All categories

3 years ago

The value of the critical Reynolds number for the flow over a flat plate is 5 x10^5. The significance of the value is:

Engineering

1 answer:

nata0808 [166]3 years ago

8 0

Answer:

(c) If the Reynolds number for the plate based on the length L is greater than this value, then the boundary layer is turbulent over the entire plate.

Explanation:

The applicability of the Reynolds number differs depending on the specifications of the fluid flow such as the variation of density (compressibility), a variation of viscosity (Non-Newtonian), being internal- or external flow, etc. The critical Reynolds number is the expression of the value to specify transition among regimes which diversifies regarding type of flow and geometry as well. Whilst the critical Reynolds number for turbulent flow in a pipe is 2000, the critical Reynolds number for turbulent flow over a flat plate, when the flow velocity is the free-stream velocity, is in a range from 10 raised to power 5 to 10 raised to power 6, .4

You might be interested in

Which basic principle influences how all HVACR systems work?

bezimeni [28]

Answer:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

Explanation:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

6 0

3 years ago

100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t

LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.

Using the given mass m, molar mass M, we can get the following equation:

pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

U=-1.848*10^6 J

Note, that the energy was taken away from the system.

5 0

3 years ago

What different tests did the team perform to come up with a workable design?

I am Lyosha [343]

They ran different shapes and materials through a wind tunnel to see which shape and material would decrease energy output so that it takes in equal COthan it puts out.

5 0

3 years ago

This problem has been solved!

lisov135 [29]

Answer: a) 135642 b) 146253

Explanation:

1- the bankers algorithm tests for safety by simulating the allocation for predetermined maximum possible amounts of all resources, as stated this has the greatest degree of concurrency.

3- reserving all resources in advance helps would happen most likely if the algorithm has been used.

5- Resource ordering comes first before detection of any deadlock

6- Thread action would be rolled back much easily of Resource ordering precedes.

4- restart thread and release all resources if thread needs to wait, this should surely happen before killing the thread

2- only option practicable after thread has been killed.

Bii) ; No. Even if deadlock happens rapidly, the safest sequence have been decided already.

5 0

2 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

2 years ago