Question

A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25ºC, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90ºC.

(a) Determine the maximum flame temperature.

(b) If 5% of the heat available for steam production is lost to the atmosphere, determine the amount of steam raised per hour when the total flow of flue gases is 1400 kmol h-1.

(c) Determine the dew point temperature assuming that the flue gas pressure is 1.00 bar and the inlet air: (i) is dry (ii) contains 0.8 kg water per kmol of air at the temperature of the inlet air.

(d) If the flue gases exiting the boiler are used to preheat the water fed to the boiler from a temperature of 28ºC to 90ºC and assuming: - a mean specific heat capacity for water over this temperature range to be 4.2 kJ kg-1 K-1 - a mean molar heat capacity for the flue gases up to 300ºC to be 31 kJ kmol-1 K-1 - 10% of the heat required to heat the water is lost in the heat exchanger - all water entering the system is converted to steam determine the final outlet temperature of the flue gas and state if the dew point will be reached in both of the cases given in part (c).

Data: Net calorific value (MJ m-3) at 25ºC of: Butane (C4H10) = 111.7 MJ m-3 Butene (C4H8) = 105.2 MJ m-3 Propane (C3H8) = 85.8 MJ m-3 Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and 76.7% nitrogen by mass.Atomic mass of C= 12, O= 16, N=14 and H= 1

Answer 1

Net calorific value (MJ m–3) at 25°C of: Butane (C4H10) = 111.7 MJ m^3 Butene (C4H8) = 105.2 MJ m^3 Propane (C3H8) = 85.8 MJ m^3 I need help on how to calculate: - the net calorific value (CV) per m^3 of the fuel/air mix at 25°C ? the net calorific value (CV) per kmol of the fuel/air mix at 25°C ? 2. Relevant equations 3. The attempt at a solution Net cv per m3 = 58.65 mJ / m^3 Net cv per kmol = 1466.25 mJ /kmol
Presumably, the flue gas is at 1 bar. How many moles of flue gas are there in 1 m^3 of the gas at 25 C? How many moles of butane, propane, and butene are there?

I make it a total of 35.35 moles with 10% excess air.
I make it : butane - 0.75mol propane - 0.1mol butene - 0.15mol oxygen - 6.3mol 10% excess air = 0.63 + 6.3 = 6.93mol nitrogen = 6.93 x 3.76 = 26.07mol
Second is What is the total number of moles here, and, using the ideal gas law, what volume does this total number of moles of gas occupy?

total number of moles is 34 moles. n = pV / RT n = no. of moles p = pressure v = volume R = constant (8.314) T = temperature n = (101300 x 1) / (8.314 x 298) n = 40.89 moles

So, if 1 m^3 of the gas contains 40.9 moles, and the total volume you calculated for the case of 1 mole of pure fuel is 34 m^3, how many moles of each species in the fuel mixture is present in 1 m^3 of mixture?
V = (n/p)RT The volume in the ideal gas equation for 34 moles of gas is 0.83 M^3
how many moles of each species in the fuel mixture is present in 1 m^3 of the mixture? So answer is  it 1.203 moles
Yes, but I want each species individually, including the oxygen and nitrogen. Oct 13, 2016 #13 GeorgeP1 butane - 0.902 mol propane - 0.12 mol butene - 0.181 mol O2 - 8.33 mol N2 - 31.36 mol Oct 13, 2016 #14 Chestermiller Staff: Mentor GeorgeP1 said: ↑ butane - 0.902 mol propane - 0.12 mol butene - 0.181 mol O2 - 8.33 mol N2 - 31.36 mol This agrees with what I got. Now, I assume that those net caloric values that they give you is for burning 1 m^3 of each of the pure three species (i.e., for burning the number of moles of each species that occupy 1 m^3 at 25 C and 1 bar). Is that correct? Oct 13, 2016 #15 GeorgeP1 great... yes the net cv is for 1 m^3 of each gas. Oct 14, 2016 #16 Chestermiller Staff: Mentor If 40.89 moles of any gas would occupy 1 m^3 under these conditions, how many m^3 would 0.902 moles of pure butane occupy? What would be net caloric value of burning this amount of butane? Oct 14, 2016 #17 GeorgeP1 I make it 0.02206 M^3 for butane does that make the net CV for butane 2.46 Mj / M^3 Last edited: Oct 14, 2016 Oct 14, 2016 #18 Chestermiller Staff: Mentor GeorgeP1 said: ↑ I make it 0.02206 M^3 for butane does that make the net CV for butane 2.46 Mj / M^3 OK. What about the other two species, and what about their combined net CV? Oct 14, 2016 #19 GeorgeP1 propane = 0.003 m^3 = 0.2574 MJ /M^3 Butene = 0.0044 M^3 = 0.463 Mj / M^3 Total = 3.1804 Mj /M^3 Oct 14, 2016 #20 Chestermiller Staff: Mentor GeorgeP1 said: ↑ propane = 0.003 m^3 = 0.2574 MJ /M^3 Butene = 0.0044 M^3 = 0.463 Mj / M^3 Total = 3.1804 Mj /M^3