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:
Explanation:
Given conditions
1)The stress on the blade is 100 MPa
2)The yield strength of the blade is 175 MPa
3)The Young’s modulus for the blade is 50 GPa
4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.
5)The temperature of the blade is 800°C.
6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)
where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K
Young Modulus, E = Stress, /Strain, ∈
initial Strain,
creep rate in the steady state
but Tinitial = 0
solving the above equation,
we get
Tfinal = 2459.82 hr
Answer:
51.4 Ohms
Explanation:
By applying voltage division rule
where v is voltage, subscripts i and f represnt initial and final, R is resistance, m is internal and l is external.Substituting 7V for final voltage, 10V for initial voltage and the external resistance as 120 Ohms then