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Heat Supplied and Heat Loss of a Furnace


Date: 05/17/2000 at 21:08:02
From: Peter Lehman
Subject: Differential Equations

A furnace is switched on at 9 a.m. Heat is supplied at a constant 
rate, but as the furnace increases in temperature, heat is lost at a 
rate determined by the difference between its temperature and the 
temperature of its surroundings. The temperature of both the furnace 
and its surroundings was 20 C at 9:00 a.m. At 9.30 a.m the furnace was 
200 C and by 10:00 a.m it was 350 C. The minimum operating temperature 
is 800 C. Use a differential equation to find the time when the 
furnace is ready for use, and the highest temperature it could reach.

It is just the differential equation I need help on. The rest is okay.


Date: 05/18/2000 at 08:16:28
From: Doctor Anthony
Subject: Re: Differential Equations

Suppose that heat is added at H units per min. It will be lost at a 
rate proportional to (T-20).

The increase in heat per unit time, dh/dt will be given by:

     dh/dt = H - k(T-20)

where k is a constant of proportionality.

For a constant mass and specific heat we can say that dT/dt is 
proportional to dh/dt and rewrite the differential equation in the 
form:

              dT/dt = a - bT

          dT/(a-bT) = dt

     -(1/b)ln(a-bT) = t + C

           ln(a-bT) = -bt + C'

               a-bT = A.e^(-bt)

                 bT = a - A.e^(-bt)

                  T = (1/b)[a - A.e^(-bt)]

at t = 0, T = 20 and so:

      20 = (1/b)[a - A]

     20b = a - A  so  a = A+20b

at t = 30, T = 200:

      200 = (1/b)[A+20b - A.e^(-30b)]

     200b = A + 20b - A.e^(-30b)

     180b = A[1 - e^(-30b)]

        A = 180b/[1-e^(-30b)]

at t = 60, T = 350:

                            350 = (1/b)[a - A.e^(-60b)]

                           350b = a - A.e^(-60b)

                           350b = A + 20b - A.e^(-60b)

                           330b = 180b[1 - e^(-60b)]/[1-e^(-30b)]

                330[1-e^(-30b)] = 180[1-e^(-60b)]

                11 - 11e^(-30b) = 6 - 6e^(-60b)

     6e^(-60b) - 11e^(-30b) + 5 = 0

putting e^(-30b) = u, this equation is a quadratic in u.

     6u^2 - 11u + 5 = 0

        (6u-5)(u-1) = 0

so u = 1 or u = 5/6

If u = 1, e^(-30b) = 1  and b = 0  (we can ignore this solution)

so u = 5/6,  e^(-30b) = 5/6

                 -30b = ln(5/6)

                    b = 0.0061

Then

     A = 180b/[1-e^(-30b)] = 180b/[1-5/6] = 180b/(1/6) = 1080b

       = 6.5636

and

     a = A+20b = 6.5636 + 0.1215 = 6.68515

Finally we have:

     T = (1/b)[a - A.e^(-bt)]

       =  164.5[6.68515 - 6.5636e^(-0.0061t)]

       =  1100.00 - 1080 e^(-0.0061t)

Clearly the highest possible temperature will be 1100 degrees.

Time to 800 degrees is given by:

                   800 = 1100 - 1080 e^(-0.0061t)

     1080 e^(-0.0061t) = 300

          e^(-0.0061t) = 5/18

              -0.0061t = ln(5/18)

                     t = 209.99 minutes

And so time to 800 degrees = 210 minutes.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Physics/Chemistry

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