Calculating amount of steam required

[abamin]

I'm trying to determine the amount of steam required to heat a water tank on our plant. How many pounds of 30 psig steam are needed to heat 1000 gallons of water from 70F to 160F?
and how do you calculate this?

I think that it's: [(pounds of water) * (change in temp)] / [total heat of evap. - latent heat of evap]

but I'm not 100% so need some clarification. Thanks.

 

[TBP]

Pick-up a copy of Spirax Sarco's "Hook-Ups", and all will be revealed. If you suggest that you might buy something like a temperature control valve or steam trap, the local rep will likely just give you one.

 

[bchoate]

abamin

Your question is about rather elementary concepts. It would seem that you should be able to do this. I'll give you a hint.

What is the duty required? (heating up the water): sensible heat
Q = m*c[sub]p[/sub]*(T[sub]2[/sub] - T[sub]1[/sub])

Where does the heat come from? (steam)
latent heat of condensing 30 psig steam to supply duty required.

 

[abamin]

haha yes these are pretty simple concepts, I'm a junior in college right now completing my BS in ChE. I'm interning with a company right now and I've been assigned to calculate the amount of steam necessary for a system conversion. I'm currently taking thermodynamics and do understand the concepts but I'm getting a little confused on this calculation. I do know how to calculate the amount of condensate this is producing and such but I can't seem to wrap my head around how to use the sensible heat to calculate the pounds of a certain type of steam necessary. This is for direct steam injection (using a sparge) and i'm trying to compare benefits of 30# steam versus 150# steam.

 

[abamin]

So I'm fairly certain that this is the way to solve the problem but my answer has me questioning the method:

q = m * cp * dT

q = (25,000 lbs h2o/hr) * ( 1 BTU / # * F) * (160-70 F)
= 2,250,000 btu

mass of steam = 2,250,000 btu / latent heat of steam pressure
= mass of steam reqd.

 

[abamin]

tthe reason i'm questioning my answer is because I'm using less steam with 30# than I would be with 150# and I assumed I would have to use less 150# steam in this situation...should I be dividing by my sensible heat instead?

 

[owg]

You better hope the tank is not losing heat to atmosphere while you are heating the water. Might be a good idea to heat the tank as well as the water. Hint - for a vertical surface insulated with 1.5 inches magnesia the heat loss will be about 2 to 2.5 BTU/hr/sq ft/deg F . That's as long as the wind does not blow. TBP gave you a good hint.

HAZOP at

http://www.curryhydrocarbons.ca

 

[abamin]

The tank is well insulated, so for calculation purposes the heat loss through convection and radiation is negligible.

 

[LucayaLive]

Abamin,

You probably use less 30 PSIG Steam because the latent heat of vaporization of 30 PSI Steam is greater than that of 150 PSIG steam (check your steam tables). Heat of vaporization is only constant at constant temperature. That means condensing a pound of 30 PSI steam transfers more heat than a pound of 150 PSIG steam.

You are talking about saturated stema, right?

 

[bchoate]

abamin

Glad you were amused as well. The problem looked like a student doing homework to which we do not respond. Based on the question as initially posted, consider the following:

1. 1,000 gal. * 8.33 lb/gal = 8,330 lbs H2O
2. Q =m*Cp*delta T = 8,330 lb * 1 BTU/lb/F * 90 F =
749,700 btu
3. Latent heat of condensation is the difference in
enthalpy between the steam and the condensate. 928.7
btu/lb for 30 psig steam and 857.4 btu/lb for 150 psig
steam. (Ref. saturated steam tables)
4. m[sub]s[/sub][sup]30[/sup] = 749,700/928.7 = 807 lb.
5. m[sub]s[/sub][sup]150[/sup] = 749,700/857.4 = 874 lb.

The latent heat of condensation for 150 psig steam is smaller because the enthalpy of the condensate increases much more than the enthalpy of the steam as the steam pressure increases.

 

[25362]

Assuming steam is saturated in both cases, and if steam and water are mixed, and the mix reaches a final temperature of 160[sup]o[/sup]F, the heat supplied by steam should be increased by the sensible heat from its saturation condensation temperature down to 160[sup]o[/sup]F.

In that case, the 150 psig-steam would supply a bit more heat per pound.