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uranmaximum [27]
4 years ago
15

A batch chemical reactor achieves a reduction in concentration of compound A from 90 mg/l to 10 mg/L in one hour. if the reactio

n is known to follow zero order kinetics, determine the value of the rate constant in (mg/L.hr) unit.
Engineering
1 answer:
inessss [21]4 years ago
8 0

Answer:

value of the rate constant is 80 mg/L-hr

Explanation:

given data

concentration of compound Cao = 90 mg/l

Ca = 10 mg/L

time = 1 hour

solution

we use here  zero order reaction rate flow

-rA = K Ca

\frac{-dCa}{dt} = K

-d Ca = k dt

now we will integrate it on the both side by Cao to ca we get

- \int\limits^{Ca}_{Cao} {dCa} \, = K \int\limits^4_0 {dt} \,  

solve it we get

cao - Ca = K t

put here value  and we get K

90 - 10 = K 1

K = 80 mg/L-hr

so value of the rate constant is 80 mg/L-hr

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Engine horsepower decreases ________% for every___________feet above sea level.
Nitella [24]

Answer:

Engine horsepower decreases <u>3.5%</u> for every <u>1,000</u> feet above sea level.

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3 years ago
As resistors are added in series to a circuit, the total resistance will
olganol [36]

The equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

<h3>
What is resistance?</h3>

Resistance is the obstruction offered whenever the current is flowing through the circuit.

So the equivalent resistance is when three resistances are connected in series. When the resistances are connected in series then the voltage is different and the current remain same in each resistance.

V eq    =    V₁    +    V₂    +    V₃

IReq    =    IR₁    +    IR₂   +    IR₃

Req    =    R₁    +    R₂   +    R₃

Therefore the equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

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8 0
2 years ago
Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
kodGreya [7K]

Answer:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

h = 20 \frac{W}{m^2 K} represent the heat transfer coefficient.

p represent the perimeter for this case and would be given by:

p = 2*0.05m +2*0.001m= 0.102m

k = 200 \frac{W}{m C} represent the thermal conductivity

w = 5cm =0.05 m represent the width

h = 1mm =0.001m represent the thickness

A= wh= 0.05m *0.001m = 0.00005 m^2

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}

Where m is a coefficient given by:

m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}

The value of x for this case represent the distance x =5 cm =0.05m

T_b =130 C represent the base temperature

T_{\infty}= 20 represent the temperature of the sorroundings or the ambient.

If we replace we have this:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

3 0
3 years ago
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