Capacity from flow in gallons and given DT

This is a question from an Canadian Exam

A water chiller operates with a 4.4°C (40°F) evaporator temperature, return water temperature of 10°C (50°F) and a 7.2°C (45°F) supply water temperature. The water flow rate is measured at 682.5 L/min (150 imp. gal./min). What is the cooling capacity?

A. 87 000 W / 300 000 Btu/h
B. 109 875 W / 375 000 Btu/h
C. 131 850 W / 450 000 Btu/h
D. 153 825 W / 525 000 Btu/h

I calculated via SI units: Q = (682.5 x 60) x 1 x 2.8 x1.16 (from cal to W) and gave in C which matches the best my answer (133 kW)
All the units in the question in SI are correct converted from the initial IP units. I checked this.

But my answer is apparently incorrect and they say that B is correct, 109 875 W / 375 000 Btu/h.

Their explanation.
Q= m.c.DT in Btu = 500 x 150 x 5 = 375.000 btu/h according to this link
https://greenenergypro.wordpress.com...water-formula/
and other links I found.

If I use on-line conversion tools to convert BTU/h to kW, then I get +/* 454.000 Btu/h

I think i Know where the pitfall is in this one.

Any help is appreciated.

Thanks

Peter

It's been over 20 years since I had to struggle with calculating stuff like this.

I figured I'd give it a try, because reading the data that was given made it seem like a poorly designed system :D

Anyway, I did all the calculations, and came up with the same answer as you did no matter how I turned the formula.

The only way to get 109Kw is to reduce the ∆T to 2.3°C



In the end I went online, and even the online calculators confirmed the findings

Attachment 15325

I Know they put "pitfalls" in assignments like this like you said, but I can't see it, and hope I didn't just become a living proof of your quote :D

Flow: 682,5/60 = 11,4 liter/s water.
This is +/- 11,4 kg/s (kcan be more precise if we use correct density, assuming 1 now)
Cooling from 7,2 down to 4,4 = 2,8ºC.
Water llooses energy : 11,4 kg/s x 4,18.10³ J/(ºCkg) x 2,8 ºC = 133,4 . 10³ J/s = 133 400 W

The question is in Imperial gallons = 4.56 l but u US gallon = 3.7854 l.
If you do the math again with 3.7854 ln, then you get 109 kW.

But,... the phormula widely used is
Q= m.c.DT
and in BTU

Btu = 500 x 150 x 5 = 375.000 btu/h

https://greenenergypro.wordpress.com...water-formula/
Q= 500 x flow rate (in gallons but which gallons???) x DT

But... for this phormula, you have to use US gallons Gallons and not Imperial gallons... I guess!!??

But BTU stands for British Thermal Unit in ...I thought ...imperial gallons...
Confusing

Tycho, what is the name of the software program you used for this?

Quote:

---

Originally Posted by Peter_1

....

But,... the phormula widely used is
Q= m.c.DT
and in BTU

Btu = 500 x 150 x 5 = 375.000 btu/h

https://greenenergypro.wordpress.com...water-formula/
Q= 500 x flow rate (in gallons but which gallons???) x DT

....

---

This phormula is only valid if you use for GPM the US gallons.

Quote:

---

Originally Posted by Peter_1

Tycho, what is the name of the software program you used for this?

---

Hi Peter,

It's from this webpage https://www.lairdthermal.com/thermal...ing-calculator

it's nice to have for quick reference when on site, haven't found any apps I'm satisfied with yet.

Quote:

---

Originally Posted by Peter_1

Flow: 682,5/60 = 11,4 liter/s water.
This is +/- 11,4 kg/s (kcan be more precise if we use correct density, assuming 1 now)
Cooling from 7,2 down to 4,4 = 2,8ºC.
Water llooses energy : 11,4 kg/s x 4,18.10³ J/(ºCkg) x 2,8 ºC = 133,4 . 10³ J/s = 133 400 W

The question is in Imperial gallons = 4.56 l but u US gallon = 3.7854 l.
If you do the math again with 3.7854 ln, then you get 109 kW.

