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Soloha48 [4]
4 years ago
14

A flat plate has a length L = 1 m and experiences air flow parallel to the surface. Transition from laminar to turbulent flow oc

curs at xc = 0.5 m. The critical Reynolds number is Rex, c = 5 × 105. (a) If the film temperature is 350 K, what’s the air velocity? (b) The local convection coefficient for laminar flow is given by

Engineering
1 answer:
tresset_1 [31]4 years ago
5 0

Answer:

velocity is 15.083 m/s

Explanation:

given data

length L = 1 m

Laminar to turbulent flow

xc = 0.5 m

Reynolds number Rex, c = 5 × 10^{5}  

solution

we take here for 350 K

\mu  = 1.81 × 10^{-5}  

and here   Reynolds number Rex, is express as

Rex = \frac{\sigma \times v \times x}{\mu }   ................1

put here value at x = 0.5

5 × 10^{5}   = \frac{1.2\times v \times 0.5}{1.81\times 10^{-5} }  

solve it we get

v = 15.083 m/s

so velocity is 15.083 m/s

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Answer:

In consolidated AB Co 300 shares.

Explanation:

Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.

6 0
3 years ago
In manufacturing a particular set of motor shafts, only shaft diameters of between 38.00 and 37.50 mm are usable. If the process
Masteriza [31]

Answer:

P(37.5

And we can find this probability with this difference:

P(-2.5

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-2.5

So then the percentage usable are 94,6%

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:

X \sim N(37.8,0.12)  

Where \mu=37.8 and \sigma=0.12

We are interested on this probability

P(37.5

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(37.5

And we can find this probability with this difference:

P(-2.5

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

P(-2.5

So then the percentage usable are 94,6%

4 0
3 years ago
On Saturday, December 31, the company's owner provided ten hours of service to a customer. The company bills $100 per hour for s
pentagon [3]

Answer:

Explanation:

Answer:

31.12. Accounts Receivable a/c              Dr.    1000

                       To Service Revenue                         1000

         10 x 100 = 1000

8 0
3 years ago
The percentage modulation of AM changes from 50% to 70%. Originally at 50% modulation, carrier power was 70 W. Now, determine th
adoni [48]

Answer:

What is percentage modulation in AM?

The percent modulation is defined as the ratio of the actual frequency deviation produced by the modulating signal to the maximum allowable frequency deviation.

3 0
3 years ago
The A-36 steel pipe has a 6061-T6 aluminum core. It issubjected to a tensile force of 200 kN. Determine the averagenormal stress
sasho [114]

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law:

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:

ε = f / k

When the force is distributed between both materials will stretch the same length:

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

5 0
3 years ago
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