Recent content by PCal

I think the flue gas coming out of the boiler is 300° C but tha's not very clear, The flow rate of the flue gas = 1400 kmol h-1, Enthalpy at 5 bar = 2018kJ/kg, the heat capacity at 90°=4.208. As far as I can tell that's all I've got

If I'm correct, just the outlet temperature from the pre-heater. The question isn't particularly clear but that's my assumption!

Fab got it! Thank you!

This is the final question on my assignment and I'm clueless as to where I should even start as the lesson seems to give no guidance. Help would be massively appreciated!

Any help would be massively appreciated please!

Is the dry basis as simple as saying... CO2 3+0.3-0.6=3.9mol N2=26.017mol O2=0.63mol

Wet basis 0.75mol C4H10 Requires 4.875 mols O2 Produces 3 mols of CO2 and 3.75 mols of H2O 0.1mol C3H8 Requires 0.5 mols O2 Produces 0.3 mols of CO2 and 0.4 mols of H2O 0.15mol C4H8 Requires 0.9 mols O2 Produces 0.6 mols of CO2 and 0.6 mols of H2O Theoretical oxygen= 6.3mol +10% excess...

You haven’t said that because I asked you to?? Haha I’m currently doing a happy dance

Butane 0.75 x 58 = 43.5g Propane 0.1 x 44 = 4.4g Butene 0.15 x 8.4g Total mass= 56.3g 0.75 x 208 = 156g 0.1 x 160 = 16g 0.15 x 192 = 28.8g Total O2 =200.8 (200.8/23.3) x 76.7 = 661 (200.8 + 661) x 1.1 = 947.97 947.97/ 56.3 = 16.83 1: 16.83 Please say this is correct?

Aaa I think I understand. So... Butane 0.75 x 58 = 43.52g Propane 0.1 x 44 = 4.4g Butene 0.15 x 8.4g Total mass= 56.32g 1895.96/56.32 = 33.664 = 1 : 33.67

Oh amazing so... Butane 4.875 x 58= 282.75g Propane 0.5 moles x 44 = 22g Butene 0.9 x 56 = 50.4g Total mass= 355.15g 1895.96/355.15 = 5.33 = 1 : 5.33 Does this look better please?

Sorry I'm confused, I'm struggling to see anything in that thread that helps with calculating the ratios? It looks like that's for the next part of my question.

By Volume: Butane 0.75x6.5= 4.875 moles of O2 Propane 0.1x5= 0.5 moles of O2 Butene 0.15x6= 0.9 moles of O2 Total O2 4.875+0.5+9= 6.275 moles Air required 6.275/0.21=29.88 moles/m^3 With 10% excess air 29.88x1.1= 32.868moles 1 mole fuel:32.87 moles of air By Mass: CO2 Moles...

Got it! I was doing F=∆Tc x LMTD as opposed to F=∆Tc/LMTD Thank you for all your help!

0.59! Why have I had to take the reciprocal though?