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/
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