Answer and explanation:
The graphical representation of the electronic eye
The state table showing
the present state
input
Next state and
the output
are shown in the attached file
After the load impedance has been transformed through the ideal transformer, its impedance is: + . Enter the real part in the fi
rst blank and the imaginary part in the second blank. If a value is negative, include the negative sign. Provide up to four digits of precision. If the exact value can be provided with fewer digits, merely provide the exact value. These instructions pertain to the following blanks as well. What is the total impedance seen by the source? + . What is the current phasor Ig (expressed in rectangular form)?
1 answer:
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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:
Explanation:
The additional power consumption of the car when v :
35 mi/h = 0.464hp
70 mi/h = 3.71hp.
For the step by step explanation of how we arrived at these answers, please go through the attached files.
