Q19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

Step-by-Step Solution

Answer

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. \({\bf{65}}{\bf{.974}}\).

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.