Question

Carbon Monoxide at 25°C and steam at 150°C are fed to a continuous water-gas shift reactor. The product gas, which contains 50.0 mole% H2, 40.0% ????????2, and the balance H2????, emerges at 500°C at a rate of 2.50m3/h and goes to a condenser. The gas and liquid streams leaving the condenser are in equilibrium at 15°C and 1 bar. The liquid may be taken to be pure w"a) Calculate the % excess steam fed to the reactor and rate of condensation of the water (kg/h) leaving the condensor.b) Calculate the rate (kW) at which heat must be removed from the condensor.c) Calculate the rate of heat transfer (kW) to or from the reactor (state which it is).

Answer 1

The final answer for the solution is Q= -8.8171*$10^{-4}$ kW

Explanation:

$CO +H_2O  ⇄  CO_2+H_2$

basic:

100 moles of product gas

$CO_2$ = 40 moles

$H_2$ = 40 moles

$H_20$ = 20 moles

for 40 moles of$CO_2$ we need 40 moles of CO and 40 moles of $H_2O$

therefore,

Inlet :

No of moles of CO = 40

No of moles of $H_20$ = 40

but $H_20$ is in the outlet stream = 20 modes

Excess = $\frac{20}{40}\ \times 100$%

= 50%

50% excess stream is fed to the reactor

a) Rate of conversion:

ρ_water (density ) =$\frac{1g}{cm^{3}}$

Given volume rate = 3.5 $\frac{cm^{3} }{hr}$

3.5 * density = 3.5 * $10^{-3}$ $\frac{kg}{hr}$

b) Rate (kW)

ΔH = ΔH_1 + ΔH_2 +ΔH_3

Rate of condensation Q° =n° ΔH

Q = n° (ΔH_1 + ΔH_2 +ΔH_3)

ΔH_1 is sensible heat

V =$\int\limits^T_S {C_p(T) } \, dT$

C_p(T) = a + bT + CT^2 +dT^3

H_2O

a=32.218

b=o.192 * 10^-2

c= 1.055 * 10^-5

d= -3.593 * 10^-9

ΔH_2 = ∫(a+bT +a^2 +dT^3)dT (of limits 100 to 500 )

= -13497.6

ΔH_3 =∫(a +bT +T^2 +dT^3)dT (of limit 95 to 100)

= -2751.331

phase change

ΔH_2 =  - ΔH_v

ΔH_v = 7.3  @ 100° C

liquid → vapor ⇒ ΔH_v

vapor → liquid ⇒ -ΔH_v

Q= n°(-16324.231)

=350 *$10^{-5}$ (-16324.231)

=$\frac{350* 10^{-5} }{18}$ (-16324.231) *$\frac{kg}{hr}$$\frac{J}{mol}$

=$\frac{350* 10^{-5} }{18}$ (-16324.231) * \frac{J}{mol}[/tex]

=$\frac{350* 10^{-5} }{18}$  (-16324.231)  *kW

Q= -8.8171*$10^{-4}$ kW