Question

CHE202 Quiz # 1/Section (08-10) 180 kmol/h of steam is to be heated at constant pressure of 5 bar from 250°C to 350°C. Estimate the required heating rate (kW) using two different methods.

Answer 1

1) heat required = n × Cp × (T2 - T1)

Where n= feed rate = 180 kmole/hr

Cp for steam = 35.928 J/mol K(u can find from internet)

T2= outlet temp. = 350 ℃ = 623 K

T1 = inlet temp. = 250 ℃ = 523 K

Heat required = 180000(mol/h) × 35.928(J/mol K) × [ 623 K - 523 K]

= 646704000 J/h

= 646704000 J/ 3600 sec

=179640 J/sec

= 179640 W

= 178.6 KW

2)for isobaric process, as first law of thermodynamic

Q= ∆U + ∆W

Where ∆U = n × Cv × ∆T

∆W = P ×∆V = n × R × ∆T

Here we take steam as ideal gas,because at high temp,it acts as an ideal gas.

Q = nCv∆T + nR∆T

= n [T2- T1] × ( Cv +R)

Cv for steam = 1.4108 j/g K (from internet)

   = 1.4108 ×18 j/ mol k ( 1g = 1/18 mol)

=25.7 j/ mol K

= 180000(mol/h) ×( 623 - 523)K × ( 25.7 +8.314)(J/mol K)

= 612252000 J/h

= 612252 KJ/3600 sec

=179 Kw