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# Problem 44897. Calculate value of Ra for three runs. Flow rates

• Created by A C

Problem 5.31, page 220, from Felder and Rousseau Elementary Principles of Chemical Processes book

The inputs are stated in the problem. You only need to have y = []; to pass.

I worked the problem out. It was fun and challenging for me. I wanted to share. Work it out any way you want and learn. My answers for Ra are [5.6267 7.4260 5.5061]. Yours answer might be different. They might be right or wrong.

Here my code.

% Read in the values of Rf, Rt, Rp, the fuel gas component mole fraction xa, xb, xc, % xd, and xe, and the percent excess air PX, and to find the required value of Ra. % Methane = a , ethane = b, propane = c, n-butane = d, isobutane = e

% The linear proportionalities between the in and out signals and the % corresponding process variables may be determined by:

% Fuel Temperature (K)

Tf1 = 25+273; Tf2 = 35+273; Rt1 = 14; Rt2 = 27;

% Fuel Pressure (atm) assume Patm = 1 atm

Pf1 = 1; Pf2 = 1 + 20000/101325; Rp1 = 0; Rp2 = 6;

% Fuel Flow rate (L/h)

Vf1 = 0; Vf2 = 2000*1000; Rf1 = 0; Rf2 = 10;

% Air Flow Rate (L(STP/h)

Va1 = 0; Va2 = 100000*1000; Ra1 = 0; Ra2 = 25;

Rf = [7.25,5.8,2.45]; Rt = [23.1, 7.5 46.5]; Rp = [7.5, 19.3, 15.8];

% data mole%

xa = [.81,.58,0]; xb = [.08, .31, 0]; xc = [.05, .06, .65]; xd = [.04, .05, .25]; xe = [.02, 0, .10];

% data percent excess air

PX = [.15,.23,.33];

% Linear equations R => x T(K)

slopeT = 1/((Rt2 - Rt1)/(Tf2 - Tf1)); Tb = Tf1 - slopeT*Rt1; %Rt = linspace(Rt1,Rt2,11);

T = Rt*slopeT + Tb;

% Linear equations R => x Pa(atm)

slopeP = 1/((Rp2 - Rp1)/(Pf2 - Pf1)); Pb = Pf2 - slopeP*Rp2; %Rp = linspace(Rp1,Rp2,11);

P = Rp*slopeP + Pb;

% Linear equations R => x V(L/h)

slopeRf = 1/((Rf2 - Rf1)/(Vf2 - Vf1)); Vb = Vf1 - slopeRf*Rf1; %Rf = linspace(Rf1,Rf2,11);

Vf = Rf*slopeRf + Vb;

% assume ideal gas behavior PV = nRT, R = constant find n(Kmol/h)

R = 0.08206; %(atm *L)/(gmol*k) nf = ((P.*Vf)./(R*T))/1000; % Kmol/h

% species' mole in the fuel mixer Kmol/h

nfa = nf.*xa; nfb = nf.*xb; nfc = nf.*xc; nfd = nf.*xd; nfe = nf.*xe;

% Theoretical O2 % combustion chemical equations:

% 1. CH4 + 2O2 > CO2 + 2H2O methane

% 2. C2H6 + 3.5O2 > 2CO2 + 3H2O ethane

% 3. C3H8 + 5O2 > 3CO2 + 4H2O propane

% 4. C4H10 + 6.502 > 4CO2 + 5H2O BUTANE

% 5. C4H10 + 6.502 > 4CO2 + 5H2O ISOBUTANE

% no2s, s = stoichiometric when its assumes the species are completely consumed.

nO2s = nfa*2 + nfb*3.5 + nfc*5 + nfd*6.5 + nfe*6.5; %(kmol O2/h)

% excess of O2

nO2_feed = PX.*nO2s + nO2s; %kmol O2/h

% assume 21 mole%O2 and 79 mole%N2

nair_feed = nO2_feed/.21; %Kmol air/h

% assume air behavior is ideally PV = nRT find V() using T(stp) =273K,P(stp) = 1atm % 1000gmol = 1 kmol

Vair = (nair_feed.*273*.08206)*(1000); %(L/h)

% Linear equations R => x V(L(STP)/h)

slopeVa = (Ra2 - Ra1)/(Va2 - Va1); Vab = Ra1 - slopeVa*Va1; Ra = Vair.*slopeVa + Vab

### Solution Stats

60.0% Correct | 40.0% Incorrect
Last solution submitted on Sep 04, 2019