A tank contains 100 gallons of water. Five gallons of brine per minute flow into the tank, each gallon of brine containing 1 pound of salt.

Question

A tank contains 100 gallons of water. Five gallons of brine per minute flow into the tank, each gallon of brine containing 1 pound of salt. Five gallons of brine flow out of the tank per minute. Assume that the tank is kept well stirred.A) find a differential equation for the number of lbs of salt in the tank (call it A)
B) assume the tank initially contains 10 lbs of salt, solve this differential equation.
C) how much salt in the tank after 10 minutes?
D) at what time will there be 199 lbs of salt in the tank?

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Anna 2 weeks 2021-11-19T07:21:06+00:00 1 Answer 0 views 0

Answers ( )

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    2021-11-19T07:22:34+00:00

    Answer:

    A) (dA/dt) = 5 – 0.05A

    B) A(t) = 100 – 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

    C) At t = 10 mins, A = 45.42 g

    D) When A = 199 lb, t = 62.1 minute

    Step-by-step explanation:

    First of, we take the overall balance for the system,

    Let V = volume of solution in the tank at any time = 100 gallons (constant because flowrate in and out is the same)

    Rate of flow into the tank = Fᵢ = 5 gallons/min

    Rate of flow out of the tank = F = 5 gallons/min

    A) Component balance for the concentration.

    Let the initial amount of salt in the tank be A₀

    The rate of flow of salt coming into the tank be 1 lb/gallon × 5 gallons/min = 5 lb/min

    Amount of salt in the tank, at any time = A

    Rate of flow of salt out of the tank = (A × 5 m³/min)/V = (5A/V) g/min

    But V = 100 gallons

    Rate of flow of salt out of the tank = 0.05A lb/min

    The balance,

    Rate of Change of the amount of salt in the tank = (rate of flow of salt into the tank) – (rate of flow of salt out of the tank)

    (dA/dt) = 5 – 0.05A

    B) dA/(5 – 0.05A) = dt

    ∫ dA/(5 – 0.05A) = ∫ dt

    Integrating the left hand side from A₀ to A and the right hand side from 0 to t

    – 20 In [(5 – 0.05A)/(5 – 0.05A₀)] = t

    In (5 – 0.05A) – In (5 – 0.05A₀) = – 0.05t

    In (5 – 0.05A) = In (5 – 0.05A₀) – 0.05t

    A₀ = 10 lb

    In (5 – 0.05A) = (In 4.5) – 0.05t

    In (5 – 0.05A) = 1.504 – 0.05t

    (5 – 0.05A) = e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

    A(t) = 100 – 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

    C) When t = 10 min

    A = 100 – 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

    1.504 – 0.05 (10) = 1.004

    A = 100 – 20e⁽¹•⁰⁰⁴⁾

    A = 100 – 54.58

    A = 45.42 g

    D) When A = 199

    A = 100 – 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

    199 = 100 – 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

    20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾ = 99

    e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾ = 4.95

    -(1.504 – 0.05t) = In 4.95 = 1.599

    0.05t = 1.599 + 1.504 = 3.103

    t = 62.1 minute

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