refrigeration capacity math

[neteffect]

I was wondering if anyone could help me with a math problem. I tried it but my answer doesn't seem right? Here's the Question. What is the capacity in tonnes of refrigeration of a system in which 1200 kg of the refrigerant are circulated per hour? The enthalpy of the vaporized refrigerant leaving the evaporator is 181 kj/kg and the enthalpy of the liquid refrigerant entering the expansion valve is 64.66 kj/kg.
Here's how i did it. The refrigerating effect is calculated by subtracting enthapy of liquid refrigerant enetering the expansion valve from the enthapy of the vaporized refrigerant leaving the evaporator. Thus i have 181 kj/kg - 64.66 kj/kg = 116.34 kj/kg( refrigerating effect)
I now need to calculate the heat of fusion. I get
1200 kg of refrigerant x 334 kj(which is the amount of heat required to melt 1 kg of a substance, of ice)= 400800 kj per 24 hrs or 400800 divided by 24 = 16700 kj/hr. I then divided further 16700 kj/hr by 60 to get 278.33 or 278 kj/min.
Therfore to get Tons of refrigeration i divided 116.34 by 278 and the answer i get is 0.418 tonnes.
For some reason this doesn't seem right. Could someone assist me with this math.
Thanks
Happy New Year

[jacob perkins]

Originally posted by neteffect
I then divided further 16700 kj/hr by 60 to get 278.33 or 278 kj/min.
Therfore to get Tons of refrigeration i divided 116.34 by 278 and the answer i get is 0.418 tonnes.
Thanks
Happy New Year

I think you should have stopped there.(It is tons per 24 hrs,right?)

Where are you from?

[neteffect]

Thanks for answering. You are right. After i reread the question , i can see that i went to far. I was going by an example in the information provided to me.( bad excuse). You were wondering where i was from. It's Canada.
Thanks again.

[CityHvac]

Originally posted by neteffect
Thanks for answering. You are right. After i reread the question , i can see that i went to far. I was going by an example in the information provided to me.( bad excuse). You were wondering where i was from. It's Canada.
Thanks again.

Net.....may I give you some advice from another Canadian..FILL OUT YOUR PROFILE. There's a ton of info here for the asking...the guys here just like to know who they're talking to.

[neteffect]

Right you are. I have updated my profile. Now back to this math question. According to the sample info provided to me, it shows in their final math equation, they have multiplied their refrigerating effect times the number of times that their sample refrigerant circulates( there's is in minutes).( This is how they got the answer for capacity of refrigeration in a system). And then have divided the refrigerating effect by tons of refrigeration. Their refrigerant used is ammonia. After figuring out the refrigerating effect in my question , (the refrigerant used is R-12). Should I be multiplying my refrigerants number of times it circulates. I do not know how many times it circulates nor can i find it anywhere in the course material. The answer i get now according to the numbers i have are .007 tonnes. i'm still not sure if this is right or not? If anyone could set me straight on this one it would be appreciated. It's probably simple, but i can't seem to wrap my head around it.

[jacob perkins]

Originally posted by neteffect
Thanks for answering. You are right. After i reread the question , i can see that i went to far. I was going by an example in the information provided to me.( bad excuse). You were wondering where i was from. It's Canada.
Thanks again.

Hold on there,Sasquactch!
You really confused me with the metrics,so I was just guessing.



Btu/h divided by 12000=tons of rerigeration.Thats what I can remember.The reason why that works is long forgotten-I guess I am getting old.

Good question,I have to look that up in the book.

And a happy New Year to you too.(Even if it is metric!)

[westcoast refer man]

Hey neteffect,

I've looked everywhere for a conversion factor (kj/kg to btu/lb)to help you out, but have had little success. The basic formula for your question is easy, but I can't work in the units you have.

You said the refrigerant is R-12. If you can tell me the liquid temperature and the evaporating temperature (degrees C or F is okay), I'd be happy to show you how to calc it out. Or, you can do it yourself.

The basic formula is the flow rate (lbs/hr) x net refrigeration effect (btu/lb) and divide by 12000 btu/hr-ton. Is it a coincindence that you screen name is part of the formula for your question?

