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Arte-miy333 [17]
3 years ago
9

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)

is 0.5 m. If the plate is at a constant temperature of 100C, find:
Engineering
1 answer:
mezya [45]3 years ago
5 0

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

q=1.7845

Explanation:

From the question we are told that:

Air Temperature T_1=40c

Length l=2m

Velocity v=7m/s

Width w=0.5

Constant temperature T_t= 100C

Generally the equation for Total heat Transfer is mathematically given by

 q=hA(T_s-T_\infty)

Where

h=Convective heat transfer coefficient

 h=29.9075w/m^2k

Therefore

 q=h(L*B)(T_s-T_\infty)

 q=29.9075*(2*0.5)(100+273-(40+273))

 q=1794.45w

 q=1.7845

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A meter stick can be read to the nearest millimeter and a travelling microscope can be read to the nearest 0.1 mm. Suppose you w
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Answer: No

Explanation:

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Now meter stick can read to nearest millimeter.

It is given that length is to be measured with a precision of 1% of 20mm= 1/100 * 20= 0.2mm

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A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside
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An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°
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Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

D =  7.5 \times 10^{-11} m^2/s

for 0.225% carbon concentration following formula is used

\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})

where, erf stand for error function

\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248

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erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248

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from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

\frac{x}{2\sqrt{DT}} = 0.815

x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)

x = 2.39\times 10^{-3} m

x = 0.002395 mm

8 0
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