Refrigeration and Air Conditioning Problems Solution Part -2

May 31, 2016 | By | Reply More

In this article we will be solving a general refrigeration and air conditioning problems solution.  We will also discuss about some common questions and answers. Please follow the link for viva questions of refrigeration and air conditioning.

Refrigeration and Air Conditioning Solved Problem 01 

Given data,

The refrigeration cycle is an ideal vapor compression cycle.

Refrigerant = R134a.

Cooling water flows through the water jacket to cool the condenser has a temp rise of 10 degree Celsius.

Pressure at evaporator side Pev = 140 kPa

Pressure at compression side Pc= 1400 kPa

Mass flow rate of the water jacket = 200 kg/s

Latent heat of melting for ice = 333 KJ/kg

To Do

  • Draw the T-s diagram and the hardware of the whole setup.
  • Determine the flow rate of the refrigerant and potable water.

Solution:

At first we will draw the hardware for all the working fluids. In this case working fluid are refrigerant, water jacket water and the water to be cooled. In the question it is mentioned that the system operates on ideal vapor compression cycle. So we will have the simple vapor compression setup for solving the problem. We will be needing R134a Property table to solve the problem. ( It is available in internet, just google it )

Hardware of the whole set-up 

For more Visit: T-S and P-h diagram with subcooling and superheating. 

ideal vapor compression cycle

T-s Diagram of the ideal vapor compression cycle setup 

ideal vapor compression cycle T-s diagram

 

In the question we need to solve the equations to get the refrigerant flow rate and the potable water flow rate.

From the diagram it is clear that heat absorbed by the water jacket water is equal to the heat supplied by the condenser. And this time the temperature rise is 10 k  or degree Celsius. So let’s assume heat of condenser and the water jacket are respectively Q cond and Q water Jacket. So,

Q cond = Q water Jacket

Mrefrigerant(h2-h1) = Mwater jacket Cw Δt  ———1

And

Qpotable water = Q ev

Mpotable water x  Lf = Mrefrigerant(h1-h4)———–2

We know

Lf = 333 kJ/kg

Mwater jacket = 200 kg/s

Cw = 4.18 KJ/Kg.K

Δt = 10 K

Now we have to use the tables to get the values of enthalpies h1, h2, h3, h4.

Now for state – 1

Pressure = 140 kPa and the condition of the refrigerant R134a is saturated vapor enthalpy

h1= 239.16 KJ/kg

and entropy s1 = 0.94456 KJ/Kg.K

For state -2

S1= S2

Pc = 1400 kPa

So by interpolation h2 = 287.38 KJ/kg (for superheated refrigerant)

State – 3

Pressure = 1400 kPa  for saturated liquid h3= 127.22 KJ/kg

And h3 = h4

Now substituting the values of h1,h2 and h 4 in the equations 1 and 2 we get

mass flow rate of the refrigerant ,Mrefrigerant =  52.27 Kg/s  (Answer)

Mpotable water   = 17.73 Kg/s (Answer)

I hope you understand how the units are canceled out and the flow rate comes into Kg/s . If you have any problems understanding the problem feel free to contact. Please contact if you don’t have the refrigerant properties table.

Category: Fluidics, Heat Transfer, Mechanical, Q & A, Refrigeration

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