Question

If hydrogen is burned at a rate of 1 kg/s, to completion, with 25% excess air to generate electricity in a modified combustion turbine, what is the dew point temperature (°C) of the flue gas?

Assume exit of the combustion chamber at 150 kPa.

Answer 1

Given : 1 Kg/s of hydrogen

Reaction: H2 +0.5O₂ ---> H₂O              .........(1)

amount of hydrogen=1 Kg/s =0.5 Kmole/s

Amount of oxygen theoretically required =0.5*0.5=0.25 Kmole/s

Amount of air required theoretically =0.25/0.21=1.19 Kmole/s

It is givent that 25 % exceess air is supplied

theerefore air supplied =1.19*(1+0.25)=1.4875 Kmol/s

Amount of oxygen supplied =0.21*1.4875=0.3124 Kmole/s

Amount of Nitrogen present in Air =0.79*1.4875=1.1751 Kmole/s

Now here we will assume that hydrogen completely reacts as per stoichometry to form Water

therfore oxygen consumed =0.5*0.5=0.25 Kmole/s

Amount of water formed =0.5 KMole/s

Amount of oxygen unreacted =0.3124-0.25=0.0624 Kmole/s

Component	Moles	Mole fraction (Yi)
Water	0.2	0.13913
oxygen	0.0624	0.0434
nitrogen	1.1751	0.81746

Now let us calculate the partial pressure of water vapour in air,

assuming ideal gas law,

partial pressure of water vapour =Yi*P =0.13913*150=20.8695 Kpa

Dew point calculation: At dew point the vapour pressure of water vapour in air is same as partial pressure of water vapour of water vapour

Hence , using antoine equation , we will calculate the temperature at the vapour pressure of 20.8695 Kpa

lop₁₀P=8.14019-(1810.94/(294.485+T))

Where P= is in mmHG and T=^OC

From above antoine equation we get dew point temperature as T=10.1 ^OC at P=156.534 mmHg=20.8695 Kpa