# Homework Help: Heat balance and steam

1. Mar 4, 2014

### maarjaaur

Hello,

Im new here, sorry for my english, but im in desperate need of help regarding my own experiment which i need to finish. I need to calculate steam kg/kg (flash steam maybe in english?). Basically i have entering steam temperature and pressure and exiting steam temp. and pressure from evaporator? and i need to find ... S/W... kg/kg

1. The problem statement, all variables and given/known data
I need to find basically 4 different answers all with different entering steam/water variables.
1) entering temp 100 Celsius,1atm. exiting 100C 2atm.
2)entering temp 15C, 1atm. exiting 100C 2 atm.
3)entering 0,8atm 100C. exiting 100C 2atm.
4) 15c 0,8 atm. exiting 100C 2atm.

2. Relevant equations
I have tried according to these calculations but something doesnt add up.
I take info from a table where i get kJ for water at different pressure but i dont understand how does temperature change the calculations. And i get minus answer for the first variable...is it ok?

3. The attempt at a solution
example at 1. variable
I take the formule given above and fill the blanks with steam table here: http://www.engineeringtoolbox.com/saturated-steam-properties-d_101.html
then i get minus answer and i think its false, i dont know where to calculate the temperature i have only taken the pressure and kJ-s from the table into account according to that formula. ( 417,5-504,7 / 2201,59 = - 0,0396 kg/kg )
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 5, 2014

### BvU

Hello Maar, and welcome to PF. Your essay is indeed a little difficult to read.
Let's start with 1). Going from 1 atm = 101.325 kPa to 2 atm is a compression, not so much a flash. Is that really intended ? In a flash you decrease the pressure and then it makes sense to calculate how much vapour you get per kilogram water coming in.
This repeats in the other three, so I thought I'd better ask.

3. Mar 5, 2014

### maarjaaur

Hey,

Probably yeah, you are right and the entering/exiting temp and pressure is in reverse. Will edit in main post.

4. Mar 5, 2014

### maarjaaur

Ok, cant change so... please see entering/exiting variables in reverse. I myself also noted that it is a bit unlogical that way.

5. Mar 5, 2014

### BvU

OK, so we flash 1 kg of water at 100 C 2 x 101.33 kPa to 1 kg of water liquid + water vapour at 100 C, 101.33 kPa and want to know the mass of the water vapour.
Relevant equations: mass conservation and enthalpy conservation (assuming the flash is adiabatic: no energy is added or removed).

mass conservation: $m_{\rm water \ in} = m_{\rm water\ out} + m_{\rm steam \ out}$

enthalpy balance: $m_{\rm water\ in} H_{\rm water\ in} = m_{\rm water\ out}H_{\rm water\ out} + m_{\rm steam\ out}H_{\rm\ steam \ out}$

Per kilogram of water in, defining x = $m_{\rm steam \ out}/ m_{\rm water \ in}$ :

$H_{\rm water\ in} =(1-x) H_{\rm water\ out} + x H_{\rm\ steam \ out} \ \Leftrightarrow$

$H_{\rm water\ in}- H_{\rm water\ out} = x ( H_{\rm\ steam \ out}- H_{\rm water\ out}) \ \Leftrightarrow$

$$x = { H_{\rm water\ in} - H_{\rm water\ out} \over H_{\rm\ steam \ out}- H_{\rm water\ out}}$$