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:
c. less than 60 mi/h
Explanation:
To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.
Total Distance Traveled = S = 100 mi + 100 mi
S = 200 mi
Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.
Total Time = t = Time from A to B + Time from B to C
t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)
t = 2 h + 1.43 h
t = 3.43 h
Now, the average speed of bus will be given as:
Average Speed = V = S/t
V = 200 mi/3.43 h
<u>V = 58.33 mi/h</u>
It is clear from this answer that the correct option is:
<u>c. less than 60 mi/h</u>
Answer:
Those products are generally called Work in Process WIP
Explanation:
Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.
These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.
Answer:
The correct option is;
c. the exergy of the tank can be anything between zero to P₀·V
Explanation:
The given parameters are;
The volume of the tank = V
The pressure in the tank = 0 Pascal
The pressure of the surrounding = P₀
The temperature of the surrounding = T₀
Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change
The exergy balance equation is given as follows;
![X_2 - X_1 = \int\limits^2_1 {} \, \delta Q \left (1 - \dfrac{T_0}{T} \right ) - [W - P_0 \cdot (V_2 - V_1)]- X_{destroyed}](https://tex.z-dn.net/?f=X_2%20-%20X_1%20%3D%20%5Cint%5Climits%5E2_1%20%7B%7D%20%5C%2C%20%5Cdelta%20Q%20%5Cleft%20%281%20-%20%5Cdfrac%7BT_0%7D%7BT%7D%20%5Cright%20%29%20-%20%5BW%20-%20P_0%20%5Ccdot%20%28V_2%20-%20V_1%29%5D-%20X_%7Bdestroyed%7D)
Where;
X₂ - X₁ is the difference between the two exergies
Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system
Therefore, the exergy of the tank can be anything between zero to P₀·V.
Answer:
public static int average(int j, int k) {
return (int)(( (long)(i) + (long)(j) ) /2 );
}
Explanation:
The above code returns the average of two integer variables
Line 1 of the code declares a method along with 2 variables
Method declared: average of integer data type
Variables: j and k of type integer, respectively
Line 2 calculates the average of the two variables and returns the value of the average.
The first of two integers to average is j
The second of two integers to average is k
The last parameter ensures average using (j+k)/2