What do you often notice after you balance a hydrocarbon combustion
2 answers:
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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.
<u>Explanation:</u>
We are given:
Rate of flow of ideal gas , n = 4 kmol/hr = (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)
Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)
We know that:
(For isothermal process)
So, by applying first law of thermodynamics:
.......(1)
Now, calculating the work done for isothermal process, we use the equation:
where,
= change in work done
n = number of moles = 1.11 mol/s
R = Gas constant = 8.314 J/mol.K
T = temperature = 475 K
= initial pressure = 100 kPa
= final pressure = 50 kPa
Putting values in above equation, we get:
Calculating the heat flow, we use equation 1, we get:
[ex]\Delta q=3.038kW[/tex]
Now, calculating the rate of lost work, we use the equation:
Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.