Yahoo Answers: Answers and Comments for Please help Mathematical Modelling first order differential equation (newtons law of heating and cooling)? [Mathematics]
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Mon, 22 Mar 2010 23:53:33 +0000
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Yahoo Answers: Answers and Comments for Please help Mathematical Modelling first order differential equation (newtons law of heating and cooling)? [Mathematics]
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https://ca.answers.yahoo.com/question/index?qid=20100322235333AA96NA9
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From hfshaw: Newton's law of cooling states that the ra...
https://ca.answers.yahoo.com/question/index?qid=20100322235333AA96NA9
https://ca.answers.yahoo.com/question/index?qid=20100322235333AA96NA9
Tue, 23 Mar 2010 07:16:56 +0000
Newton's law of cooling states that the rate of rate of temperature change of a object is proportional to the difference in temperature between the object and the surroundings. In this case, the temperature of the surroundings varies with time, and there is an internal source of thermal energy in the building, so this is more complicated than your usual problem.
The temperature of the surroundings at time t is given by:
Ts(t) = r*t, where r = 2 deg/hr
If T(t) is the temperature of the building, then the rate at which the temperature of the building changes, according to Newton's law of cooling is:
dT(t)/dt = k*(T(t)  Ts(t))
and we are told that k = 1/(2hr)
The rate of heat loss from the building is therefore dq_loss/dt = (1/β) dT(t)/dt, where β is the "heat capacity" given in the problem. (Note that in reality, heat capacity has units of energy/temperature, so what you are given here is the inverse of the heat capacity.)
Thermal energy is being added to the interior of the building at a constant rate of H = 60,000BTU/hr because of the furnace. The net heat balance is then given by:
dq/dt = H  dq_loss/dt = H + (1/β) dT(t)/dt
dq/dt = H  (1/β)*(k*(T(t)  Ts(t)))
Multiplying through by β converts us back to temperature instead of heat:
dT(t)/dt = H*β  k*T(t) + k*Ts(t))
Plugging in the expression for Ts(t):
dT(t)/dt = H*β  k*T(t) + k*r*t
This is a firstorder linear equation. In standard form we have:
dT/dt + k*T = H*β + k*r*t
This can easily be solved using an integrating factor, p(t):
p(t) = exp(INTEGRAL of {k dt})
p(t) = exp(k*t)
Then:
T(t) = exp(k*t) * INTEGRAL of {(H*β + k*r*t)*exp(k*t) dt}
T(t) = exp(k*t) * [(H*β/k)*exp(k*t) + r*t*exp(k*t)  (r/k)*exp(k*t) + c]
where c is the constant of integration.
T(t) = [r*t  r/k + H*β/k + c*exp(k*t)]
Plugging in the values for this problem:
r = 2 deg/hr
k = 1/2hr, so r/k = 4 deg
H = 60,000 BTU/hr and β = 0.25*10^3 BTU/hr, so:
H*β/k = (60,000 BTU/hr)(0.25*10^3 deg/BTU)/(1/2hr) = 30 deg
So:
T(t) = 34 deg  (2deg/hr)*t + c*exp(t/(2hr))
Now use the initial condition to solve for the constant.
T(0) = 18 deg = 34 deg + c
c = 16 deg
So the particular solution for this case is:
T(t) = 34 deg  (2deg/hr)*t  (16 deg)*exp(t/(2hr))
The temperature is a maximum when dT/dt = 0. Differentiating this solution gives:
dT/dt = 2 deg/hr + (8 deg/hr)*exp(t/(2hr))
Setting this to zero:
2 = 8*exp(t/(2hr))
ln(1/4) =  t/(2hr)
2hr*ln(4) = t
2.773 hr = t
The maximum temperature occurs at 2.773hrs. At this time, the temperature is:
T_max = 34 deg  (2deg/hr)*(2.773hr)  (16 deg)*exp(2.773/2)
T_max = 24.45 deg