Question

111 m³/min of hydrogen measured at normal conditions and nitrogen at a 300% excess enter an isobaric reactor. The inlet gases at 490°C are reacted in the presence of a catalyst to synthesize ammonia. The process has an efficiency of 33%. The exhaust gases are cooled to -40°C to condense the ammonia. Assume that the Cp of the gas phase species can be used in the temperature range, and that Cp NH3 (l) = 0.071938 kJ/mol K. Calculate the overall heat flow of the system in kW.

Answer 1

The reaction occuring is

H2 is 111 m3/min at NTP

NTP conditions

T = 293 K

P = 1 atm = 1.013×10⁵ Pa

n = PV/(RT)

n = (1.013×10⁵×111)/(8.314×293)

n= 4615.882 mol/min

Moles of H2 = 4615.882 mol/min

According to stiochiometry amount of nitrogen required = 4615.882/3 = 1538.627 mol/min

N2 is 300% excess

N2 supplied = 1538.627 + (1538.627×3)= 6154.5093 mol/min

Efficiency of process =33%

Conversion = 33%

Limiting reactant = H2

Products should be cooled to -40°C

Enthaply reference is 25°C

H(25°C)= 0

Reaction occurs = 490°C

Products average temperature = (490-40)/2 = 225°C = 498 K

Reactants average temperature = (25+490)/2 = 257.5°C = 530.5 K

Cp is taken from handbook

Cp(liquid ammonia )= 0.071938 KJ/mol

B.p of ammonia = -33°C

Component	inlet moles	Cp(KJ/mol°C)	Hi(KJ/min)	outlet moles	Cp(KJ/mol°C)	Ho(KJ/min)
N​​​​​​2	6154.5093	0.029738	85105.60	6154.5093-(4615.882(0.33)/3)= 5646.762	0.029573	-88505.59
H​​​​​​2	4615.882	0.029278	62841.8638	4615.882(1-0.33)= 3092.640	0.02925	-47943.6516
NH​​​​​3	-	-	-	1.5(4615.882)(0.33)= 2284.861	0.04196	2284.861((0.04196)(-33-490)+0.0719(-40+ 33)= -51291.42
Total			147947.46			-187740.6616

Also from handbook

Latent heat of vaporization = 23.4 kJ/mol

Latent heat taken out = 2284.861(-23.4)= -53465.747 KJ/min

From handbook

At T = 25°C

∆Hf(NH3)= -45.94 KJ/mol

∆H(N2 ) and ∆H(H2)= 0

∆H_r = ∆H_products - ∆H_reactants

∆Hr (25°C)= -45.95(2)= -91.88 KJ/mol

According to Kirchoffs law

∆H(490°C)= ∆H(25°C)+ ∆Cp(490-25)

∆Cp = (nCp) _{product s} -(nCp)_reactants

Cp is taken at average temperature of 530.5 K

Cp(NH3)(g)= 0.043014 KJ/mol °C

Cp(N2) and Cp(H2) are given in table

∆Cp = (2×0.043014)-(3×0.029278)-(0.029738)= -0.031544

∆H(490°C)= -91.88+(-0.031544)(490-25)

∆H(490°C)= -107.328 KJ/mol

Moles formed = 2285.861 mol

∆H(490°C)= 2285.861(-107.328)= -245336.88

Q = ∆H(490°C)+ latent heat + H(reactants + H(product)

Q = -245336.88-53465.747+ 147947.46- 187740.6616 = -338595.82 KJ/min

Q =-5643.263 KW

Please upvote if helpful