Refrigerant - mass flow, power, COP

[sci0x]

Homework Statement

Brewery operates closed circuit vapour compression refrigeration system using ammonia as refrigerant. The process heat to be removed at the evaporator is 120 kW. The vapour enters the compressor at 300 kPa with 5 deg C superheat. It is compressed isentropically to pressure of 1100 kPa, condensed at constant pressure and sub-cooled by 3 deg C. Finally it is expanded at constant enthalpy to 300 kPa at which pressure evaporation occurs. Using the pressure-enthalpy diagram calculate:

Calculate the required mass flow rate of refrigerant

Relevant Equations

M = Q/Cp(T2-T1)

This is a past exam Q, I don't have the attached pressure enthalpy diagram.

Q = 120kW
Cp = Refrigerant enthalpy = I've looked this up, for ammonia enthalpy = 35.06 kJ / mol
T2 = 5 degrees C = (5 deg C) (1100 kPa/300 kPa) = 18.33 deg C +273.15 K = 291.48 K
T1 = 2 degrees C = (2 deg C)(1100 kPa/300 kPa) = 7.33 deg C + 273.15 = 280.48 K

Mass Flow = Q / Cp (T2-T1)
= 120 kW / (35.06 kj / mol)(291.48 K - 280.15 K)
= 0.30 kg /s

Am I on the right track here
The expected answer is 0.115 kg/s

 

[sci0x]

Im after looking up that the enthalpy will change depending on temp and pressure:
Ive used this calc: https://www.peacesoftware.de/einigewerte/nh3_e.html

So at 300 kPa at 5 degrees, the enthalpy is 1495.25 kj/kg
At 1100 kPa at 2 deg C, the enthalpy is 209.74 kj/kg
Enthalpy change is 1285.51 kj/kg

Can you tell me what values you get from an Ammonia pressure-enthalpy chart
https://theengineeringmindset.com/r717-ammonia-pressure-enthalpy-chart/5

So mass flow is:
= 120 kW / (1285.51 kj / mol)(291.48 K - 280.15 K)
= 8.23 x 10^-3 kg/s

 

[Chestermiller]

Your equation for the mass flow rate is wrong. First of all, it is /kg, not /mol. Secondly, why are you using the temperature difference in the denominator? If you evaluated the enthalpies correctly from the diagram or tables, then you should get the right answer.

 

[sci0x]

Okay hopefully correct formula Q = m(h1 - h2)

Using this pressure enthalpy chart for Ammonia https://www.google.com/url?sa=t&sou...Vaw15UV5yyscEbE7v40SY8cv9&cshid=1588279401068

Q= 2kW

300kPa is 0.3 bar - at 5 degrees superheat
Now do i draw the along from 0.3 bar until i come to the 5 degree curve coming down in the superheated area - in which case i get 516.65 kj/kg

1100kPa is 1.1 bar I've drawn this over until i hit the black line (which comes approx the T=30) point. And we're at 2 degrees after compressing and sub-cooling.
Now can i draw a line straight down which would be 499.99 kj/kg ? Or do i draw a line up from the 2 degrees to meet the 1100 kPa and draw the line down to get the enthalpy.

 

[sci0x]

This is after giving me more insight.


Ive to draw the pressure bars 1.1 and 0.3 bar across to the temperature and use superheating and subcooling temperatures to calc h1, h2, h3, h4 like he does in the video.

 

[Chestermiller]

The only provided data that is relevant to determining the evaporator heat load is the inplet condition to the evaporator, 300 kPa=0.3 MPa with 3 degrees sub cooling, and the outlet condition, 300 kPa = 0.3 MPa with 5 degrees superheating. What is the equilibrium temperature at 300 kPa? What is the inlet temperature and pressure? What is the outlet temperature and pressure? What is the inlet enthalpy? What is the outlet enthalpy?

 

[sci0x]

What is the equilibrium temperature at 300 kPa?
Drawing line straight across from 0.3MPa gives me -16 deg C approx

What is the inlet temperature and pressure?
5 degrees superheat and 0.3Mpa
You asked me the equilibrium temp so I am guessing 5 deg superheat means inlet temp is -11 deg C

What is the outlet temperature and pressure?
0.3 Mpa and 3 deg subcooling
So temp is -19 deg C

What is the inlet enthalpy?
At superheated right side of bell curve following -19 deg curve down gives me 500 kj/kg enthalpy

What is the outlet enthalpy?
At subcooled liquid left side of graph drawing a line from -11 deg over to left side of bell curve and dropping gives me -800 kj/kg

120÷(500-(-800)) = 0.0923 kg/s

Ans should be 0.115 kg/s

If I've gone completely wrong here id appreciate if you can show me how to get 0.115 kg/s

 

[Chestermiller]

I agree with your answer.

 

[sci0x]

Okay great, thanks.

The next Q is to calc the electrical power required by the compressor set if it is 80% effecient

Pressure and temp before compressor = 0.3 Mpa & 5 degrees superheat
Enthalpy before compressor = 500 kj/kg

Pressure and temp after compressor = 1.1Mpa & 5 deg superheat
Enthalpy after compressor = 615 kj/kg

Compressor power = Enthalpy at compression x flow rate x effeciency
(615 kj/kg - 500 kj/kg) x 0.0923 kg/s = 10.61 kW
80% effecient so need 13.26 kW

Would you agree with this?

 

[Chestermiller]

Okay great, thanks.

The next Q is to calc the electrical power required by the compressor set if it is 80% effecient

Pressure and temp before compressor = 0.3 Mpa & 5 degrees superheat
Enthalpy before compressor = 500 kj/kg

Pressure and temp after compressor = 1.1Mpa & 5 deg superheat
Enthalpy after compressor = 615 kj/kg

Compressor power = Enthalpy at compression x flow rate x effeciency
(615 kj/kg - 500 kj/kg) x 0.0923 kg/s = 10.61 kW
80% effecient so need 13.26 kW

Would you agree with this?

Your condition coming out of the compressor is incorrect. You are supposed to start at the inlet condition and follow an isentrope to 1.1 MPa (and whatever temperature this corresponds to). Do you know how to follow a constant entropy line?

 

[sci0x]

Yes i get you, the constant entropy lines are the lines going up to the right.

So starting at inlet compressor following this line up to where it intersects with the outlet compressor has enthalpy of 700 kj/kg approx.

(700kj/kg - 500 kj/kg) x 0.0923 kg/s = 18.46 kW
80% effecient so need 23.075 kW

And finally; calculate the coeffecient of performance (COP)

= heat removed at evaporator / compressor power

= 120kW / 23.075 kW = 5.2 kW

 

[Chestermiller]

Yes i get you, the constant entropy lines are the lines going up to the right.

So starting at inlet compressor following this line up to where it intersects with the outlet compressor has enthalpy of 700 kj/kg approx.

(700kj/kg - 500 kj/kg) x 0.0923 kg/s = 18.46 kW
80% effecient so need 23.075 kW

And finally; calculate the coeffecient of performance (COP)

= heat removed at evaporator / compressor power

= 120kW / 23.075 kW = 5.2 kW

I don't have the PH diagram any more, but it sounds like you did this correctly.