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Vaporization of Methanol

Vaporization of Methanol

Vaporization of Methanol

(OP)
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

RE: Vaporization of Methanol

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

Best regards

Morten

RE: Vaporization of Methanol

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.

RE: Vaporization of Methanol


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

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

RE: Vaporization of Methanol

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

-Mike

RE: Vaporization of Methanol

(OP)
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

RE: Vaporization of Methanol


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.

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