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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.

Short Answer

Expert verified

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 by step solution

01

Important hint.

To get the result apply Euler’s formula.

02

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.

03

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

04

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.

05

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

06

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.

07

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.

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