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REY [17]
3 years ago
9

Air at 1 atm and 20°C is flowing over the top surface of a 0.2 m 3 0.5 m-thin metal foil. The air stream velocity is 100 m/s and

the metal foil is heated electrically with a uniform heat flux of 6100 W/m2. If the friction force on the metal foil surface is 0.3 N, determine the surface temperature of the metal foil. Evaluate the fluid properties at 100°C.
Physics
1 answer:
Paraphin [41]3 years ago
3 0

Answer:180.86^{\circ}C

Explanation:

Properties of Fluid at 100^{\circ}C

P_r=0.711

\rho =0.9458 kg/m^3

c_p=1009 J/kg/k

Flux =6100 W/m^2

Drag force F_d=0.3 N

A=0.2\times 0.5=0.1 m^2

drag force is given by

F_d=c_f\cdot A\rho \frac{v^2}{2}

c_f=\frac{2F_d}{\rho Av^2}

c_f=\frac{2\times 0.3}{0.9458\times 0.1\times 100^2}

c_f=\frac{0.6}{945.8}

c_f=0.000634

we know average heat transfer coefficient is

h=\frac{c_f}{2}\times \frac{\rho vc_p}{P_r^{\frac{2}{3}}}

h=\frac{0.000634}{2}\times \frac{0.9458\times 100\times 1009}{(0.711)^{\frac{2}{3}}}

h=37.92 W/m^2-K

Surface Temperature of metal Foil

\dot{q}=h(T_s-T{\infty })

T_s=\frac{\dot{q}}{h}

T_s is the surface temperature and T_{\infty }[/tex] is ambient temperature

T_s=\frac{6100}{37.92}+20=180.86^{\circ}C

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\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

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