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andrezito [222]
2 years ago
15

For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the vari

ation of the local heat transfer coefficient with position along the plate is
Engineering
1 answer:
Aliun [14]2 years ago
3 0

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

Given the data in the question;

We write the expression for the local Nusselt number for Laminar flow over the flat plate;

Nu = C(Re_x)^{0.5 (Pr)^{1/3

Nu = C(\frac{Vx}{v})^{0.5} (Pr)^{1/3

\frac{h_xx}{k} = C(\frac{V}{v})^{0.5}  (Pr)^{1/3  (x)^{0.5

h_x = \frac{1}{x^{1/2}}

Next we write down the expression for the local heat flux from the plate with  uniform surface temperature;

q = h_xA( T_s - T∞ )

q ∝ h_x

∴

q ∝  \frac{1}{x^{1/2}}

The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.

Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

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Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

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where P_1 = P_2 = 0

V1 AND V2  =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV  =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f  = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

4 0
2 years ago
The pressure drop across a valve through which air flows is expected to be 10 kPa. If this differential were applied to the two
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Answer:hit that soulja boy

Explanation:

6 0
2 years ago
1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a - 2b + c, 5a + 8b -70​
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Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

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1 year ago
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a
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Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = 2*10^{8}  N/m^{2}

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/(2*10^{8}) = 10^{-5} m^2

recall that area = \pi d^{2} /4

10^{-5} = \frac{3.142*d^{2} }{4} = 0.7855d^{2}

d^{2} = \frac{10^{-5} }{0.7855} = 1.273*10^{-5}

d = \sqrt{1.273*10^{-5}  } = 3.57*10^{-3} m = 3.57 mm

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4 0
2 years ago
A platinum resistance temperature sensor has a resistance of 120 Ω at 0℃ and forms one arm of a Wheatstone bridge. At this tempe
oksian1 [2.3K]

Answer : 9.36ohms/ temperature

Explanation:

Expression for the variation of resistance of platinum with temperature

Rt= Ro(1+*t)

Rt= resistance @ t°C

Ro= resistance @ 0°C

*= temperature coefficient of resistance

Calculate the change in resistance by putting 120ohms for Ro,

0.0039/K for *

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Using this formula:

Rt = Ro(1+*t)

Rt- Ro = Ro*t

= (120ohms)(0.0039/K)(20°C)

= 9.36ohms/K

8 0
2 years ago
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