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4 years ago

Select the correct answer.

Engineering

1 answer:

uysha [10]4 years ago

3 0

Answer:

Lazermicrometer 100%

Explanation:

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Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

Take Re=100,000:

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

Take Re=500,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

Take Re=1,000,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

7 0

3 years ago

Time management is a learned behavior. True False

larisa [96]

Answer:

true

Explanation:

3 0

3 years ago

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The insulator is the connection between the grounded circuit conductor and the equipment grounding conductor at the service.

Arada [10]

The answer to your question is correct/ true

4 0

3 years ago

20 points and brainliest is it A, B, C, D

andre [41]

I think its c maybe..

6 0

3 years ago

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IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The follow

Irina-Kira [14]

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, $h_0=2.2$ inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

$\Delta h = H_0-h_f=\mu^2R$

$=0.2^2 \times 14 = 0.56$ inches

$h_f = 2.2 - 0.56$

= 1.64 inches

Roll strip contact length (L) = $\sqrt{R(h_0-h_f)}$

$=\sqrt{14 \times 0.56}$

= 2.8 inches

Absolute value of true strain, $\epsilon_T$

$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$

Average true stress, $\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$ Psi

Roll force, $L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$

= 788,900 lb

For SI units,

Power = $\frac{2 \pi FLN}{60}$

$=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$

= 10399.81168 W

Horse power = 13.9357

6 0

3 years ago