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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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4A. How high is a 12 kg monkey in a tree if it has 509 J of gravitational potential Energy?
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4A. PE = MxGxH. (You can consider g as 9.8 / 10m/s as well)

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H = 509/120

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8 0
3 years ago
. A ball is thrown downward at a speed of 20 m/s. Choosing
a_sh-v [17]

The final velocity is v = 20 - gt

The distance traveled by the ball at time t is y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is 0 = (20)^2 - 2g(y-y_0)

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

v = 20 - gt

The distance traveled by the ball at time t;

y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is calculated as;

v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)

Learn more here: brainly.com/question/16878713

3 0
2 years ago
Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long
nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

R = \dfrac{\rho L }{A}

where;

A = πr²

R = \dfrac{\rho L }{\pi r ^2}

For the shorter cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{ 4 \rho L }{\pi \ D   ^2}

For the longer cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{32\rho L }{\pi \ (4 D)   ^2}

R = \dfrac{2\rho L }{\pi \ (D)   ^2}

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0
3 years ago
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