The following problem deals with the open three-tank system. Fresh water flows into tank 1; mixed brine flows from tank 1 into tank 2, from tank 2 into tank 3, and out of tank 3; all at the given flow rate r gallons per minute. The initial amounts x1(0)= x0 (lb), x2(0) = 0, and x3(0) = 0 of salt in the three tanks are given, as are their volumes V1, V2, and V3 (in gallons). First solve for the amounts of salt in the three tanks at time t , then determine the maximal amount of salt that tank 3 ever contains. Finally, construct a figure showing the graphs of x1 (t) , x2(t) , and x3(t)r = 60, x0 = 45, V1 = 15, V2 = 10, V3 =30

Answered: The following problem deals with the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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The following problem deals with the open three-tank system. Fresh water flows into tank 1; mixed brine flows from tank 1 into tank 2, from tank 2 into tank 3, and out of tank 3; all at the given flow rate r gallons per minute. The initial amounts x₁(0)= x₀ (lb), x2(0) = 0, and x₃(0) = 0 of salt in the three tanks are given, as are their volumes V₁, V₂, and V₃ (in gallons). First solve for the amounts of salt in the three tanks at time t , then determine the maximal amount of salt that tank 3 ever contains. Finally, construct a figure showing the graphs of x₁ (t) , x₂(t) , and x₃(t)

r = 60, x₀ = 45, V₁ = 15, V₂ = 10, V₃ =30

Expert Solution

Step 1

Given:

Open three tank system

Water flows from tank 1. Mixed brine flows from tank 1 to tank 2, tank 2 to tank 3, and out of tank 3.

Initial amounts $x_{1} (0) = x_{0} (l b), x_{2} (0) = 0, x_{3} (0) = 0$

$r = 60, x_{0} = 45, V_{1} = 15, V_{2} = 10, V_{3} = 30$

To find:

(a) Amount of salt in three tanks.

(b) Maximal amount of salt that tank 3 ever contains.

Step 2

By definition of problem,

${(x_{1})}^{'} = - k_{1} x_{1}, {(x_{2})}^{'} = k_{1} x_{1} - k_{2} x_{2}, {(x_{3})}^{'} = k_{2} x_{2} - k_{3} x_{3} w h e r e, k_{i} = \frac{r}{V_{i}}; i = 1, 2, 3 H e r e, r = 60, V_{1} = 15, V_{2} = 10, V_{3} = 30 T h e r e f o r e, k_{1} = 4, k_{2} = 6, k_{3} = 2 . N o w, {(x_{1})}^{'} = - 4 x_{1}, {(x_{2})}^{'} = 4 x_{1} - 6 x_{2}, {(x_{3})}^{'} = 6 x_{2} - 2 x_{3} . [\begin{matrix} {x_{1}}^{'} \\ {x_{2}}^{'} \\ {x_{3}}^{'} \end{matrix}] = X^{'} = [\begin{matrix} - 4 & 0 & 0 \\ 4 & - 6 & 0 \\ 0 & 6 & - 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] N o w l e t u s f i n d i t s e i g e n v a l u e [\begin{matrix} - 4 - λ & 0 & 0 \\ 4 & - 6 - λ & 0 \\ 0 & 6 & - 2 - λ \end{matrix}] = 0 o r, λ = - 4, - 6, - 2 .$

For $λ_{1} = - 4,$

Eigen vector corresponding to $λ_{1} = - 4,$

$(A - λ_{1} I) B_{1} = 0 o r, [\begin{matrix} 0 & 0 & 0 \\ 4 & - 2 & 0 \\ 0 & 6 & 2 \end{matrix}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] O n s i m p l i f y i n g, w e g e t a = k, b = 2 k, c = - 6 k S o, B_{1} = [\begin{matrix} - 1 \\ - 2 \\ 6 \end{matrix}]$

For $λ_{2} = - 6,$

Eigen vector corresponding to $λ_{2} = - 6$ ,

$(A - λ_{2} I) B_{2} = 0 o r, [\begin{matrix} 2 & 0 & 0 \\ 4 & 0 & 0 \\ 0 & 6 & 4 \end{matrix}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] O n s i m p l i f y i n g, w e g e t a = 0, b = - 2 k, c = 3 k S o, B_{2} = [\begin{matrix} 0 \\ - 2 \\ 3 \end{matrix}]$

For, $λ_{3} = - 2,$

Eigen vector corresponding to $λ_{3} = - 2,$

$(A - λ_{3} I) B_{3} = 0 o r, [\begin{matrix} 2 & 0 & 0 \\ 4 & - 4 & 0 \\ 0 & 6 & 0 \end{matrix}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] O n s i m p l i f y i n g, w e g e t a = 0, b = 0, c = k S o, B_{3} = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}]$

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