Air from a workspace enters the air conditioner unit at 30°C dry bulb and 25°C wet bulb temperatures. The air leaves the air con
ditioner and returns to the space at 25°C dry bulb and 6.5°C dew point temperature (ϕ2≠100%). If there is any, the condensate leaves the air conditioner at the same temperature of the air leaving the cooling coils (i.e., different from our previous assumption that condensate has same temperature of the cooling coils). The volume flow rate of the air returned to the workspace is 1000 m3 /min (Rair=0.287 kJ/kg∙K). Atmospheric pressure is 101.325 kPa. Compute
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Answer:
The conversion in the real reactor is = 88%
Explanation:
conversion = 98% = 0.98
process rate = 0.03 m^3/s
length of reactor = 3 m
cross sectional area of reactor = 25 dm^2
pulse tracer test results on the reactor :
mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2
note: space time (t) =
t = Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor
therefore (t) = = 25 s
since the reaction is in first order
X = 1 -
= 1 - X
kt = In
k = In / t
X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then
K = 0.156
Calculating Da for a closed vessel
; Da = tk
= 25 * 0.156 = 3.9
calculate Peclet number Per using this equation
0.65 =
therefore
solving the Non-linear equation above( Per = 1.5 )
Attached is the Remaining part of the solution
