Vaporization of Methanol

[mjpetrag]

Hi,

I am trying to do a calculation to find how much length of pipe is needed to completely vaporize methanol from room temperature to ~110 F.

The flow rate of methanol is 5.27E-5 kg/s
Methanol's heat of vaporization is 1100 kJ/kg

So... Qevap = (mass flow) * (heat of vap) = 5.27E-5 * 1100 kJ/kg ~ 58 W

The pipe supplying the methanol will be coming from a 1/16" diameter stainless pipe at room temperature and into an oil bath at 110 F (43 C).

I keep coming up with very high answers for the length of pipe I need (~ 18 m)

Am I doing something wrong? Can someone help me out?


Thanks,

Mike

-Mike

 

[IRstuff]

How did you determine the required length?

TTFN

FAQ731-376

 

[MortenA]

its also a very low flow MeOH stream less than 0.2 kg/l

Best regards

Morten

 

[chicopee]

Since the methanol is evaporating as it travel thru the tubing, the ambient temperature which is not discussed by you may require that much tube length for methanol to reach saturated vapor state.

 

[25362]

Apart from the above queries and remarks, I noted that:

[•] BP of methanol at atmospheric pressure is 64.5^oC.
[•] Enthalpy change from liquid at 25^oC to vapor at normal BP is 1208 kJ/kg.

 

[mjpetrag]

sorry, I meant the temperature of the heating oil bath is 110 degrees celsius, not fahrenheit.

-Mike

 

[IRstuff]

And, again, how did you determine the length required?

TTFN

FAQ731-376

 

[mjpetrag]

I assumed constant wall temperature...and I'm not sure if I used the right equation since the temperature changes.

I think I have to use an equation with the log mean temperature difference, but I used...

Q = K A (T2-T1)
58 W = (25 W/m^2) * (Pi * d * L) * (110-25)

I solved for L.

-Mike

 

[IRstuff]

Where did you get the 25 W/m^2-ºC?

TTFN

FAQ731-376

 

[25362]

Recheck your figures. Although I disagree with the values used, and if I didn't err, your figures result in a length of about 5.5 m.