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loris [4]
2 years ago
13

The homogeneous, reversible, exothermic, liquid phase reaction: A근R Is being carried out in a reactor system consisting of an id

eal PFR followed by an ideal CSTR. The volume of the PFR is 75,000 L, and the volume of the CSTR is 150,000 L. The PFR operates at 150 °C, and the CSTR operates at 125 °C. The volumetric flow rate entering the PFR is 55,000 L/h, the concentration of A in this stream is 6.5 M, and the concentration of R is zero. The reaction is first order in both directions. The rate constant of the forward reaction at 150 °C is 1.28 h and the equilibrium constant based on concertation at 150 °C is 2.5. The rate constant at 125 °C is 0.280 h and the equilibrium constant is 3.51. (a) What is the rate of R leaving the CSTR (mol/h)? (b) The flowrate to the PFR is increased so that the fractional conversion of A is 0.5 in the efluent from the CSTR. All other parameters remain unchanged. What is the new production rate of R?

Engineering
1 answer:
baherus [9]2 years ago
3 0

Answer:

attached below

Explanation:

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Answer:

second-law efficiency  = 62.42 %

Explanation:

given data

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solution

as we know that thermal efficiency of reversible heat engine between same  temp reservoir

so here

efficiency ( reversible ) η1 = 1 - \frac{T2}{T1}      ............1

efficiency ( reversible ) η1  = 1 - \frac{293}{1473}  

so efficiency ( reversible ) η1  = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = \frac{thernal\ efficiency}{0.801}

second-law efficiency = \frac{50}{0.801}  

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