Calculating Final Outlet Temp of Flue Gases

[Mitch1]

Homework Statement

1. If the flue gases exiting the boiler are used to preheat the water fed to the boiler from a temperature of 28 C to 90 C and assuming:
   

• a mean specific heat capacity for water over this temperature range to be 4.2 kJ kg–1 K–1

• a mean molar heat capacity for the flue gases up to 300oC to be 31 kJ kmol–1 K–1
• 10% of the heat required to heat the water is lost in the heat exchanger
• all water entering the system is converted to steam
  
  determine the final outlet temperature of the flue gas
  

Homework Equations

pw = (mole fraction)water × total pressure

The Attempt at a Solution

I'm not sure where to start off on this once as it does not state any pressure the steam is at, I thought you needed a pressure to multiply by the mole fraction to then obtain a temperature using steam tables?
Any help/suggestions on equations would be much appreciated -thanks

 

[Bystander]

preheat the water fed to the boiler from a temperature of 28 C to 90 C

The boiler feed is in the liquid phase.

 

[Mitch1]

The boiler feed is in the liquid phase.

Thanks for your reply, so is there a different equation that is used? Because you do not need to use specific heat capacity when using the method I described I believe

 

[Bystander]

This is the second (nth?) part of the flame temperature question isn't it? Vapor pressure of water at 90 C is less than 1 atm, and the steam generation occurs in the boiler.

There are some parameters missing here --- water feed rate, flue gas generation/flow rate?

 

[Mitch1]

This is the second (nth?) part of the flame temperature question isn't it? Vapor pressure of water at 90 C is less than 1 atm, and the steam generation occurs in the boiler.

There are some parameters missing here --- water feed rate, flue gas generation/flow rate?

Yes it is the final part there is one after the flame temp question that I believe to have completed just the notes are not very clear,
The flow rate of flue gas is 1400 kmol h^-1
And it does not state any water feed rate, what is the equation that you are intending on using please

 

[Bystander]

• a mean specific heat capacity for water over this temperature range to be 4.2 kJ kg–1 K–1

From the earlier part, "steam at 5atm?" Gauge or absolute? So, you've got 1400 kmol/h at whatever heat content from 300 C to flame T (again from earlier) generating however much steam it takes to soak up that much heat (liquid heat capacity from 90 C to 5atm steam T plus enthalpy of vaporization at 5 atm.), be it gauge or absolute, giving you the water feed rate.

'Nuff of a start?

 

[Mitch1]

From the earlier part, "steam at 5atm?" Gauge or absolute? So, you've got 1400 kmol/h at whatever heat content from 300 C to flame T (again from earlier) generating however much steam it takes to soak up that much heat (liquid heat capacity from 90 C to 5atm steam T plus enthalpy of vaporization at 5 atm.), be it gauge or absolute, giving you the water feed rate.

'Nuff of a start?

Thanks for the explanation would you be willing to give me a formula to use I might be able to tackle it and give it a go that way
Thanks

 

[Bystander]

Total heat in (flue gas flow rate at flame T x heat content) = Total heat out (sum of pre-heat/0.9, plus heat content of steam (or, heat content of flue gas x flue gas flow rate between flame T and 300 C, which is the steam generated, minus losses if they mention it's not a perfect boiler) ) --- I think I got everything, but double-check it.

 

[Mitch1]

Total heat in (flue gas flow rate at flame T x heat content) = Total heat out (sum of pre-heat/0.9, plus heat content of steam (or, heat content of flue gas x flue gas flow rate between flame T and 300 C, which is the steam generated, minus losses if they mention it's not a perfect boiler) ) --- I think I got everything, but double-check it.

Thanks mate I will give it ago using this!

 

[Mitch1]

What would the usual difference be between inlet and outlet temperatures with regards to flue gas as I have worked it out although it barely differs from the inlet temp (around4 degrees) surely I have gone wrong?

 

[Bystander]

This is for the feed water pre-heat? What did you get for feed rate?

