Heat Exchanger Efficiancy 1

[mkh012]

How do I calculate the efficiancy for crude heat echanger ( Shell-tube type):

I have the following data:
1- Tube fluid = Crude
2- Shell Fluid = Kero.
3- Inlet temp. ( tube ) = 204.9 Deg C.
4- Outlet Temp. (tube) = 216.5
5- Inlet Temp. (shell) = 325 Deg C
6- Outlet Temp. (shell) = 313.6 Deg C
7- Flow rate for crude = 190,000 bbl/Day

 

[TD2K]

How do you define efficiency for a heat exchanger? We need to know that to comment.

For me, if I was asked "how is that exchanger performing" I would normally model the heat exchanger using say HTRI to determine the predicted clean and service overall heat transfer coefficient and then compare it to my calculated heat transfer coefficient using the field data. One wrinkle to this are the fouling factors. Those are shown on the current design data sheet for the heat exchanger but are essentially a fudge factor based on experience. If your service doesn't foul as much as was originally allowed for, the exchanger can be working better than design but still worse than it was at an earlier time.

You can also use the service heat transfer coefficient from the data sheet in evaluating the heat exchanger's performance but changes in process flow rates or physical properties (flow rate, heat capacity, thermal conductivity and viscosity) will result affect the applicability.

If you calculate a heat transfer coefficient just after it's been cleaned, you'll get a different value than say 1 year later all things being equal.

So we are back to my question "How do you define efficiency for a heat exchanger?"

 

[25362]

It is usual to speak about the heat exchanger "effectiveness-NTU" method, defined as:

Effectiveness = actual heat transfer/maximum possible heat transfer

The maximum possible heat transfer Q_max = (m.c)_min (T_{h, inlet} – T_{c, inlet})

Where (m.c)_min refers to the fluid with minimum m.c.

m =mass flow rate
c = specific heat
T_h = temperature of the hot fluid
T_c = temperature of the cold fluid

See, for example, Heat Transfer by J.P. Holman, McGraw-Hill.

 

[mkh012]

Thank you for your help, but I am looking for Heat Exchanger in % unit not as Qmax

 

[mkh012]

I mean can I calculate the Heat Exchanger efficiancy in % ?

can I use the difference between the temperature in tube and shell side following equ. for calculating the Efficiancey :
Efficiency = (Tout-Tin)for shell /(Tout-Tin) for tube

 

[cloa]

mkh012
That's not right for two fluids.

See page 3

http://www.aip.org/tip/INPHFA/vol-2/iss-4/p18.pdf

 

[sheiko]

mkh012,

25362 has already given you the effectiveness as %. Read his post more carefully.

"We don't believe things because they are true, things are true because we believe them."

 

[TD2K]

mhk012, you can of course define "efficiency" as you want but it's not going to be a very useful parameter if you think about it. You'd also get negative efficiencies with your formula since one side is heating up so Tout - Tin is positive while the other side is being cooled so Tout - Tin is negative. However, that's not the problem with your suggestion how to define efficiency and I'll just use positive efficiencies from now on.

Let's look at your system. For your system, the tube side dT is 11.6 deg C. The shell side dT is 11.4. Crude and kerosene are going to have similar heat capacities at the same temperature so I expect your crude flow rate is about the same as the kerosene flow rate in this exchanger (conservation of energy says what one side gains the other side must give up ignorning heat losses to the surroundings). Using your suggestion, the efficiency is 98.3%.

Let's say the exchanger fouls so the crude isn't heated up as much. In fact, it only gains 1/2 the heat so the dT goes to 11.6/2 or 5.8 deg C. The kerosene side isn't cooled as much so the shell side dT is also about 1/2 or -5.7 deg C, this assumes the mass flow * Cp on one side is approximately equal to the other side which I estimate is the case based on your original data. Efficiency is still 98.3%. Would you say the exchanger is performing as well as previously when the heat transfer is 1/2?

How about a case where the dT on each side isn't nearly the same? Let's take your crude being heated from 200 to 210C tubeside by another hydrocarbon stream that is much smaller so it's cooled from 300C to 225C. Efficiency is 10 / 75 or 13.3%. Again, let's say the exchanger fouls so only 1/2 the heat is being exchanged. The crude is now heated from 200C to 205C. The other side is cooled from 300C to 262.5C. Efficiency is now (205 - 200) / (300 - 262.5) = 13.3%, no change. Does that mean the crude/kerosene exchanger is more than 8 times as efficient?

Think about the case where one side has no temperature change, condensing steam or boiling water, refrigerant, etc. dT on that side is zero so you efficiency is going to always be either 0 or infinite.

