Heat exchanger overall heat balance check

[knapee]

Dear
The BEM type heat exchanger utility flow data (hot oil) is changed by licenser,and I have to decide which utility flow quantity data (old or new) for the overall heat balance is correct.
OLD data
shell side (hot oil)
flow quantity : 126000 (in-out all liquid)
density: in=725 kg/m3, out =755 kg/m3
specific heat:in=0.48kcal/kgC, out =0.47
temperature: in=250, out=230
pressure drop =1 kg/cm2
duty = 2214000 kcal/hr
tube side (chlorosilane)
flow quanity : in = 560000 kg/hr (liquid) ,out=110000 (vapor)
specific heat in=0.32, out=0.24 (vapor) 0.33 (liquid)
latent heat = 18 kcal/kg
pressure drop = 1kg/cm2
temperature : in=210, out=210
MTD = 24C
NEW data
Only shell side oil flow mass increase to 155000 kg/hr
heat duty is the same.
I used Q(hot) = m*cp*delta T =
Q(cold) = m * latent heat
to check the hot side / cold side energy balance.
But, it seems the calculation by hand can’t get a good result, and our poor EPC company already used the old hot oil flow quantity data to order the hot oil pump ,but now we can’t get any reason about the hot oil mass change from licenser, and owner hope EPC to reorder the new hot oil pump to satisfy our licenser’s new hot oil quantity data. Strangely, the hot oil quantity increases to 2 tons, but the shell side pressure drop / heat duty still keep the same value in the new case.
Could any master help me out in this terrible situation?
Thank you very much.

 

[25362]

Kindly recheck the temperatures, or the flow rate, on the hot oil shell side; it appears there is a discrepancy on the estimated heat duty of 2214000 kcal/h.

 

[knapee]

Sorry.

The duty is 2213000 kcal/hr.

And, the second question is
if the case is the vapor condensed to liquid and vapor, do we use the mass of liquid to multiply the latent heat or the mass of vapor multiply the latent heat, then add the mass of liquid times the CP and delta T ?


Thank you.

 

[knapee]

Dear
I would like to check duty balance for the heat exchanger shell & tube side.
(please see the attached for my calculation). But, I confuse that

case 1) When vapor condense to liquid, the duty balance is used M(liquid)*latent heat or M(vapor)*latent heat ?

case 2) there is no latent heat provided in the case. Can I just use M(liquid)*Cp*delta T + M(vapor)*Cp*delta T
for the tube side duty?

Thank you very much for your generous help.

 

[25362]

Still, from the given data, the heat lost by the hot oil assuming flow rates are in kg/h, is:

126,000[(0.48)(250)-(0.47)(230)] [≠] 2,213,000 kcal/h​

Kindly revise your data.

 

[knapee]

I am sorry to bother you all this trivial. I put my question clear.

Ex. hydrocarbon (30kg at 30C) boil to vapor (10kg at 110C) and liquid(20kg at 110C)

I guesss the following is right ?

The delta H = Mass(liquid 20kg) * Cp *(110-30) + Mass(vapor 10kg) * latent heat.


Cause 10kg mass of liquid boil to 10kg mass of vapor so the 10kg mass of this liquid will

go through this phase change of latent heat release.


Please give me some reply if I am wrong.


Thank you so much.

 

[25362]

1. Why don't you refer to my question on the heat duty calculated on the thermal oil side? Your "old" data doesn't seem to be right.

2. Regarding the example in your last posting, the heat duty on the vaporizing fluid would be ~ OK assuming:

a. The C_P of the liquid is about constant; otherwise the sensible heat gain should be 20 (C_p,110[×]110-C_p,30[×]30). Use O.33 and 0.32, as given.

b. The latent heat gain would be ~ OK provided it is taken at the prevailing pressure.

c. The heated fluid doesn't undergo a chemical reaction; for example, an endothermic bond breaking.(What type of chlorosilane or chlorosilane solution is it?)

Be careful and follow instructions for safe handling of chlorosilanes.

 

[knapee]

thank you for your kindly reply
sorry not answer your first question

and the flow rate should be 126600 instead of 12600

Indeed the old data, 1266,000[(0.48)(250-230)]=1804050 < 2,213,000 kcal/h

,and licenser give new oil flow rate is 1556600kg/hr
and heat balance become 1566,000[(0.48)(250-230)]=2255040 > 221300 kcal/hr

Still don't make sense to me.

 

[25362]

Still, both equations don't suit, because of assuming a constant C_P = 0.48, which, to my grasping, is erroneous.

Old: 1,266,000[(0.48)(250-230)]=1,215,360 not 1,804,050 as reported

New: 1,566,000[(0.48)(250-230)]=1,503,360 not 2,205,040 as reported

The original heating duty: 110,000 [&times;] 18 = 1,980,000

C_P values for thermal fluids at the reported temperatures generally differ by more than 0.01 kcal/(kg.K)

Now, using a slightly modified thermal oil C_P value:

1,566,000 [(250[&times;]0.483)-230[&times;]0.47)] = 1,980,000 (as needed)​

Meaning that the discrepancies stem, probably, from rounding off the thermal fluid C_P values. If so, it would be logical to verify the thermophysical properties of the thermal fluid.

 

[knapee]

Very Sorry again

the oil flow rate(old data) =126600 kg/hr
and the flow rate (new data)=156600kg/hr.
The temperature is from 250 to 220 C.
I am blamed for common sense error that this duty in balance result, and the duty for utility side should be larger than process side.