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Yuki888 [10]
3 years ago
13

If a torque of M = 300 N⋅m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD t

o prevent the flywheel from rotating. The coefficient of static friction between the friction pad at B and the flywheel is μs = 0.4.

Engineering
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

<em>866.1 N</em>

Explanation:

The torque on the flywheel = 300 N-m

The force from the hydraulic cylinder will generate a moment on CA about point A.

The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m

we know that moment = F x d

where F is the force, and

d is the perpendicular distance from the turning point = 1 m

Equating, we have

300 = F x 1

F = 300 N   this is the frictional force that stops the flywheel

From F = μN

where F is the frictional force

μ is the coefficient of static friction = 0.4

N is the normal force from the hydraulic cylinder

substituting, we have

300 = 0.4 x N

N = 300/0.4 = 750 N

This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship

N = R sin (90 - 30)

750 = R sin 60°

750 = 0.866R

R = 750/0.866 = <em>866.1 N</em>

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<h3>What is lot number?</h3>

A lot number can be defined as an alpha-numeric designator which is systematically designed and assigned at the time of manufacture, so as to identify the manufacturer, month, year, location, and batch.

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8 0
2 years ago
An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc
Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : \overline{x}=2.45

Standard deviation : \sigma=0.07

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, z=\dfrac{x-\mu}{\sigma}, the z-value corresponds to 2.39 will be :-

z=\dfrac{2.39-2.45}{0.07}\approx-0.86

z-value corresponds to 2.60 will be :-

z=\dfrac{2.60-2.45}{0.07}\approx2.14

Using the standard normal table for z, we have

P-value = P(-0.86

=P(z

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0
3 years ago
Since the passing of the Utah GDL laws in 1999:
wlad13 [49]
The answer is b, I hope this helps you
7 0
3 years ago
A 132mm diameter solid circular section​
Ganezh [65]

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5 0
3 years ago
Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at
a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

c)

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

8 0
3 years ago
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