9 replies to this topic

[goggles30]

I'm designing a heat exchanger to heat methanol from 25C to 80C. Should I use hot water or saturated steam to heat the methanol? If I use saturated steam, my methanol flowrate would be 27kg/s and steam 2.4 kg/s. Do these values look ok?

[breizh]

Hi ,

If you have a doubt show your calculation .

energy balance :

methanol                steam

( m*cp *Delta T )  = M * Latent heat

 

Good luck,

 

Breizh

[goggles30]

Breizh:

 

Thanks for replying.  I don't really have a problem with the calculations.  But I was wondering if using steam is justifiable from an economic point of view, or is hot water sufficient.  I believe that using steam would reduce the heat exchanger size but increase the operating costs?

[Saml]

It all depends on the context.

- If the steam is excess low pressure steam from turbine exhausts and you are condensing it to recover the BFW, then steam may be a good optiion

- If you need to add fire to the steam boiler, then you have a cost that is (heat)/(steam boiler effciency) x Cost of your fuel.

- If your hot water is a process surplus and you are using cooling water to cool it down and reuse, the water may be a good option.

- If you have to install a hot water heater and circulation pumps that uses gas or electricity, then you are probably better with steam.

[goggles30]

Hello again. I'm planning to heat methanol at 45 bar from 25C to 80C using saturated steam. What pressure of steam should I consider?

[Pilesar]

To heat your process to 80C, the steam should have a condensing temperature above 80C. Practically, you will probably have utility steam in the plant maintained at a specific pressure used for heating. The available steam levels are set early in the plant design for a new plant. For an existing plant, you should use the steam levels that are already available. If you have a clean slate for an academic project, then you get to choose a reasonable low pressure steam utility on the order of 2.1 barg or 3.5 barg. Document the utility steam levels in your project write-up and assume this steam pressure level is available throughout the plant.

[srfish]

Good answer Pilesar.

[goggles30]

I have calculated my shell-side heat transfer coefficient and the number seems a bit too large (16,000W/m^2K)

Nu = hd/k = j*Re*Pr^(1/3)

 

These are my values

Hydraulic diameter, d=15.8mm

Heat transfer factor, j = 0.039

Reynolds number, Re=52651

Prandtl number, Pr=0.00053

Thermal conductivity of water, 0.6 W/mK

 

From what I have read online, the heat transfer coefficient is supposed to be around 1000-2000W/m^2K. Is there something wrong?

Edited by goggles30, 08 April 2017 - 01:15 AM.

[srfish]

That value of 16,000  for the heat transfer coefficient is a little high but possible. 

[Pilesar]

The overall heat transfer coefficient will be much smaller. Your calculation is only for the steam. Resistance is the inverse. As you recognize, the inverse of a large number is very small and not very significant. This is why you do not need to calculate a coefficient for condensing steam on the shell side and can just use a standard value such as 2000 BTU/hr without concern. You will also have resistance from the outside fouling, the tube wall, the inside fouling, and the inside coefficient to account for. The largest resistance in your case will be from the heat transfer coefficient of the hydrocarbon and will be the controlling factor for the overall resistance.