Answer:
a) 0.684
b) 0.90
Explanation:
Catalyst
EO + W → EG
<u>a) calculate the conversion exiting the first reactor </u>
CAo = 16.1 / 2 mol/dm^3
Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst
Vo = 7.24 dm^3/s
Vm = 800 gal = 3028 dm^3
hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins
next determine the value of conversion exiting the reactor ( Xai ) using the relation below
KIm =
------ ( 1 )
make Xai subject of the relation
Xai = KIm / 1 + KIm --- ( 2 )
<em>where : K = 0.311 , Im = 6.97 ( input values into equation 2 )</em>
Xai = 0.684
<u>B) calculate the conversion exiting the second reactor</u>
CA1 = CA0 ( 1 - Xai )
therefore CA1 = 2.5438 mol/dm^3
Vo = 7.24 dm^3/s
To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below
XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )
<em> where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )</em>
XA2 = 0.90
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Answer:
7.94 ft^3/ s.
Explanation:
So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.
Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.
Therefore; kp/ks = 1/6.
Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.
Reynolds scaling==> Hp/ 700 = (1/6)^2.5.
= 7.94 ft^3/ s
Answer:
Q = -68.859 kJ
Explanation:
given details
mass 
initial pressure P_1 = 104 kPa
Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K
final pressure P_2 = 1068 kPa
Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K
we know that
molecular mass of 
R = 8.314/44 = 0.189 kJ/kg K
c_v = 0.657 kJ/kgK
from ideal gas equation
PV =mRT






WORK DONE

w = 586*(0.1033 -0.514)
W =256.76 kJ
INTERNAL ENERGY IS



HEAT TRANSFER

= 187.902 +(-256.46)
Q = -68.859 kJ
Answer:
of 5 lb/ft and a concentrated service live load at midspan. .... length = 12 feet) to support a uniformly distributed load. Taking ... w 7..'{ 'f.- ~ s-·. 344 ft-kip. Fy : s-o ks I. 299 ft-kip. Li.. ::::- I 2.. }-t-. 150 ft-kip ..... The concrete and reinforcing steel properties are ... Neglecting beam self-weight . and based only on the ...... JI : Lf, 2. l.. ;VI.
Explanation: