An expandable container is filled with 45 in.3 of air and is sitting in ice water that is 32°F. At what volume in in.3 will the
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Answer:
hello your question has some missing information attached to the answer is the missing component
Answer : αaxial,p = -6.034 ksi ( compressive )
αbend,p = 19.648 ksi ( tensile )
Explanation:
αaxial, p = equation 1
αbend, p = equation 2
P = load = 35 kips
A = area of column = 5.8
d = column cross section depth = 9.5 in
= 55.0
Hence equation 1 becomes
αaxial,p = -35 / 5.8 = - 6.034 ksi ( compressive )
equation 2 becomes
αbend, p = = + 19.648 ksi ( tensile )
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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