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

Hydrogen fed = 111 m³/min

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

H₂ fed = 4615.882 mol/min

The reaction occuring is

N₂ required according to stiochometry =

4615.882/3 = 1538.276 mol/min

N₂ fed is 300% in excess

N₂ supplied = 1538.276+ (1538.276) (3) = 6153.104 mol/min

Process efficiency = 33%

Here efficiency of process is asummed as conversion

Conversion = 33%

H₂ reacted = (4615.882) (0.33)

= 1523.241 mol/min

N₂ reacted according to stiochometry = 1523.241/3 = 507.747 mol/min

The reaction occurs at 490°C

From handbook

At T = 25°C

∆Hf(NH3) (g) = -45. 94 KJ/mol

∆Hf(N2) and ∆Hf(H2) = 0

∆H_r = ∆H_products - ∆H_reactants

∆H_r = (-45.94×2) -0= -91. 88 KJ/mol

According to kirchoffs law

∆H(490°C) = ∆H(25°C) + ∆Cp(490-25)

∆Cp= (nCp) _products -(nCp) _reactants

Cp is taken from handbook at average temperature of

(25+490) /2 = 257.5°C

Cp(NH3)(g) = 0.043015 KJ/mol°C

Cp(N2) = 0.029738 KJ/mol°C

Cp(H2) = 0.02927 KJ/mol°C

∆Cp = (2×0.043015) -(0.029738) -(3×0.02927) = -0. 031518 KJ/mol°C

∆H(490°C) = ∆H(25°C) + ∆Cp(490-25)

∆H(490°C) = -91. 88+(-0.031518) (490-25)

∆H(490°C) = -106. 535 KJ/mol

Moles of amminia formed = (2×507.747) = 1015.494 moles

∆Hr= -106. 535(1015.494) =

-108185. 6533 KJ/min

Enthalpy reference temperature = 25°C

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

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

B. P of ammonia = -33°C

Cp(liquid) = 0.071938 KJ/mol°C

∆H = nCp(Hf- Hi)

Component	inlet moles/min	Cp(KJ/mol°C)	Hi(KJ/min)	outlet moles (mol/min)	Cp(KJ/mol°C)	Ho(KJ/min)
N2	6153.104	0.029738	85086.168	6153.104- (4615.882×0.33) /3)= 5645.356	0.02957	-88474.583
H2	4615.882	0.02927	62824.692	4615.882(1-0.33) =3092.640	0.029258	-48484.554
NH3	-	0.043015	-	1015.494	0.04196	1015.494((0.04196(-33-490) +(0.071938(-33+40)) = -21773. 728
Total			147910.8608			-158732. 865

NH3 condenses at T = -33°C

Latent heat of ammonia = 23.5 KJ/mol

Latent heat removed =- 23.5(1015.494) = -23864.109 KJ/min

Q = latent heat + ∆H(r) + H_reactants + H_products

Q = -23864. 109 - 108185.6533 +147910.8608 -158732.865 = -142871. 766 KJ/min

Q= -2381. 196 KW

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