Exposicion Ejercicio 3-3 y 3-5.
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Transcript of Exposicion Ejercicio 3-3 y 3-5.
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8/12/2019 Exposicion Ejercicio 3-3 y 3-5.
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EXERCISE 3-3 AND 3-5 Jos David Arzuaga Garrido
Edgar De Jess Campo Castrilln
Eliana Paola Castillo Vergara
Maury Esther Romero Cabarcas
Marcela Patricia Torres Len
UNIVERSITY OF CARTAGENA
CHEMICAL ENGINEERING
2014
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EXERCISE 3-3
A storage tank has a diameter of 20ft and a height of 10ft. The output volumetric flow
tank is given by = 2
Where is the height of the liquid in the tank. At a particular time the tank is at ste
with an input flow of 10
.
a) What is the steady -state liquid height in the tank?
b) If the input flow is ramped up at the rate of 0.1
, how many minutes will it take fo
to overflow?
D
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From exercise statements
= 20
= 10
The output volumetric flow from the tank is given by:
= 2
= 10
a)
Mass balance in the tank in steady state
=
() () =
Considering =
=
= 0
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= 0
= (1)
2
= 10
() =10
2
= 5
b)Considering that input flow is ramped up at the rate of 0,1
= 0,1 ()
Mass Balance
() () =
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() () =
But
=
() () =
=
2
Its known that = 2
Substituting in equation (2)
2 =
(3)
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Now, Mass balance in steady- state (ss)
2 =
(4)
Applying deviation variable
= ()
=
Subtracting equation (3) and (4)
= 2
Dividing by 2
2
=
1
2
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=0,1
2
2+ 1
=0.05
157 + 1(5)
Multiplying equation (5) by 1/157
=3.18 10
+ 6.36 10
Applying Partial fractions
3.18 10
+ 6.36 10=
+
+
( + 6.36 10 )
Then
3.18 10
+ 6.36 10=
( + 6.36 10) + + 6.36 10 +
+ 6.36 10
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Solving for A, B and C
= 7.861
= 0.05
= 7,861
Then
= 7.861
+
0.05
+
7.861
+ 6.36 10
Applying Inverse Laplace Transforms
=
7.861
+
0.05
+
7.861
+ 6.36 10
= 7.861 1
+ 0.05
1
+ 7.861
1
+ 6.36 10
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Then
= 7.861 + 0.05 + (7.861 (.))
By definition
=
Solving for
= +
Then
= 7.861 + 0.05 + (7.861 (.)
) + 5
Solving for t, when = 10
7.861 + 0.05 + (7.861 (.)) + 5 = 10
= 217.896
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EXERCISE 3-5
Hot water at a rate of 2 liters per minute (constant) and temperature is mixed w
cold water at a constant rate of 3 liters per minute and a constant temperature of 20
Both streams flow into a bathtub, but because of carelessness, the water is overflow
and keeping the bathtub full of water. The volume of the bathtub is 100 liters. Assum
the water in the bathtub is perfectly mixed, derive the differential equation relating
temperature in the bathtub, ,to the temperature of the hot water, , Obtain
transfer function , and calculate its gain and time constant.
= 3
;
=100 L
= 2
;
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Energy balance
+ =
Where
= + + And K and P is negligible; = =
+ =
=
+ =[ ]
= 2
; = 3
; = 20 ; = 100
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Assuming constant and equal the heat capacity and Density, in all line, we hav
+ =
(1)
Steady-state balance: + =
(2)
Subtracting the equation (2) to the equation (1)
=
(3)
Keep in mind that = + divides the equation (3) between
+ =
+
=
+ ;
=
+
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=
Apply Laplace
() = [ (0)]
And solving for , we have:
() =
(1 + )()
Where
=
+ =
100
2 / + 3 /= 20 ;
= + = 2 /
2 / + 3 /= 0,4
()
()=
0,4
(1 + 20)
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THANKS
