (ρ=1.14 kg/m3 ,cp=1.001 kJ/kg.K). 

For our purpose, we open the engine of the fridge and realise that it has ammonia as a working fluid with the following diagram for the engine: 

If the ammonia at point 1 and point 3 is saturated vapor and saturated liquid at -20 and 40 oC. To find our objective which is to show the trend of room temperature against the mass of water in the fridge, first you need assume specific mass for bottle of water in the fridge (let say; 5 10 15 20 kg) (Cp 4186 J/kg). Then, you may need to follow the below procedure

1) Find the conditions of points 1, 2 ,3 and 4 using steam tables for ammonia.

2) Draw a graph representing the trend of room temperature against the mass of water in the fridge.

3) Calculate the entropy generation in the ammonia cycle as a system and room as an environment. For this part, do the calculations for every mass of water in the fridge and assume the room temperature is constant at whatever it should be.

4) Draw P-V, V-T and S-T diagram for ammonia using steam tables. Afterwards, show the position of point 1-4 in the diagram and discuss how the mass of water in the fridge could influence the condition of ammonia in the engine of fridge.

5) From the trend you have obtained, discuss how the performance of fridge in under the influence of water. For how much water in the fridge, you expect that fridge stops working. For this part, you should consider the temperature of condenser and evaporator constant at 40 and -20 oC and use the fact the compressor is not able to compress any liquid meaning that the output of the condenser could be exceptionally saturated liquid at 40 oC.

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