Explanation Given: Quantity of water in the tank is 500 gallons , quantity of salt per gallon of brine is 2 pounds and the rate at which brine is pumped into the tank is 5 gal / min . The mixed solution is pumped out at the same rate. Calculation: Calculate the rate of salt entering into the tank. R in = ( 2 lb gal ) ⋅ ( 5 gal min ) = 10 lb min Calculate the rate of outflow of salt. R out = ( A 500 lb gal ) ( 5 gal min ) = A 100 lb / min Consider the equation for the mixing of two fluids. d A d t = R i n − R o u t d A d t = 10 − A 100 d A d t + A 100 = 10 (1) Compare Equation (1) with d A d t + p ( t ) A = Q ( t ) . The value of p ( t ) is 1 100 and Q ( t ) is 10 . Calculate the integrating factor. I F = e ∫ p ( t ) d t = e ∫ 1 100 d t = e t 100 Solve the differential equation using the integrating factor e t 100 . d A d t ( e t 100 A ) = 10 e t 100 e t 100 A ( t ) = 10 ∫ e t 100 d t e t 100 A ( t ) = 1000 e t 100 + c A ( t ) = 1000 + c e − t 100 (2) Since the tank contains pure water initially, A ( 0 ) = 0 . Substitute 0 for t and A ( 0 ) = 0 in equation (2)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.7, Problem 24E

To determine

The concentrations of salt in the tank at time t, t=5 min and at t=∞, and the time at which the concentration of the salt in the tank is equal to one-half of the limiting value.

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Chapter 2.7, Problem 23E

Chapter 2.7, Problem 25E

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The concentrations of salt in the tank at time t , t = 5 min and at t = ∞ , and the time at which the concentration of the salt in the tank is equal to one-half of the limiting value.

Solution Summary: The author explains the concentrations of salt in the tank at time t, and the rate at which the salt is pumped out.

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The concentrations of salt in the tank at time t, t=5 min and at t=∞, and the time at which the concentration of the salt in the tank is equal to one-half of the limiting value.

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Chapter 2 Solutions