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Calculate the rate of cooling (kW) required to bring 300 kg/min of carbon monoxide from 450C...

Calculate the rate of cooling (kW) required to bring 300 kg/min of carbon monoxide from 450C to 50C.

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Answer #1

The specific heat capacity of CO is given as

Cp° = A + B*t + C*t^2 + D*t^3 + E/t^2

Cp° in J/mol

Heat required to change the temperature from

t1 = 450+273 = 723 K to

t2 = 50 + 273 = 323 K

dH = Cp°dt

Integrate the equation and we get

H = A(t2 - t1) + (B/2)(t22 - t12) + (C/3)(t23 - t13) + (D/4)(t24 - t14) - (E) (1/t2 - 1/t1)

= 25.56759(323 - 723) + (6.096130/2)(323^2 - 723^2) + (4.054656/3)(323^3 - 723^3) - (2.671301/4)(323^4 - 723^4) - (0.131021) (1/323 - 1/723)

= - 10227.036 - 1275310.396 - 465251232.4096 + 175211345293.1068 - 0.0002244

= 174744808523.26497 J/mol

= 1.747 x 10^11 J/mol x 1mol/28g x 300*1000 g/min x 1min/60s

= 3.12 x 10^13 J/s x 1kJ/1000J

= 3.12 x 10^10 kJ/s

= 3.12 x 10^10 kW

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