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?
1 answer:
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Answer:
Load carried by shaft=9.92 ft-lb
Explanation:
Given: Power P=4.4 HP
P=3281.08 W
<u><em>Power: </em></u>Rate of change of work with respect to time is called power.
We know that P=
rad/sec
So that P=
So 3281.08=
T=13.45 N-m (1 N-m=0.737 ft-lb)
So T=9.92 ft-lb.
Load carried by shaft=9.92 ft-lb
Answer:
So the fraction of proeutectoid cementite is 44.3%
Explanation:
Given that
cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.
We know that
Fraction of proeutectoid cementite phase gievn as
Now by putting the values
So the fraction of proeutectoid cementite is 44.3%