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:
P = 80.922 KW
Explanation:
Given data;
Length of load arm is 900 mm = 0.9 m
Spring balanced read 16 N
Applied weight is 500 N
Rotational speed is 1774 rpm
we know that power is given as

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm
angular speed
Therefore Power is

P = 80.922 KW
Wait why do you want me to
Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.
Answer:

Explanation:
Given data:
Ammonia Nitrogen 30 mg/L
pH = 8.5
-log[H +] = 8.5
[H +] = 10^{-8.5}

Rate constant is given as
...........1

Total ammonia as NItrogen is given as 30 mg/l
![\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}](https://tex.z-dn.net/?f=%5C%25NH_4%5E%7B%2B%7D%20%3D%20%5Cfrac%7B%20%5BNH_4%5D%20%5Ctimes%20100%7D%7B%5BNH_4%5E%7B%2B%7D%5D%20%2B%20%5BNH_3%5D%7D)
= 
.....2
from equation 1 we have
{10^{8.5}}
plug this value in equation 2 we get

Total ammonia as N = 30 mg/lt
