Calculating amount of steam required
Calculating amount of steam required
(OP)
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.
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.





RE: Calculating amount of steam required
RE: Calculating amount of steam required
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*cp*(T2 - T1)
Where does the heat come from? (steam)
latent heat of condensing 30 psig steam to supply duty required.
RE: Calculating amount of steam required
RE: Calculating amount of steam required
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.
RE: Calculating amount of steam required
RE: Calculating amount of steam required
HAZOP at www.curryhydrocarbons.ca
RE: Calculating amount of steam required
RE: Calculating amount of steam required
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?
RE: Calculating amount of steam required
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. ms30 = 749,700/928.7 = 807 lb.
5. ms150 = 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.
RE: Calculating amount of steam required
Assuming steam is saturated in both cases, and if steam and water are mixed, and the mix reaches a final temperature of 160oF, the heat supplied by steam should be increased by the sensible heat from its saturation condensation temperature down to 160oF.
In that case, the 150 psig-steam would supply a bit more heat per pound.