[neteffect]

Thanks guys. Sorry i haven't replied sooner, Had to work you know. Feed the family and all that. Lots to think about in this question. I believe i'm going to use the same number that the sample question has used when they multiply by the number of times the refrigerant circulates (3) even though their refrigerant is r717. I'm going to assume this is what they are looking for. When i get the results back i'll let you know if it was right or not. Yes my screen name was derived from a question i had earlier on refrigerating effect. thanks again.

[mattm]

Here is one of the best conversion sites available. You can convert just about anything on here. http://www.onlineconversion.com/

[jacob perkins]

Originally posted by jacob perkins

Good question,I have to look that up in the book.



[/B]

Ton of refer=288,000 btu/24 hours

Or 12000btu/1 hour

Good luck.

[clydemule]

Given:
h = enthalpy
h1 = 181 kj/kg
h2 = 64 kj/kg
1 kW = 1 kilojoule / sec
mass flow "em dot" = 1200 kg/hour
dh = 181-64 = 117 kj/kg
1 kw = 3412 BTU/HR
12,000 BTU/HR = 1 ton

Required:

Refrigerating Capacity

Equations:

Q = mdot * dh

Solution

Q = 1200 kg/hour * 117 kj/kg

***WE CANCEL OUT kg***

Q = 140400 kj/hour

***CANCEL HOURS***

Q = 140400 kj/hour * 1 hour/3600 sec = 39 kj/sec

*** 1 kw = 1 kj/sec***

Therefore Q = 39 kw.

***CONVERT TO I-P (normal) UNITS***

Q = 39 kw * 3412 BTU/KW = 133,000 BTU/HR

Q = 133,000 BTU/HR / 12,000 BTU/HR/TON = 11 TONS


Generally it is easier to just work with the metric units, and get an answer in the same form as what you are looking for. The units may be unfamiliar, but screwing with conversions in the middle of the problem invites problems.

In addition, it really is no harder to visualize the flow of XYZ kj/kg more than BTU/lb, or kg/hr over lb/hr. How many of us can really visualize mass flows anyway?

Clyde

[thehumid1]

Originally posted by clydemule
Given:
h = enthalpy
h1 = 181 kj/kg
h2 = 64 kj/kg
1 kW = 1 kilojoule / sec
mass flow "em dot" = 1200 kg/hour
dh = 181-64 = 117 kj/kg
1 kw = 3412 BTU/HR
12,000 BTU/HR = 1 ton

Required:

Refrigerating Capacity

Equations:

Q = mdot * dh

Solution

Q = 1200 kg/hour * 117 kj/kg

***WE CANCEL OUT kg***

Q = 140400 kj/hour

***CANCEL HOURS***

Q = 140400 kj/hour * 1 hour/3600 sec = 39 kj/sec

*** 1 kw = 1 kj/sec***

Therefore Q = 39 kw.

***CONVERT TO I-P (normal) UNITS***

Q = 39 kw * 3412 BTU/KW = 133,000 BTU/HR

Q = 133,000 BTU/HR / 12,000 BTU/HR/TON = 11 TONS


Generally it is easier to just work with the metric units, and get an answer in the same form as what you are looking for. The units may be unfamiliar, but screwing with conversions in the middle of the problem invites problems.

In addition, it really is no harder to visualize the flow of XYZ kj/kg more than BTU/lb, or kg/hr over lb/hr. How many of us can really visualize mass flows anyway?

Clyde

You just make me feel inferior...I think i followed you as far as H=enthalpy ...Do u have a website ?

or just a thread in the wall of pride with pics of your equipment?

Either one can put the link up here I want to bookmark in case one day I "need to bring the big guns" .

[icemeister]

I'm glad you got that straight. I can remember doing stuff like that until it was coming out of my ears back forty years ago.....on a slide rule!!!

The trick is to keep track of your units (ie BTU/HR, lbs/min, etc). My thermodynamics professor back at Wentworth Institute in Boston stressed that if you can get the units to come out right, you likely have done it correctly.

I just still don't have a clue what a kilojoule is.