 

[Mitch1]

This is for the feed water pre-heat? What did you get for feed rate?

2113 kg/hr? I used the liquid heat cp 90c + 5bar + enthalpy of vaporisation 5 bar
4.208+2.32+2107=2113

 

[Bystander]

Lemme see --- 2113 x 4.2 x 62 = 555 kJ/h; and (31kJ/kmol x 1400 kmol/h)/.9 = 48.2 kJ/h; goes into 555 11.4 times for an 11.4 K delta T. And, you evaporated water in the pre-heat, I'll bet ---- pre-heat T gets to 90 C, at 5 atm., so it can't evaporate.

"Are we there yet?"

 

[Mitch1]

Lemme see --- 2113 x 4.2 x 62 = 555 kJ/h; and (31kJ/kmol x 1400 kmol/h)/.9 = 48.2 kJ/h; goes into 555 11.4 times for an 11.4 K delta T. And, you evaporated water in the pre-heat, I'll bet ---- pre-heat T gets to 90 C, at 5 atm., so it can't evaporate.

"Are we there yet?"

I'm not sure hahah! So what is the 11.4 difference in temperature?

 

[Bystander]

That's how much the flue gas exhausted from the boiler at 300 C is further cooled by running it through the pre-heater.

 

[Mitch1]

Ah so it is simply 300- 11.4 to give the outlet temperature of the flue gas?

 

[Bystander]

That's the way I'd play it.

 

[Kyle Smith]

Sorry to drag up a dead post, but I'm stuck on this question and am dying to get this module out of the way!

I can't figure out at all how to get the mass flow rate of the water feed?

And this sum I can't see where any of the figures are coming from?
"2113 kg/hr? I used the liquid heat cp 90c + 5bar + enthalpy of vaporisation 5 bar
4.208+2.32+2107=2113"

 

[PaxFinnica96]

Lemme see --- 2113 x 4.2 x 62 = 555 kJ/h; and (31kJ/kmol x 1400 kmol/h)/.9 = 48.2 kJ/h; goes into 555 11.4 times for an 11.4 K delta T. And, you evaporated water in the pre-heat, I'll bet ---- pre-heat T gets to 90 C, at 5 atm., so it can't evaporate.

"Are we there yet?"

Hi @Bystander
I was wondering if you could help me with post No. 12 which I am really struggling with.
I've figured out most of the values to calculate feed rate - liquid heat capacity of water is a constant of 4.2 kJ/kg and, latent heat of steam (enthalpy of vaporisation) at 5 bar = 2108 kJ/kg (from steam tables) but cannot figure out where the 2.32 comes from.

I think I can work the rest out once I've got the value of feed rate so any help with this would be very much appreciated!
Cheers

 

[Bystander]

Hi @Bystander
I was wondering if you could help me with post No. 12 which I am really struggling with.
I've figured out most of the values to calculate feed rate - liquid heat capacity of water is a constant of 4.2 kJ/kg and, latent heat of steam (enthalpy of vaporisation) at 5 bar = 2108 kJ/kg (from steam tables) but cannot figure out where the 2.32 comes from.

You'll have to give me a moment to review the other "original post;" @Mitch1 broke a heat transfer problem into two threads.

 

[PaxFinnica96]

You'll have to give me a moment to review the other "original post;" @Mitch1 broke a heat transfer problem into two threads.

Cheers, I had a go at it yesterday from what I think is right...

Specific heat of steam (@ 5 bar) = 2.32 kJ/kg K
Water feed rate = liquid HC at 90deg + Specific heat of steam at 5 bar + enthalpy of vaporisation at 5 bar
= 4.2 + 2.32 + 2108
= 2114.52

So...
Total Heat In = (2114 x 4.2 x (90 - 28)) = 550 kJ/h
Total Heat Out = (31 kJ/kmol x 1400 kmol/h)/0.9 = 48.2 kJ/h

550/48.2 = 11.4 K delta T

Therefore final outlet temp. of flue gas = 300 - 11.4 = 288.6 degC

...does this look right to you?