Question help please

Hi

Would anyone be able to point me in the right direction for answering this question? I don't really want the answer, I would just like some help on how to attempt the question (formula etc).

A SOFC based Micro-CHP is designed to produce 3 kW of AC electrical power. It has fuel-to-electricity efficiency equal to 60% and the efficiency of its DC-to-AC conditioning system is 75%. Electrical power output from each cell is 210 W DC. This Micro-CHP system is fuelled by natural gas. The molar mass of natural gas is 18.1 kg/mole, LHV=46.7 MJ/kg. Pressure and temperature in gas mains is 17.5 mbar and 180C, respectively. The total efficiency of the Micro-CHP unit is 85% and temperature of the water at the inlet and exit of the boiler-radiator is 20 and 500C. For water cp= 4.18 kJ/kgK.

Determine the number of tubular cell in the system

Thanks

Reply 1

Willa

As with all questions, break it down, step by step. Each new piece of information you are given, think about what you can calculate from it

LCR

A SOFC based Micro-CHP is designed to produce 3 kW of AC electrical power.

So set Q = 3000 (say)

It has fuel-to-electricity efficiency equal to 60%

set R=0.6 (this will b the fuel-to-electricity efficiency which you use later)

and the efficiency of its DC-to-AC conditioning system is 75%.

So to have Q = 3000, we need DC power, P, such that P*0.75 = 3000.

P = 4000W; (look we did our first calculation!!!)

Electrical power output from each cell is 210 W DC.

So the number of cells needed is:

N = P/210 = 19.04 = 20

This Micro-CHP system is fuelled by natural gas. The molar mass of natural gas is 18.1 kg/mole, LHV=46.7 MJ/kg. Pressure and temperature in gas mains is 17.5 mbar and 180C, respectively. The total efficiency of the Micro-CHP unit is 85% and temperature of the water at the inlet and exit of the boiler-radiator is 20 and 500C. For water cp= 4.18 kJ/kgK.

I dont understand why this is here. I am guessing it is for a later part of the question. Becuase as far as my understand goes, we finished the question with the first half of the information.

Reply 2

Theorist

Willa

set R=0.6 (this will b the fuel-to-electricity efficiency which you use later)

Where did you use this in your calculation? I agree with your workings completely but I don't see where you use this figure?

Willa

I dont understand why this is here. I am guessing it is for a later part of the question. Becuase as far as my understand goes, we finished the question with the first half of the information.

Probably how much gas is needed to make this much power, especially as you're given fuel-to-electricity efficiency, which I don't think you use in the first question. :wink:

Reply 3

Willa

Dharma

Where did you use this in your calculation? I agree with your workings completely but I don't see where you use this figure?Probably how much gas is needed to make this much power, especially as you're given fuel-to-electricity efficiency, which I don't think you use in the first question. :wink:

i wrote down that R thing because I was going through the question in order, so you're right, I never used that value.