steam and variation of t

[mir]

Hi.
I think I'm lost in a glass of water. Let's see.
I have saturated steam at 180 °C (19 bar) passing in a heat exchanger where it yields 420000 kcal.

how to calculate the output temperature? I know that qvapore = cl * mass ... means that 420000 kcal simply reduce one
mass = qvapore / cl = 420000 [Kcal] / 543 [kcal/kg] = 773.5 kg of steam condensate in water?

so if my flow rate is 773.5 kg/h of steam at the outlet of the heat exchanger I will have 773.5 kg/h of water at 99 °C.

right my reasoning?

Thank you.
♪

 

[luigi.paiano]

Hi.
I think I'm lost in a glass of water. Let's see.
I have saturated steam at 180 °C (19 bar) passing in a heat exchanger where it yields 420000 kcal.

how to calculate the output temperature? I know that qvapore = cl * mass ... means that 420000 kcal simply reduce one
mass = qvapore / cl = 420000 [Kcal] / 543 [kcal/kg] = 773.5 kg of steam condensate in water?

so if my flow rate is 773.5 kg/h of steam at the outlet of the heat exchanger I will have 773.5 kg/h of water at 99 °C.

right my reasoning?

Thank you.
♪

So, thermodynamics isn't very fresh in mind, but I'd like to challenge that 543 kcal/kg and that 19 bar pressure. that I know, steam is saturated at 180 degrees at the pressure of 10 bar and something. So he's probably overheated??? (and this changes the calculation as the heat must first saturate the steam and then there must be phase change). Moreover, having to deal with courses, more than the total heat would interest the heat power exchanged.

 

[mir]

I read the table wrong... it's 10 bar the pressure so it's just saturated steam.

substantially the steam passes into a serpentine tube immersed in 6000 lt of water and must vary the temperature from 50 to 85 °c of water in half an hour.

(tf-ti) = 4184 j/kg k * 12000 kg * (85-50) = 419799 kcal/h
I doubled the mass to get the necessary power in half an hour.

the cl I use found it here: http://it.wikipedia.org/wiki/calore_latente in the tabellina where there is latent heat of boiling.

 

[luigi.paiano]

I hope the pictures are clear. in practice all steam mass condenses at 180 degrees and then the remaining part of heat is used to cool water.