Assignment 1 Solution

1.

The reaction

A→ B

is to be carried out isothermally in a continuous-flow reactor. Calculate both CSTR and PFR reactor volumes necessary to consume 80% of A (Hint: CA = 0.2 CA0) when entering molar flow rate is 4 mol/hr and entering volumetric flow rate is 20 dm3/hr assuming the reaction rate –rA is kCA with k = 0.0001 s-1. CSTR We know that, Volume of CSTR can be calculated by the equation F X V = A0 A where X A = 0.8, CA = 0.2CA 0 , FA 0 = 4 mol / hour , k = 0.0001sec −1 −rA FA 0 4 mol = = 0.2 v0 20 lit F X 4 × 0.8 V = A0 A = = 222.2 dm3 −rA 0.0001× 3600 × 0.2 × 0.2

C A0 =

PFR Mole balance for a differential element of volume dV Input = Out put +Disappearance of A due to reaction Input = FA Output = FA + dFA

Disappearance of A = ( −rA ) dV

FA = FA + dFA − rA dV −dFA = −rA dV

, where − rA = kCA

− dFA = dV kC A −

d ( C Av0 ) = dV kC A

−

v0 dC A = dV k CA

CA

∫

−

CA0

−

v0 dC A V = dV k C A ∫0

v0 C A ln =V k C A0

V =−

2.

FA = C A v0

20 ln 0.2 = 89.413 dm3 3600 × 0.0001

Repeat Q1 to calculate the time necessary to consume 80% of species A in a 1000

dm3 constant volume batch reactor with CA0 = 0.2 mol/dm3. −rA = − −

dC A = kCA dt

dC A = kdt CA

CA

t

dC ∫C − CAA = ∫0 kdt A0 CA

t

− ln

CA = kt C A0

− ln

0.2C A0 = kt C A0

dC A −∫ = kdt C A ∫0 CA 0

k t = 1.609 t = 16090 Sec = 4.47 hr

3.

An isothermal PFR is used for the gas phase reaction A → 2B+C. The feed,

flowing at 2.0 L/s, contains 50% mol A and 50% mol inert species. The rate is first order with respect to A; the rate constant is 2.0 s-1. a) Determine the reactor volume required to consume 90% of A? b) If the reactor is replaced with CSTR, what is the reactor volume required to consume 99% of A? C A = 0.1CA 0 PT RT Assuming PT = 1 atm T=298.15 K

C A 0 = y A0

y A0 =

0.5 = 0.5 1

1 mol = 0.02 0.082 × 298.15 lit mol C A = 0.1CA 0 = 0.1× 0.02 = 0.002 lit C A0 = 0.5 ×

−dFA = −rA dV

, where − rA = kCA

− dFA = dV kC A −

d ( C Av0 ) = dV kC A

−

v0 dC A = dV k CA

FA = C A v0

CA

v0 dC A V ∫ − k C A = ∫0 dV CA0 −

v0 C A ln =V k C A0

V =−

2.0 0.002 ln = 2.303 dm3 2.0 0.02

b.

V=

FA0 X A −rA

V=

C A0 × v0 × X A CA 0 × v0 × X A CA0 × 2 × 0.99 = = = 9.9 dm3 −rA kCA 2.0 × 0.1× CA 0

where X A = 0.99, CA 0 = 0.02

mol mol , CA = 0.1CA 0 = 0.1× 0.02 = 0.002 , lit lit