Answer:
Gs = 2.647
e = 0.7986
Explanation:
We know that moist unit weight of soil is given as

where,
= moist unit weight of the soil
Gs = specific gravity of the soil
S = degree of saturation
e = void ratio
= unit weight of water = 9.81 kN/m3
From data given we know that:
At 50% saturation,
puttng all value to get Gs value;

Gs - 1.194*e = 1.694 .........(1)
for saturaion 75%, unit weight = 17.71 KN/m3

Gs - 1.055*e = 1.805 .........(2)
solving both equations (1) and (2), we obtained;
Gs = 2.647
e = 0.7986
Answer:
2062 lbm/h
Explanation:
The air will lose heat and the oil will gain heat.
These heats will be equal in magnitude.
qo = -qa
They will be of different signs because one is entering iits system and the other is exiting.
The heat exchanged by oil is:
qo = Gp * Cpo * (tof - toi)
The heat exchanged by air is:
qa = Ga * Cpa * (taf - tai)
The specific heat capacity of air at constant pressure is:
Cpa = 0.24 BTU/(lbm*F)
Therefore:
Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)
Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))
Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
Answer:
4 number answer is correct.
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:
a) The rate of heat transfer will be 19.58 Watts.
b) The temperature drop of the hot water will be 0.024 Degree Celcius.
Explanation: