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grandymaker [24]

3 years ago

7

In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical

example of the appropriate area to use?

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

Explanation:

We know that Drag force $F_D$

$F_D=\dfrac{1}{2}C_D\rho AV^2$

Where

$C_D$ is the drag force constant.

A is the projected area.

V is the velocity.

ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

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Vending machine controller (adapted from Katz, "Contemporary Logic Design") Design and implement a finite state machine that con

babunello [35]

Answer:

Check the explanation

Explanation:

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6 0

3 years ago

A liquid jet vj of diameter dj strikes a fixed cone and deflects back as a conical sheet at the same velocity. find the cone ang

dmitriy555 [2]

Answer:

lol i cant real it sorry

Explanation:

7 0

1 year ago

Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with

Scilla [17]

Answer:

DIAMETER = 9.797 m

POWER = $\dot W = 28.6 kW$

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

$v =\frac{RT}{P}$

where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

$v = \frac{0.287\times 293}{100}$

v = 0.8409 m^3/ kg

from continuity equation

$A_1 V_1 = A_2 V_2$

$\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2$

$D_2 = D_1 \sqrt{\frac{V_1}{V_2}}$

$D_2 = 8 \times \sqrt{\frac{12}{8}}$

$D_2 = 9.797 m$

mass flow rate is given as

$\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}$

$\dot m = 717.309 kg/s$

the power produced $\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]$

$\dot W = 28.6 kW$

8 0

3 years ago

• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the

Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$ ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

$\sigma a$ = $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$

$\sigma a$ = 12 kpsi

and

$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$ ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$

$\sigma m$ = 17.34 kpsi

now we apply here goodman line equation here that is

$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$ ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

$\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}$

solve it we get

FOS = 1.5432

6 0

2 years ago

Identify and describe the three stages of a life cycle analysis.

sesenic [268]

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Hope this is helpful

5 0

2 years ago