4 replies to this topic

[roy2377]

Hi everyone,

 

I was recently doing a calculation on the use of DIESEL-50 to produce steam.  My supervisor gave me this number:

 

1 Liter of Diesel/Hr = 16 kg of Steam/Hr

 

I asked him how did he get this value and he couldn't answer and he said its experience.

 

Grateful if anyone can explain me how we can come up with such value as I wanted to make sure I know the equations that gave me 16kg/Hr .

 

Is there any graph about diesel vs Steam ??

 

Thanks

Roy.

[sam03]

hello roy,
Answer of ur question lie in the heat or energy blance equation...from very fundamental if we compare the
energy supply by the diesel=energy absorbed by water
let consider per kg of diesel then energy released by burning of one kg diesel equal to 43500kj/kg.
43500=energy absorbed by water to conver it into 100°c of water+ latent absorbed by steam
43500kj/kj =m*4.2*100+m*2256
on solving you will get m= 15.8819
which approximate equal 16kg......

[paulhorth]

Sam,

A few corrections are needed to your heat balance calculation.

 

• Roy quoted the figure 16 kg steam per litre of diesel but you did the calc for a kg of diesel, which is more..
• You have not allowed for the heat lost in the flue gases from the boiler, which can typically be 30% or more of the total heat released by burning.
• You have not allowed for the heat losses from the boiler itself .
• You have used the higher heating value for diesel when you should have used the lower heating value, since you will not be condensing the water vapour of combustion
• You have assumed the steam is raised at atmospheric pressure, which is never the case in an industrial boiler, but Roy has not bothered to tell us what pressure he wants his steam.

All of these factors will reduce the steam raised to significantly less than 16 kg per litre of diesel.

 

Paul

[S.AHMAD]

The formula is..

1. kg steam production per liter of diesel = Heat of combustion per liter of diesel/heat of vaporiation of steam at conditions x efficiency of boiler.

2. The above value divided by 100 if efficiency is in %.

Edited by S.AHMAD, 15 November 2013 - 07:24 PM.

[roy2377]

Sam, Paul and Ahmad, thank you very much for those explanation.  Really really appreciate the help.