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So if we have: S + O2 → SO2 S=0.61 moles thus 1:1 ratio so 0.61 moles of O2 2H + 1/2 O2 → H2O H=111.2 moles thus 2:1/2 ratio so 27.8 moles O2 C + O2 →CO2 here we know 95% of the C went to CO2 and 1:1 ratio so 66.92 moles O2 C + O →CO here 5% of C went to CO so 1:1 ratio so 3.52 moles O2 So total...

If we start with mole ratios I would then have: 846 g of C / 12.01 g/mol = 70.44 moles C 111.2g of H / 1.00 g/mol = 111.2 moles H 19.5g of S / 32.07 g/mol = 0.61 moles S 23.3g of inerts -> don't know moles because we don't know what the inerts contain. the 846 g of C or 70.44 moles would mean...

Okay I think I got this! We use q=m*Cp*deltaT to find our heat that would be removed thus: q = (1.5kg/s)*(5200 J/kgdegreesC)*(60-35) q=195 000 J/s then to get the area, we would use A= q / h(or U)*LMTD so A= 195000 J/S / (520 W/m^2degreesC * 18.20) A=20.60 m^2 Seems right now?

Wait I just realized now that you pointed that out... we are dealing with wt% so it is the mass percent and would not be on the mole ratio right? So if we assumed 1kg of fuel we know that 77% of the air is consumed in theory. We don't know what the starting mixture is of the fuel we just know...

Homework Statement The chemical analysis of a fuel is determined to be: 84.60 wt% C, 11.12 wt% H, 1.95 wt % S and the balance, inerts. The fuel is consumed in a steam generating furnace with 33% excess theoretical air. The air is preheated to 190degreesC and enters the furnace at a gauge...

So the heat load, using the formula given would then be: LMTD = {(Ti-ti)-(To-to)} / ln{(Ti-ti)/(To-to)} = {(15-60)-(30-35)}/ln{(15-60)/(30-35)} = {(-45)-(-5)} / ln(-45/-5) = {(-40) / (2.197)} = -18.20 So I know: -flow rate is 1.5 kg/s -specific heat is...

Homework Statement [FONT=Times New Roman] [FONT=Times New Roman]A hydraulic fluid initially at 60◦C is cooled to 35◦C using a co-current heat exchanger. Cooling water is available at 15◦C, which must not exceed 30◦C on leaving the heat exchanger. The flow rate of the hydraulic fluid is 1.5...