Calculating Mass Flow Rate of Steam

[sjh94]

Homework Statement

If 5% of the heat available for steam production is lost to the
atmosphere, determine the amount of steam raised per hour when the
total flow of flue gases is 1400 kmol h-1.

qmH = 1400 kmol h-1

Flue gas temperatures:
Inlet = 1832°C
Outlet = 300°C

Water temperature:
Inlet = 90°C
Steam produced at 5 barg = 159°C

Latent Heat of Steam @ 5 barg = 2085 kJkg-1 (hfg)

Specific heat capacity of water = 4.18 kJ kg-1 k-1 (Cpc)

2. The attempt at a solution:

The specific heat capacity of the flue gas is not given, therefore I am missing 'cph' and cannot use the equation below:

If I could work out the heat transfer rate then I would be able to use the equation below to find the the mass flow of the steam??

Any help would be much appreciated.
Thanks.

 

[BvU]

Hello sjh,

Seems to me the problem statement you were given is incomplete indeed. Otherwise your expressions are fine.
Perhaps there is a context ? Or else you are supposed to look it up somewhere and use an approximate value ? Somewhat strange, when the input temperature of the flue gas is given with such a high accuracy (on the other hand the 5% looks like a crude estimate) .

 

[sjh94]

Thanks for the reply.

I should have mentioned that a previous question asked me to calculate the flame temperature, this is why the temperature is so precise.

I emailed my tutor with the following question:

'To my understanding I would require the specific heat capacity of the flue gases to use in this equation?

Is this correct? as the question doesn't state what the specific heat capacity is for the flue gas?'

His reply:

'That sounds sensible, yet as you mentioned the question doesn't state what the specific heat capacity is for the flue gases. So you need to think of other way to calculate the heat available, maybe you can use the steam table.'

Don't know if this helps further, thanks.

 

[BvU]

HI,
I'm afraid I don't see any inroads and the steam table doesn't help me for the process side.
@Chestermiller : am I overlooking something ?

Otherwise your expressions are fine

After reading a bit better: Note that

is a bit deficient on the steam side: it only describes the phase transition; you also have to heat the water from 90°C to 159°C

 

[Chestermiller]

HI,
I'm afraid I don't see any inroads and the steam table doesn't help me for the process side.
@Chestermiller : am I overlooking something ?
After reading a bit better: Note that

is a bit deficient on the steam side: it only describes the phase transition; you also have to heat the water from 90°C to 159°C

This can't be done unless you know the composition or the heat capacity of the fluid gas. Is this the second part of a problem?

 

[sjh94]

Hi,

Yes I have already calculated the flue gas composition, below is what I got:

There is 10% excess air provided in combustion, this is why I have a small amount of oxygen in flue gas.

 

[BvU]

Do your 'steam tables' also have properties for these gases ? As in Cengel, for instance ?

 

[sjh94]

I have been provided with this table, which shows enthalpy of combustion gases.

I used these values when determining my flame temperature by calculating the heat content for each component = number of moles * enthalpy of component
Then added the heat content per combustion gas to give me my total

I calculated the total heat input per kmol of fuel to be 2349 kJ (maybe be right, maybe wrong).

I then used this value to determine my flame temperature, I started with a flame temperature of 1800°C which gave me a total heat content of 2305 kJ.

So I knew my flame temperature was higher, therefore I did the same with a flame temperature of 1900°C.
This gave me a heat content of 2448kJ which is higher than my calculated heat input of 2349 kJ. So my flame temperature was between 1800 and 1900, from these 2 results I produced a straight line graph and ended up with 1832°C.

Make sense?

The chart below should be kJ kmol-1 not MJ, I think.

 

[Chestermiller]

Hi,

Yes I have already calculated the flue gas composition, below is what I got:

View attachment 111351

There is 10% excess air provided in combustion, this is why I have a small amount of oxygen in flue gas.

So look up the heat capacities of these gases at the two extreme temperatures, and take their temperature average. Then take their molar average.

 

[Chestermiller]

I have been provided with this table, which shows enthalpy of combustion gases.

I used these values when determining my flame temperature by calculating the heat content for each component = number of moles * enthalpy of component
Then added the heat content per combustion gas to give me my total

I calculated the total heat input per kmol of fuel to be 2349 kJ (maybe be right, maybe wrong).

I then used this value to determine my flame temperature, I started with a flame temperature of 1800°C which gave me a total heat content of 2305 kJ.

So I knew my flame temperature was higher, therefore I did the same with a flame temperature of 1900°C.
This gave me a heat content of 2448kJ which is higher than my calculated heat input of 2349 kJ. So my flame temperature was between 1800 and 1900, from these 2 results I produced a straight line graph and ended up with 1832°C.
So forget what I said in my previous post. Get the enthalpy change per mole of the mixture between the inlet and outlet temperatures using this table.

Make sense?

The chart below should be kJ kmol-1 not MJ, I think.

View attachment 111354

 

[BvU]

The specific heat capacity of the flue gas is not given, therefore I am missing 'cph'

So instead of $c_p \Delta T$ you use $\Delta H$ -- much better.

So much for answering your original post #1. The exercise is growing a bit bigger with your post #8 !

The chart below should be kJ kmol-1 not MJ, I think

You can compare with Cengel...

 

[evoke1l1]

May I just ask what delta H is, please?

 

[BvU]

May I just ask what delta H is, please?

Enthalpy difference

 

[evoke1l1]

Ahh I see! It makes sense to me now. Many thanks.