Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

please help Mathematical Modelling first order differential equation (newtons law of heating and cooling)?

the temperature outside the building decreases at a constant rate of 2 degree per hour. The inside of the building is heated by furnace, and there is no other source of heating or cooling. The furnace was switched on at time t=0, when the temperature inside was T* = 18, and the temperature outside was 0 degC. Assume that the furnace generates a constant heat of H= 60,000 Btu/hr when it is working, the heat capacity of the building is β = ¼ degrees per thousand Btu, and the time constant for heat transfer between the outside and the inside of the building is τ = 2 hr. On the basis of Newton’s law of cooling, find the upper value of the temperature in the building.

I tried solving this Newton’s law of cooling with one heating source and wrote out the ODE for the temperature x(t) in the building....then I added initial values to get an IVP, after I solved the IVP, applied calculus and found the time when x(t) reaches a maximum...the problem is that The answer I got is no where close to the estimate...please help...thanks a lot

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  • hfshaw
    Lv 7
    1 decade ago
    Favorite Answer

    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 first-order 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

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