The ester of an organic base is hydrolyzed in a CSTR. The rate of this irreversi
ID: 1009306 • Letter: T
Question
The ester of an organic base is hydrolyzed in a CSTR. The rate of this irreversible reaction is first-order in each reactant. The liquid volume in the vessel is 6500 L. a jacket with coolant at 18 Celsius maintains the reactant mixture at 30 celsius.
Additional data:
Ester feed stream: 1 M, 30 oC, 20 L/s
Base feed stream: 4M, 30 Celsius, 10 L/s
Rate constant= 1014 exp(-11,000/T) (Lmol-1s-1, T in K )
Hr= - 45 kcal/mol ester.
The average heat capacity (based on volume of the mixture) is approximately constant at 1.0 kcalL-1 oC-1.
a. What is the conversion of ester in the reactor?
b. Calculate the rate at which energy must be removed to the jacket to maintain 30 oC in the reactor. If the heat transfer coefficient is 15 kcal s-1m-2K-1, what is the necessary heat transfer area?
c. If the coolant supply fails, what would be the maximum temperature the reactor could reach?
Explanation / Answer
The ester of an organic base is hydrolyzed in a CSTR. The rate of this irreversible reaction is first-order in each reactant. The liquid volume in the vessel is 6500 L. a jacket with coolant at 18 Celsius maintains the reactant mixture at 30 celsius.
Additional data:
Ester feed stream: 1 M, 30 oC, 20 L/s
Base feed stream: 4M, 30 Celsius, 10 L/s
Rate constant= 1014 exp(-11,000/T) (Lmol-1s-1, T in K )
Hr= - 45 kcal/mol ester.
The average heat capacity (based on volume of the mixture) is approximately constant at 1.0 kcalL-1 oC-1.
a)What is the conversion of ester in the reactor
the conversion of ester in the reactor is
higer the concentration is expected in PFR.a PFR opeates at a conditions that change down the length of the reactor.a cstr opertes at the exit conditionsand would allow the reaction to program to the formation of cat the max rate of consumption of B(at the max,B concentration)
the first order reaction is xA=kt/(1+kt)
no shift ,mole balance does not depend on concentration for first order reaction
At exit of the reactor
dt/dv=ua(ta-t)=30-18
t=ta=12c
b)Calculate the rate at which energy must be removed to the jacket to maintain 30 oC in the reactor. If the heat transfer coefficient is 15 kcal s-1m-2K-1, what is the necessary heat transfer area?
Heat transfer is all about the transfer of heat from one point to another. If we consider any system which will be at higher temperature compared to surroundings, there will be transfer of heat from system to the surroundings.
Heat transfer is given by
Q=m*c* T
Where,m is the mass,
C is the specific heat and
T is the temperature difference in K.
Q=7000*15*30
=3150000j
c)If the coolant supply fails, what would be the maximum temperature the reactor could reach?
the coolant supply fails then heating will raise in the reactor
30-18=22
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