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Found in: Page 131

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

Building Temperature. In Section 3.3 we modeled the temperature inside a building by the initial value problem (13)$$\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{M}}\,{\bf{(t) - T}}\,{\bf{(t)}}} \right]{\bf{ + H}}\,{\bf{(t) + U}}\,{\bf{(t),}}\,\,{\bf{T}}\,{\bf{(}}{{\bf{t}}_{\bf{o}}}{\bf{) = }}{{\bf{T}}_{\bf{o}}}$$ , where M is the temperature outside the building, T is the temperature inside the building, H is the additional heating rate, U is the furnace heating or air conditioner cooling rate, K is a positive constant, and $${{\bf{T}}_{\bf{o}}}$$ is the initial temperature at time $${{\bf{t}}_{\bf{o}}}$$ . In a typical model, $${{\bf{t}}_{\bf{o}}}{\bf{ = 0}}$$ (midnight),$${{\bf{T}}_{\bf{o}}}{\bf{ = 6}}{{\bf{5}}^{\bf{o}}}$$, $${\bf{H}}\left( {\bf{t}} \right){\bf{ = 0}}{\bf{.1}}$$, $${\bf{U(t) = 1}}{\bf{.5}}\left[ {{\bf{70 - T(t)}}} \right]$$ and $${\bf{M(t) = 75 - 20cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}$$ . The constant K is usually between$$\frac{{\bf{1}}}{{\bf{4}}}{\bf{and}}\frac{{\bf{1}}}{{\bf{2}}}$$, depending on such things as insulation. To study the effect of insulating this building, consider the typical building described above and use the improved Euler’s method subroutine with$${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}}$$ to approximate the solution to (13) on the interval $$0 \le {\bf{t}} \le 24$$ (1 day) for $${\bf{k = 0}}{\bf{.2,}}\,{\bf{0}}{\bf{.4}}$$, and 0.6.

The temperature at midnight when $${\bf{k = 0}}{\bf{.2}}$$ is approx. $${\bf{68}}{\bf{.385}}$$.

The temperature at midnight when $${\bf{k = }}\,{\bf{0}}{\bf{.4}}$$ is approx. $${\bf{67}}{\bf{.050}}$$.

The temperature at midnight when $${\bf{k = 0}}{\bf{.6}}$$ is approx. $${\bf{65}}{\bf{.974}}$$.

See the step by step solution

Step 1: Important hint.

To get the result apply Euler’s formula.

Step 2: Find the value of temperature when $${\bf{K = 0}}{\bf{.2}}$$.

The given equation is

$$\begin{array}{c}\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{M}}\,{\bf{(t) - T}}\,{\bf{(t)}}} \right]{\bf{ + H}}\,{\bf{(t) + U}}\,{\bf{(t),}}\,\,{\bf{T}}\,{\bf{(}}{{\bf{t}}_{\bf{o}}}{\bf{) = }}{{\bf{T}}_{\bf{o}}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = K}}\,\,\left[ {{\bf{75 - 20}}\,{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - T}}\,{\bf{(t)}}} \right]{\bf{ + 0}}{\bf{.1 + 1}}{\bf{.5(70 - T}}\,{\bf{(t))}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = 75}}\,{\bf{K + 105}}{\bf{.1 - 20}}\,{\bf{K}}\,{\bf{cos}}\,\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - (K + 1}}{\bf{.5)}}\,{\bf{T}}\,{\bf{(t)}}\\{\bf{T(0) = 65}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = 120}}{\bf{.1 - 4}}\,{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.7}}\,{\bf{T}}\,{\bf{(t)}}\end{array}$$

Now apply improved Euler’s method subroutine with $${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}} \approx 0.6667$$ and $${\bf{N = 36}}$$.

$$\begin{array}{c}{\bf{f}}\,{\bf{(t,T) = 120}}{\bf{.1 - 4}}\,{\bf{cos}}\,\,\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.7}}\,{\bf{T}}\,{\bf{(t)}}\\{\bf{F = f}}\,{\bf{(t,T)}}\\{\bf{ = 120}}{\bf{.1 - 4}}\,{\bf{cos}}\,\,\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.7T}}\,{\bf{(t)}}\\{\bf{G = f}}\,{\bf{(t + h,T + h}}\,{\bf{F)}}\end{array}$$

Apply initial conditions$${\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,T = }}{{\bf{T}}_{\bf{o}}}{\bf{ = 65}}$$.

$$\begin{array}{c}{\bf{F(0,65) = 5}}{\bf{.6}}\\{\bf{G(0,65) = 0}}{\bf{.6862}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.6667}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 66}}{\bf{.638}}\end{array}$$

Therefore at 0.6667h after midnight which is 12.40 AM, the temperature is approx. 66.638.

Now apply the same procedure for the period of 24h.

Step 3: Get the value for 36 steps.

Since there are 36 steps so by construct a table to get the required result.

 Time $${{\bf{t}}_{\bf{0}}}$$ $${{\bf{T}}_{\bf{o}}}$$ Midnight 0 65 12:40 AM 0.667 66.638 2:00 AM 2 68.073 4:00 AM 4 69.073 6:00 AM 6 70.301 8:00 AM 8 71.484 10:00 AM 10 72.437 12:00 PM 12.001 72.909 2:00 PM 14.001 72.775 4:00 PM 16.001 72.071 6:00 PM 18.001 70.985 8:00 PM 20.001 69.809 10:00 PM 22.001 68.857 MIDNIGHT 24.001 68.385

Step 4: Determine the value of temperature when $${\bf{K = 0}}{\bf{.4}}$$.

The given equation is

$$\begin{array}{l}{\bf{T(0) = 65}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = }}135.{\bf{1 - }}8{\bf{cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}.9{\bf{T(t)}}\end{array}$$

Now apply improved Euler’s method subroutine with $${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}} \approx 0.6667$$ and $${\bf{N = 36}}$$.

$$\begin{array}{c}{\bf{f(t,T) = 135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9T(t)}}\\{\bf{F = f(t,T) = 135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9T(t)}}\\{\bf{G = f(t + h,T + hF)}}\\{\bf{ = 135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi (t + 0}}{\bf{.6667)}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9(T + 0}}{\bf{.6667(135}}{\bf{.1 - 8cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 1}}{\bf{.9T))}}\end{array}$$

Apply initial conditions $${\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,T = }}{{\bf{T}}_{\bf{o}}}{\bf{ = 65}}$$.

$$\begin{array}{c}{\bf{F(0,65) = 3}}{\bf{.6}}\\{\bf{G(0,65) = - 0}}{\bf{.838678}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.6667}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 65}}{\bf{.920}}\end{array}$$

Thus, at 0.6667h after midnight which is 12.40 AM, the temperature is approx. 65.920.

Now apply the same procedure for the period of 24h.

Step 5: Get the value for 36 steps.

Since there are 36 steps so by construct a table to get the required result.

 Time $${{\bf{t}}_{\bf{0}}}$$ $${{\bf{T}}_{\bf{o}}}$$ Midnight 0 65 12:40 AM 0.667 65.92 2:00 AM 2 67.01 4:00 AM 4 68.565 6:00 AM 6 70.561 8:00 AM 8 72.667 10:00 AM 10 74.349 12:00 PM 12.001 75.161 2:00 PM 14.001 74.885 4:00 PM 16.001 73.597 6:00 PM 18.001 71.641 8:00 PM 20.001 69.541 10:00 PM 22.001 67.861 MIDNIGHT 24.001 67.05

Step 6: Evaluate the value of temperature when $${\bf{K = 0}}{\bf{.6}}$$.

The given equation is at

$$\begin{array}{c}{\bf{T(0) = 65}}\\\frac{{{\bf{dT}}}}{{{\bf{dt}}}}{\bf{ = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T(t)}}\end{array}$$

Now apply improved Euler’s method subroutine with $${\bf{h = }}\frac{{\bf{2}}}{{\bf{3}}} \approx 0.6667$$ and $${\bf{N = 36}}$$.

$$\begin{array}{c}{\bf{f(t,T) = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T(t)}}\\{\bf{F = f(t,T) = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T(t)}}\\{\bf{G = f(t + h,T + hF)}}\\{\bf{ = 150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi (t + 0}}{\bf{.6667)}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1(T + 0}}{\bf{.6667(150}}{\bf{.1 - 12cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - 2}}{\bf{.1T))}}\end{array}$$

Apply initial conditions $${\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,T = }}{{\bf{T}}_{\bf{o}}}{\bf{ = 65}}$$.

$$\begin{array}{c}{\bf{F(0,65) = 1}}{\bf{.6}}\\{\bf{G(0,65) = - 0}}{\bf{.1483}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.6667}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 65}}{\bf{.381}}\end{array}$$

Hence, at 0.6667h after midnight which is 12.40 AM, the temperature is approx. $${\bf{68}}{\bf{.38}}1$$.

Now apply the same procedure for the period of 24h.

Step 7: Find the value for 36 steps.

Since there are 36 steps so by construct a table to get the required result.

 Time $${{\bf{t}}_{\bf{0}}}$$ $${{\bf{T}}_{\bf{o}}}$$ Midnight 0 65 12:40 AM 0.667 65.381 2:00 AM 2 66.199 4:00 AM 4 68.13 6:00 AM 6 70.825 8:00 AM 8 73.668 10:00 AM 10 75.919 12:00 PM 12.001 76.978 2:00 PM 14.001 76.563 4:00 PM 16.001 74.784 6:00 PM 18.001 72.119 8:00 PM 20.001 69.282 10:00 PM 22.001 67.032 MIDNIGHT 24.001 65.974

Hence, this is the required result.