# Steam produced heat transfer heat exchangers

1. Nov 15, 2014

### Mitch1

1. The problem statement, all variables and given/known data

Dry saturated steam at a temperature of 180oC is to be produced in a fire tube boiler from the cooling of 50 000 kg h–1 of flue gases from a pressurised combustion process. The gases enter the tubes of the boiler at 1600oC and leave at 200oC. The feed water is externally preheated to 180oC before entering the boiler.

The mean specific heat capacity of the flue gases is 1.15 kJ kg–1 K–1. The latent heat of vaporisation of the water at 180oC is 2015 kJ kg–1. Feed water temperature = 180oC.

Determine the amount of steam produced per hour, if the total heat loss is 10% of the heat available for steam raising.

2. Relevant equations
Mass steam= heat available / heat required to produce 1kg steam
Qmc=heat transfer rate/ Cpc(tc2-tc1)
Heat transfer rate=Qmh hfg
3. The attempt at a solution
Heat required would be 2257 from steam table at 100C evaporation
Heat available
Qmh hfg
50000*2015=10075x10^4 kW
Ht rate=10% of 10075x10^4 = 90675x10^3
Mass steam=heat available / heat req 1kg
Mass steam = 90675x10^3/2257
=40175 kg h-1
Can anyone check over this as I am unsure if it is correct
Thanks

2. Nov 15, 2014

### SteamKing

Staff Emeritus
Why? The feed water is being supplied already pre-heated to 180 C
The units here are not kW. 1 W = 1 J/s, and 1 hour ≠ 1 second

Show the units here and carry thru with the multiplication. That's how to avoid mistakes.
As mentioned above, the feed water is already pre-heated to 180 C when it enters the heat exchanger. The problem statement tells you that the latent heat of vaporization is 2015 kJ/kg for water at 180 C.

The enthalpy of steam at 100 C is irrelevant here.

3. Nov 15, 2014

### Mitch1

Ah so if I work out the correct units and carry it through I will divide it by 2015 which is at 180C ? Does the rest Seem ok?

4. Nov 15, 2014

### SteamKing

Staff Emeritus
The heat supplied by the flue gas seems to have been overlooked. The gas enters at a temp. of 1600 C and leaves at a temp. of 200 C. The specific heat capacity of the flue gas is 1.15 kJ/kg-K. You've got to work out how much heat is available to be transferred from the gas to the feed water to make steam.

5. Nov 15, 2014

### Mitch1

Hi again,
To find out this is this the spec heat*temp difference
I'm not that clued up on this subject
Once this is found is this part of the heat available

6. Nov 15, 2014

### SteamKing

Staff Emeritus
Yes. You are transferring the heat from the flue gas to the feed water to turn it into steam. Remember to include the 10% heat loss in your calculations of steam production.

7. Nov 15, 2014

### Mitch1

So it would be 1400*1.15*.9 for the ten percent heat loss which gives 1449 kW I'm assuming?
So is this the heat available? Then we divide it by 2015 the enthalpy at 180C ?
I believe I am missing a step out as this seems too low of a value

8. Nov 15, 2014

### SteamKing

Staff Emeritus
Again, you need to show and carry thru the units in these calculations. You are just assuming what units appear in the results.

You've also neglected to include the flow rate of the flue gases in your heat calculations, which is why the result appears so low.

For the heat available:

H = eff. * flow * cp * ΔT, which has units

H = kg/hr * kJ/kg-K * K

H = kJ/hr, which is not the same units as kW, so stop calling the heat rate kW

9. Nov 17, 2014

### Mitch1

So, (1600-200)*1.15*.9*50,000=72450000kj/hr
7245x10^4/2015=35955kg/hr ?

10. Nov 17, 2014

### SteamKing

Staff Emeritus
This looks better.

11. Nov 17, 2014

### Mitch1

Finally,

12. Oct 8, 2015

### lycee

On the same question: The overall heat transfer coefficient based on outside area of the tubes is given as 54 Wm K .Determine the area of heat transfer required to perform this duty

13. Oct 9, 2015

### insightful

Are these units correct?

14. Oct 9, 2015

### lycee

Should be Wm^-2 K^-1

15. Oct 9, 2015

### insightful

Good. Now, what's the equation for the heat transfer in a heat exchanger?

16. Oct 10, 2015

### lycee

I'm new to this forum and not sure yet how to write down the symbols, but heat transfer= Phi = U A (LMTD).So finding A would be phi/U(LMTD).Finding the LMTD is proving difficult as I'm not sure of the outlet temp of the feed water.

17. Oct 10, 2015

### SteamKing

Staff Emeritus
It's not clear what you are trying to analyze here.

The original problem involved a fire tube boiler. The conditions at which the boiler produced steam were not disclosed in the OP. The only thing you know for certain is that the feed water entered the boiler at 180° C.

18. Oct 10, 2015

### lycee

If you refer to the first post from Mitch1 all the details of the first part are there.The second part of the question is asking for the area required to perform this duty when the overall heat coefficient is 54 Wm^-2 K^-1

19. Oct 10, 2015

### insightful

Good. You have a pool of water boiling at 180oC producing steam at 180oC. What might the water temperature be everywhere outside the tubes?

20. Oct 10, 2015

### lycee

Of course it's obvious when you look at it logically.Thanks for the heads up.I can know find the LMTD and ultimately find the required area