Compared with space operations specialists, intelligence officers are which of the following?

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

Conduct online research and write a short report on the origin and evolution of the meter as a measurement standard. Discuss how

valina [46]

Answer:

People have come up with all sorts of inventive ways of measuring length. The most intuitive are right at our fingertips. That is, they are based upon the human body: the foot, the hand, the fingers or the length of an arm or a stride.

In ancient Mesopotamia and Egypt, one of the first standard measures of length used was the cubit. In Egypt, the royal cubit, which was used to build the most important structures, was based on the length of the pharaoh’s arm from elbow to the end of the middle finger plus the span of his hand. Because of its great importance, the royal cubit was standardized using rods made from granite. These granite cubits were further subdivided into shorter lengths reminiscent of centimeters and millimeters.

piece of black rock with white Egyptian markings

Fragment of a Cubit Measuring Rod

Credit: Gift of Dr. and Mrs. Thomas H. Foulds, 1925

Later length measurements used by the Romans (who had taken them from the Greeks, who had taken them from the Babylonians and Egyptians) and passed on into Europe generally were based on the length of the human foot or walking and multiples and subdivisions of that. For example, the pace—one left step plus one right step—is approximately a meter or yard. (On the other hand, the yard did not derive from a pace but from, among other things, the length of King Henry I of England’s outstretched arm.) Mille passus in Latin, or 1,000 paces, is where the English word “mile” comes from.

And thus, the meter has and likely will remain so elegantly defined in these terms for the foreseeable future.

Explanation:

is this short enough

5 0

2 years ago

Two wastewater treatment plant workers (one male and one female) are exposed to hydrogen sulfide in confined spaces in the treat

Vlada [557]

Answer:

Go to explaination for the details of the answer.

Explanation:

In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:

CDI = Chronic Daily Intake

C= Chemical concentration

CR= Contact Rate

EFD= Exposure Frequency and Distribution

BW= Body Weight

AT = Average Time.

Having names our variables lets create the equations that will be used to derive our answers.

Please kindly check attachment for details of the answer.

5 0

3 years ago

Which of the following factors does not promote safety in the shop?

sergey [27]

“Thinking about pleasant things to pass the time” would not promote safety in the shop because it would be taking the focus away from important tasks, which in turn decreases safety.

6 0

3 years ago

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.7. If, afte

luda_lava [24]

Answer:

It would take approximately 305 s to go to 99% completion

Explanation:

Given that:

y = 50% = 0.5

n = 1.7

t = 100 s

We need to first find the parameter k from the equation below.