he figure shows the boundary temperatures (in degrees Celsius) of an insulated t

Question

he figure shows the boundary temperatures (in degrees Celsius) of an insulated thin metal plate. The steady state temperature at an interior junction is approximatly equal to the mean of the temperatures at the four surrounding junctions. Use a system of linear equations to mate the Interlor temperatures Tr Ti, Ta T, and T The tigure shows the boundary temperatures (in degrees Celsius) ot an insulated thin metal plate. The steady-state temperature at an intenior junction is approximately equal to the mean of the temperatures at the tour surrounding junctions. Use a system of linear equations to approximate the interior temperatures ,, J2, and Jc- 0° 0°

Explanation / Answer

a)
We can make the following four equations :
    1. T1 = (60+20+T2+T3)/4
    2. T2 = (60+50+T1+T4)/4
    3. T3 = (20+10+T1+T4)/4
    4. T4 = (50+10+T2+T3)/4

adding 1. and 4.,
   i) 4*T1 + 4*T4 = 140 + 2*T2 + 2*T3

adding 2. and 3.,
   ii) 4*T2 + 4*T3 = 140 + 2*T1 + 2*T4

2*i) + ii) :
    6*T1 + 6*T4 = 280
    Or T1 + T4 = 46.66
    Put this in 1. and 4.
Answer :
    T1 = 135.82 / 4 = 33.95
    T2 = 156.66 / 4 = 39.16
    T3 = 66.66 / 4 = 16.66
    T4 = 50.82 / 4 = 12.7

b)
We can make the following four equations :
    1. T1 = (25+50+T2+T3)/4
    2. T2 = (25+50+T1+T4)/4
    3. T3 = (25+0+T1+T4)/4
    4. T4 = (25+0+T2+T3)/4

adding 1. and 4.,
   i) 4*T1 + 4*T4 = 100 + 2*T2 + 2*T3

adding 2. and 3.,
   ii) 4*T2 + 4*T3 = 100 + 2*T1 + 2*T4

2*i) + ii) :
    6*T1 + 6*T4 = 200
    Or T1 + T4 = 33.33
    Put this in 1. and 4.
Answer :
    T1 = 116.66 / 4 = 29.16
    T2 = 108.33 / 4 = 27.08
    T3 = 58.33 / 4 = 14.58
    T4 = 16.66 / 4 = 4.164

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