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3 years ago

Exceeding critical mach may result in the onset of compressibility effects such as:______.

Engineering

1 answer:

klio [65]3 years ago

4 0

Answer:

Sound barrier.

Explanation:

Sound barrier is a sudden increase in drag and other effects when an aircraft travels faster than the speed of sound. Other undesirable effects are experienced in the transonic stage, such as relative air movement creating disruptive shock waves and turbulence. One of the adverse effect of this sound barrier in early plane designs was that at this speed, the weight of the engine required to power the aircraft would be too large for the aircraft to carry. Modern planes have designs that now combat most of these undesirable effects of the sound barrier.

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A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where

Arturiano [62]

Answer:

$\mathbf{F_D \approx 1.071 \ lbf}$

Explanation:

Given that:

The height of a triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = $1.2075 \times 10^{-5} \ ft^2/s$

the density of water ρ = 62.36 lb /ft³

$Re_{max} = \dfrac{Ux}{v}$

$Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}$

$Re_{max} = 414078.6749$

$Re_{max} = 4.14 \times 10^5$ which is less than < 5.0 × 10⁵

Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

$dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2$

where;

$(2-x) dy$ = strip area

$Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}$

Therefore;

$dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})$

$dF_D = 1.136 \times(2-x)^{1/2} \ dy$

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy$

$\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy$

Let U = (2-2y)

-2dy = du

dy = -du/2

$F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}$

$F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du$

$F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2$

$F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2$

$F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ]$

$F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ]$

$F_D = 1.071031071 \ lbf$

$\mathbf{F_D \approx 1.071 \ lbf}$

8 0

3 years ago

Consider steady heat transfer through the wall of a room in winter. The convection heat transfer coefficient at the outer surfac

AveGali [126]

The heat transfer which is in steady state, the heat transfer rate to the wall is equal to the wall.

<u>Explanation:</u>

The convection transfer of heat to the wall is

$Q=h A\left(T_{s}-T_{f}\right)$

Here, $T_{S}$ is the temperature of solid surface, $T_{f}$ is the temperature of moving fluid stream which is adjacent of solid surface, h is the heat transfer coefficient.
The coefficient of convection heat transfers outer surface contains 3 times to the inner surface which experience smaller drop of temperature for 3 times that compares to inner surface.
Hence, the temperatures outer surface get close to the surroundings of air temperature.

6 0

3 years ago

2.5 Critically discuss how community needs may influence your choice of a career

GaryK [48]

Different communities will have a demand for different careers. Areas with a large number of school will have a higher demand for teachers. The community needs professors, pays them well, you might decide to choose that path

3 0

3 years ago

Consider a mild steel specimen with yield strength of 43.5 ksi and Young's modulus of 29,000 ksi. It is stretched up to a point

mezya [45]

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, $\sigma _c$ = 43.5 ksi

Young's modulus of elasticity, ∈ = 29,000 ksi

The total strain, $\epsilon _c$ = 0.2% = 0.002

The inelatic strain $\epsilon_c^{in}$ is given as follows;

$\epsilon_c^{in}$ = $\epsilon _c$ - $\sigma _c$ /∈

Therefore, we have;

$\epsilon_c^{in}$ = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, $\epsilon_c^{in}$ = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

7 0

3 years ago

Citations must be contested within_____working days of the notice of proposed penalty. a)-15 b)-10 c)-30 d)-7

seropon [69]

Answer:

Explanation:

Legally that's when you have to respond

8 0

3 years ago