Heat tranfer formula 2

[kear]

I have to calculate the approximate heat tranfer in KW for a tube heat exchanger in KW.(for a hydraulic power pack)
I have the formula for calculating it in BTU's
Q= change in temp x thermal duty value8 GPM

The formula i thought should work was
Q= change in temp x specific heat capacity x flow rate

the medium is hydraulic oil

 

[ProfSporlan]

Either works, depending on your approach. With the former equation, and using I-P units, your thermal duty value would be in units of Btu/gal-°F, given q is in Btu/min

The latter equation would be more traditional in determining overall heat transfer. Hydraulic oil should have a specific heat of around 0.49 Btu/lb-°F. Calculating heat transfer then becomes a simple matter.

 

[kear]

then i must be missing something.
When I use a example thus.
Flow rate of 20GPM
Change in temp 20deg F
Thermal duty value of 500 (water)

change in temp x thermal duty x GPM

20 x 500 x 20 = 20000 btu's = 5.85Kw

using the other formula and using SI units

20gpm = 75.7 litres/min
20 deg F = 12deg C
500 btu's/gal-deg F = ???

This is really confusing I am from the UK where we normally use SI unit's. I just can't get the 2 equations to equal the same....

 

[Whoops]

Q = m x c x dT

m is the mass flow rate (kg/sec)
c is the specific heat capacity (4.184 kJ/kg degC for water)
dT is the change in temperature
Q is the heat transferred (kJ/sec or kW)

m = 75.7/60 kg/sec = 1.1333kg/sec
dT = 12 degC

Q = (1.1333 x 4.184 x 12) kW
= 56.9 kW

 

[kear]

exactly,
The one calculation ends up with 5.85KW and the later ends up with 56.9KW, what am i missing....

 

[quark]

Kear!

Thermal duty is the amount of heat transfered per unit time and in your equation it appears on the LHS and is the parameter which is to be calculated.

You should follow the calculation given by Whoops. And in the case of a heat exchanger compare it with UxAxLMTD.

Perhaps your two equations, you specified in your first post are same except that units are not matched.

 

[Whoops]

Correction to previous post
75.7 litres/min = 75.7/60 litres/sec = 1.26166 kg/sec

Q = (1.26166 x 4.184 x12) kW = 63.34 kW

Trying the imperial units (I too am from the UK, so correct me if I go wrong here)

1 btu will raise 1 lb of water 1 degF, I think

1.26166 kg/sec = 2.7815 lb/sec

Q = (2.7815 x 20) = 55.63 btu/sec = 200,266 btu/hr

1 kW = 3412 Btu/hr

Q = (200,266 / 3,412) kW = 58.69 kW.

7% error from the previous result. I'm not familiar with the thermal duty concept, but it's obviously a conversion factor.

 

[ProfSporlan]

specific heat of water is 1 btu/lb-°F

1 gal = 8.34 lb water (39°F)

Thermal duty for water as defined originally should then be:

8.34 Btu/gal-°F, not 500

Then: q = 20 x 8.34 x 20 = 3336 Btu/min = 56.7 kW

Close enough!

 

[ProfSporlan]

The 500 value for thermal duty for water is valid if we multiply 8.34 by 60. Therefore, units for thermal duty become: Btu-min/gal-hr-°F.

Therefore, q becomes Btu/hr instead of Btu/min

 

[kear]

spot on, thanks guys,
Wish we could all use one set of units....

 

[smb168]

Hello Kear,

there is an error in your calculation:
20 x 500 x 20 = 200000 btu's NOT 20000 btu's. that is why you only get 5.85kw.

 

[Zeroman]

WOW