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Gala2k [10]
3 years ago
8

Consider a section of muscle tissue of a cylindrical shape with a radius of 1.5 cm. During highly rigorous exercise, metabolic p

rocesses generate 15 kW/m3 of bulk tissue. The outer surface of the tissue is maintained at 37oC. The thermal conductivity of the muscle tissue is k =0.419 W/m*KDevelop a mathematical model to predict the steady-state temperature distribution, T(r), in the tissue along the radial position r. What is the temperature of the tissue at a distance of 0.75 cm from the surface?

Engineering
1 answer:
Tems11 [23]3 years ago
5 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

a)The mathematical model to predict the steady temperature distribution is

       =  T + \frac{q^{*}r^{2}}{4k} = 312.014

b)The mathematical model to predict the maximum temperature inside the muscle tissue is

      = \frac{\delta T}{\delta r}  = 0 \implies  r = 0

c) The heat flux leaving the muscle is

      = 1.06 W

d) The new maximum temperature within the muscle is

     = 38.59°C

Explanation:

The explanation is shown on the second, third , fourth and fifth image

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2 years ago
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

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3 years ago
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