Co-current Heat Exchanger

  • #1

Homework Statement


A hydraulic fluid initially at 60◦C is cooled to 35◦C using a co-current heat exchanger. Cooling water is available at 15◦C, which must not exceed 30◦C on leaving the heat exchanger. The flow rate of the hydraulic fluid is 1.5 kg/s, which has a specific heat of 5.2 J/gํ C. Calculate the heat to be removed from the hydraulic fluid and the area required for the heat exchanger. Assume no fouling takes place. U = 520 W/m2°C

Homework Equations


Co-current heat exchanger: LMTD = {(Ti-ti)-(To-to)} / ln{(Ti-ti)/(To-to)}
Ti = temp of cooling / heating water in
ti = temp of liquid to be cooled / heated in

The Attempt at a Solution


I have no idea where to begin with this!
 

Answers and Replies

  • #2
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"Start at the beginning. Go on to the end. Then, stop." ___ Caterpillar to Alice(?)

You really do need to take a stab at it before we jump in.
1. (snip)
A hydraulic fluid initially at 60◦C is cooled to 35◦C ... (snip) The flow rate of the hydraulic fluid is 1.5 kg/s, which has a specific heat of 5.2 J/gํ C. Calculate the heat to be removed from the hydraulic fluid and the area required for the heat exchanger. Assume no fouling takes place. U = 520 W/m2°C
(snip)
 
  • #3
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From the information in the first two sentences, you can calculate the heat load of the heat exchanger. You know the flow rate of the fluid, its heat capacity, and its temperature change. So, what's the heat load?

Chet
 
  • #4
So the heat load, using the formula given would then be:
LMTD = {(Ti-ti)-(To-to)} / ln{(Ti-ti)/(To-to)}
= {(15-60)-(30-35)}/ln{(15-60)/(30-35)}
= {(-45)-(-5)} / ln(-45/-5)
= {(-40) / (2.197)}
= -18.20
So I know:
-flow rate is 1.5 kg/s
-specific heat is 5.2 J/g degree C
-LMTD is -18.20

Would I then put this into the Q = Qin - Qout = flowrate*heat_capacity*(Tout - Tin)
But substitute as Q= flowrate*heat capacity*LMTD?
 
  • #5
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You seem to be very confused. I am going to recommend that you go back to your text and re-read the chapter on this again.

The heat load is defined as the rate at which heat gets removed from the hot stream, and you correctly gave one equation for it as: Q = = flowrate*heat_capacity*(Tout - Tin)

But, there is another equation involving the heat transfer coefficient, the heat transfer area, and the LMTD. You need to write this equation down also, and then you have to set the two heat load relationships equal to one another.

Chet
 
  • #6
Okay I think I got this!
We use q=m*Cp*deltaT to find our heat that would be removed
thus: q = (1.5kg/s)*(5200 J/kgdegreesC)*(60-35)
q=195 000 J/s

then to get the area, we would use A= q / h(or U)*LMTD
so A= 195000 J/S / (520 W/m^2degreesC * 18.20)
A=20.60 m^2

Seems right now?
 
  • #7
21,488
4,864
Okay I think I got this!
We use q=m*Cp*deltaT to find our heat that would be removed
thus: q = (1.5kg/s)*(5200 J/kgdegreesC)*(60-35)
q=195 000 J/s

then to get the area, we would use A= q / h(or U)*LMTD
so A= 195000 J/S / (520 W/m^2degreesC * 18.20)
A=20.60 m^2

Seems right now?
Yes. Your methodology is correct now. I haven't checked your arithmetic.

Chet
 

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