But,... the phormula widely used is
Q= m.c.DT
and in BTU

Btu = 500 x 150 x 5 = 375.000 btu/h

https://greenenergypro.wordpress.com...water-formula/
Q= 500 x flow rate (in gallons but which gallons???) x DT

But... for this phormula, you have to use US gallons Gallons and not Imperial gallons... I guess!!??

But BTU stands for British Thermal Unit in ...I thought ...imperial gallons...
Confusing

---

If the question was asked with 150 imperial gallons, then 133Kw is the correct answer.

Apparently the guy who checked the answers used US gallons and for him the answer became 109Kw...

Seems like the guy who made the exam mixed up his gallons.

I ran the numbers through our in house engineer and he also got 133 Kw

so at least I'm not stupid :D

Quote:

---

return water temperature of 10°C (50°F) and a 7.2°C (45°F) supply water temperature.

---

Quote:

---

Cooling from 7,2 down to 4,4 = 2,8ºC.

---

From the first statement, I understood that the inlet temperature of the chiller was 10C and the outlet was 7.2C

When calculating the cooling capacity of a chiller, you need the flow + in and outlet temperature of whatever is being cooled. the evaporating temperature is irrelevant, other than telling you if the chiller is efficient or ineficient.

The chillers we use has a ∆T between evaporating temp and outlet temp of 4-5K anything higher than that and you will know that the compressor is sucking more gas than the chiller can provide, and it can tell you a few things:

-poor heat transfer on the pipes (oil film on the refrigerant side or fouling on the water side)
-mismatch between chiller and compressor (usually never happens)
-Low flow on water side
-in case of a 2-pass chiller, divider plate between in/out-let may be corroded and short the waterflow
-refrigerant pump problem, blocked liquid line filter (on spray chillers)
-Low refrigerant charge

Quote:

---

Originally Posted by Tycho

From the first statement, I understood that the inlet temperature of the chiller was 10C and the outlet was 7.2C

When calculating the cooling capacity of a chiller, you need the flow + in and outlet temperature of whatever is being cooled. the evaporating temperature is irrelevant, other than telling you if the chiller is efficient or ineficient.

The chillers we use has a ∆T between evaporating temp and outlet temp of 4-5K anything higher than that and you will know that the compressor is sucking more gas than the chiller can provide, and it can tell you a few things:

-poor heat transfer on the pipes (oil film on the refrigerant side or fouling on the water side)
-mismatch between chiller and compressor (usually never happens)
-Low flow on water side
-in case of a 2-pass chiller, divider plate between in/out-let may be corroded and short the waterflow
-refrigerant pump problem, blocked liquid line filter (on spray chillers)
-Low refrigerant charge

---

Agree for 100 % Tycho... Thanks for the link

BTU is british thermal unit but america seems to use it for cooling capacities but given its a canadian exam why are they using us gallons? One unseen pitfall for europeans working there.

The us gallon used to catch us out when ordering oil from the states

Quote:

---

Originally Posted by al

BTU is british thermal unit but america seems to use it for cooling capacities but given its a canadian exam why are they using us gallons? One unseen pitfall for europeans working there.

The us gallon used to catch us out when ordering oil from the states

---

You misread, the censors used IMPERIAL gallons in the question. and then used US gallons when they calculated the answer.

Using IMPERIAL gallons, the answer is 133 Kw (455234 BTU)
Using US Gallons, the answer is 109 Kw (378472 BTU)


The problem is that the sensor has used two different meassures of flow in the task given.
He has used IMPERIAL gallons in the question, and used US gallons in his answer...


the way I see it

Totally agree with your vision Tycho...

The sooner they get rid of imperial and just use metric the better, too confusing.

I worked it out another way but now I’m doubting if my way is correct,

(682.5/60) x 4.19 (specific heat capacity of water not allowing a glycol correction factor) x 2.8 (DT)

Which gives us 133kw