3 replies to this topic

[gilad]

I am trying to calcualte how much less natural gas I will be using if I add an economizer on to an exisiting boiler.
If I add an economizer based on the current heat value of the fluegas, the amount of natural gas reduction is lower than actual, and I get a very attractive payback period. However, because less natural gas is needed to produce the same amount of steam, the heat avilavble for recovery by the economizer is lower than before. Hence, I need more Natural gas than the first calculation implied. It is sort of an endless loop.

I have considered using numerical methods to find the amount of fuel I will actually need so I can to an accurate economic analysis.
Does anyone have suggestions on how to apporach this problem?

[Zauberberg]

Actually it shouldn't be that much complicated. The required heat input is a constant, and it is divided between:

- Fuel gas combustion
- Heat recovery from the flue gas

The amount of flue gas is directly proportional to the amount of fuel gas - assuming you're maintaining given (known) air-to-fuel ratio and fairly constant fuel gas composition. Each pound of fuel requires certain number of pounds of air (flue gas flow : fuel gas flow = K). So each pound of fuel gas will have corresponding no. of pounds of flue gas with constant temperature at the outlet of combustion zone. Putting it that way, the only unknown variable is the fuel gas flow.

[gilad]

I agree with everything you say except the temperature of the flue gas being constant. Since the heat input is constant, reducing the amount of fuel gas will require higher enthalphy from the heat recovered from the flue gas.

Using your heat balance:

Fuel gas combustion - flow rate decrease->available heat decrease

Therefore, this implies the heat recovered from the flue gas must increase, which means flue gas temperature will increase until it equilibirate.

[Zauberberg]

I would still say the temperature of flue gas exiting radiant section will be the same, regardless of the amount of fuel gas burned. Radiant heat transfer is proportional to T^4 and not the function of flow/velocity. For practical purposes, this should be sufficient.

On the right side of your equation, we are missing the term 'heat recovery from flue gas' - this will make both sides equal.
No matter how much fuel you burn, the ratio between fuel/flue gas will remain constant and that's something which simplifies the calculation.

Best regards,