 

[mkh012]

TD2K.......thank you for your interest to help, there is a big difference between the flow rate of crude and kerosene, therfore,I did the calculations as per 25362 equation, please I need your feedback on the following results :

Heat Exchanger (E-3AB) in 80GD12
--------------------------------------------------

Hot Liq. = Kerosene
Cold Liq. = Crude

Tin hot = temperature of the hot fluid (Kerosene) (Inlet Shell) = 171.3 Deg C = 340.34 F
Tout hot = temperature of the hot fluid (Kerosene) (outlet Shell) = 118.3 Deg C = 244.94 F

Tin cold = temperature of the cold fluid ( Crude ) (Inlet tube) = 109.3 Deg C = 228.74 F
Tout cold = temperature of the cold fluid ( Crude ) (Inlet tube) = 121.3 Deg C = 250.34 F

Volume Flow rate Kerosene = 95.44 m3/h
Volume Flow rate Crude = 406.9 m3/h

1,000kg/m³ is the density of water (The m³ cancel out)

Sp Gr of Kerosene = .81715 @ 60 F
Sp Gr of Crude = .897 @ 60 F

M ( Hot ) = Mass Flow Rate Kerosene = 95.44 M3/hr* 1000 Kg/M3 * .81715 = 77988.8 Kg/hr
M ( Cold ) = Mass Flow Rate Crude = 406.9 M3/hr* 1000 Kg/M3 * .897 = 364989.3 Kg/hr


Note for Specific heat calculation
----------------------------------

Cp = {0.388 +0.00045 * Tout (F) }/(Sp Gr )^.5

S.G. (Crude)= .897
Tout (Crude) = 250.34 deg. F


Cp Crude = {.388 + (.00045 * 250.34)}/(.897)^.5
= .500653/(.897)^.5
= .500653/0.9471

= .528616

Cp Kerosene = {.388 + (0.00045 * 244.94)}/(.81715)^.5
= 0.498223 /0.90396
= 0.551154



(M * Cp) Kerosene (hot) = ( 77988 * 0.551154 ) = 42983.4
(M * Cp) Crude (cold) = ( 364989.3 * 0.528616 ) = 192939.2

(M * Cp) Min = 42983.4


Tin hot = temperature of the hot fluid (Kerosene) (Inlet Shell) = 171.3 Deg C = 340.34 F
Tout hot = temperature of the hot fluid (Kerosene) (outlet Shell) = 118.3 Deg C = 244.94 F

Tin cold = temperature of the cold fluid ( Crude ) (Inlet tube) = 109.3 Deg C = 228.74 F
Tout cold = temperature of the cold fluid ( Crude ) (Inlet tube) = 121.3 Deg C = 250.34 F




Eff = { (M * Cp)hot * (Tin - Tout)hot} / { (M * Cp)min * (Tin hot - Tout cold)} ....(1)
= { (M * Cp)cold * (Tout - Tin)cold} / { (M * Cp)min * (Tin hot - Tin cold)} ....(2)

Eff as equation (1)
-------------------------
= { ( 42983.4 ) * ( 340.34-244.94 ) } / { 42983.4 * ( 340.34 - 250.34 ) }
= (42983.4 * 95.4 ) / (42983.4 * 90.0 )
= 4100616.36 / 3868506
= 1.06

Eff as equation (2)
------------------


Eff = { 192939.2 * (250.34 - 228.74) } / { 42983.4 * ( 340.34 - 228.74 ) }
= ( 192939.2 * 21.6 ) / ( 92983.4 * 111.6 )

= 4167486.72 / 10376947.44
= 0.4016


Which one is correct ?
Is the Cp calculations are right ?
How I can get the result in % ?

 

[siretb]

Hi

Read carefully what 25362 writes.

Your Cps must be about right, the only check I performed what that mCp(cold) detaT cold and mCp(hot) deltat(hot) agrees

With hot side 42983*(340.34-244.94)=4,100,000 heat duty I am unsure about the units, it's irrelevant.
With cold side 192939*(250.34-228.74)=4,170,000
about the same OK
Qmax (NTU-Eff) = 42983*(340.34-228.74)= 4,800,000

Eff= 0.86 --> 86%

 

[MJCronin]

Heat Exchanger Efficiency is defined also in this paper:

http://www.gewater.com/pdf/Technical Papers_Cust/Americas/English/TP1010EN.pdf

 

[mkh012]

Thank you Siretb,
one more question , is the following equation is the right equation to calculate the Specific heat (Cp) for any oil product ?

Cp = {0.388 +0.00045 * T (F) }/(Sp Gr )^.5

If it is the right equation to calculate Cp which is part of calculating the heat exchanger efficiency , which temperature I should use , T(inlet) or T(outlet) ?

 

[siretb]

I have no idea about the validity of your equation. I am not in the petrochemical business.
If you deal a lot with heat exchangers, having good physical properties is critical.

Use average btween inlet and average.
B SIRET

 

[25362]

All those formulas are rough estimations of specific heats. I found that for lubes, taken from Engineering Tribology by Stachowiak and Bachelor, Cp expressed in kJ kg^-1K^-1, is given by the following formula:

C = (1.63 + 0.0034[×]T)/s^0.5​

Where:

T is the temperature of interest in ^oC
s = sp. gravity at 13.6^oC

This formula gives about a 2% difference with the one submitted by mkh012, at 100^oC.