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Naddik [55]
3 years ago
5

1. A flywheel is suspended by resting the inside of the rim on a horizontal knife edge so that the wheel can swing in a vertical

plane. The flywheel has a mass of 350Kg. The knife edge is parallel to and making 350mm from the axis of the wheel. The time for making one small oscillation is 1.77 seconds. Assuming that the centre of mass is in the axis of the wheel, determine:
(I).The radius of gyration of the wheel about the axis.
(II). The torque to increase the speed of the flywheel at uniform rate from 240 rev/min to 250 rev/min in 0.75seconds when the flywheel is revolving about its axis.
2. A solid metal cylinder 450mm diameter is suspended with its axis vertical by means of a wire coaxial with the cylinder and rigidly attached to it. The stiffness of the wire is 22Nm per radian of twist. Find the necessary mass of the cylinder so that when it is given a small angular displacement about its axis, it will make 40 vibrations per minute.​
Engineering
1 answer:
sammy [17]3 years ago
8 0

Answer: A fly wheel having a mass of 30kg was allowed to swing as pendulum about a knife edge at inner side of the rim as shown in figure.

Explanation:

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4. The instant the ignition switch is turned to the start position,
geniusboy [140]

Answer:

D. Both pull-in and hold-in windings are energized.

Explanation:

The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.

The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.

4 0
3 years ago
Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1
babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A     ⇒    P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E)    ⇒  ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒  ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0
3 years ago
An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied
lyudmila [28]

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

4 0
3 years ago
Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
3 years ago
The homogeneous, reversible, exothermic, liquid phase reaction: A근R Is being carried out in a reactor system consisting of an id
baherus [9]

Answer:

attached below

Explanation:

3 0
3 years ago
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