Respuesta :
Answer:
a) q = 0.927 W
b) q = 0.73 W
Explanation:
Our assumptions are that:
- there is a steady state condition
- No heat loss during insulation
- No radiation
- Heat flux is uniform
[tex]u_{\infty} = 20 m/s\\[/tex]
[tex]T_{\infty} = 24^{0} C[/tex]
Chip temperature, [tex]T_{c} = 80^{0} C[/tex]
Temperature of the film, [tex]T_{f} = \frac{T_{\infty}+ T_{c} }{2} \\[/tex]
[tex]T_{f} = \frac{24 + 80}{2} \\T_{f} = 52^{0} C[/tex]
[tex]T_{f} = 52 + 273 K \\T_{f} = 325 K[/tex]
Atmospheric air is used to insulate the chip. i.e the film is an atmospheric air
1 atm of 325 K of air has the following properties:
viscosity, v = 18.4 × 10⁻⁶ m²/s
k = 0.0282 w/mk
Pr = 0.703
Nuselt number, Nu = [tex]\frac{hL }{k}[/tex]..............(i)
Nu = [tex]0.453Re^{0.5} Pr^{0.33}[/tex]................(2)
The Reynold number, Re, is calculated by the equation
[tex]Re = \frac{u_{\infty L} }{v}[/tex], L = 12.5 mm = 12.5 * 10⁻³ m
Re = [tex]\frac{20 * 12.5 * 10^{-3} }{18.4 * 10^{-6} }[/tex]
Re = 13586.96
Equating (1) and (2) and substituting necessary values
[tex]0.453* 13586.96^{0.5} 0.703^{0.33} = \frac{h * 12.5 * 10^{-3} }{0.0282}[/tex] ..........(3)
47.006 = h * 0.4433
h = 47.006/0.4433
h = 106.05
The maximum allowable power is given by the equation, [tex]q = hA(T_{c}-T_{\infty} )[/tex]
Area, A = length * Breadth = 0.0125*0.0125
A = 0.000156 m²
[tex]q = 106.05*0.000156(80-24)\\q = 0.927 W[/tex]
b) When the lenght, L = 20 mm = 0.02 m
Reynold number, Re = [tex]\frac{20 * 0.02 }{18.4 * 10^{-6} }[/tex]
Re = 21739.13
By replacing Re and L, Equation (3) becomes:
[tex]0.453* 21739.13^{0.5} 0.703^{0.33} = \frac{h * 0.02 }{0.0282}\\59.46 = 0.709h\\h = 59.46/0.709\\h = 83.84[/tex]
[tex]q = hA(T_{c}-T_{\infty} )\\q = 83.84*0.000156(80-24 )[/tex]
[tex]q = 0.73 W[/tex]
