Energy Balance on heat exchanger

[thelorax]

I need to do an energy balance on a heat exchanger where an aziotropic mixture of 90% methanol and 10% water at 21°C is being heated to 261°C. The heat source is high pressure saturated steam at 31 bar and it is assumed that the steam leaves as condensate at the same pressure (31bar).

Calculating the steam required at the end is relatively straightforward using m = Q/ΔHv, as the steam merely goes from saturated vapour to saturated liquid.

However, calculating the actual heat required to raise the temperature of the methanol feed solution from 21°C to 261°C at 2.2 bar is where I get lost.

In this calculation we have to take into account the amount of energy gained m*Cp*dT and the amount of energy required to vaporize the water m*ΔHv. Which isn't too hard, except I don't know which value of Cp to use. Normally I would use Cp = 4.18kJ/kg K and calculate Q = m*(4.18)*(261-21) + m*ΔHv but this Cp value would only apply for water at 21°C, not steam at 261°C, which would have a different Cp.

The other option would be to use Q = m*∫CpdT + m*ΔHv, except I don't have an equation for Cp in terms of T.

I guess my question is:

What value(s) of Cp do I use when calculating the the total energy required to heat a mixture of 90% Methanol/10%water solution from 21°C to 261°C?

There is probably a very straightforward explanation for this and I would very much appreciate it if someone could please explain it to me.

Thank you for your time.

TheLorax

 

[anorlunda]

You didn't mention the big factor, the heat of vaporization of the steam.

At 31 Bar, the boiling point of water is 236C, and the heat of vaporization is 1785 kJ/kg