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kvasek [131]
3 years ago
5

A(n) _____________ is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when

the motor reaches approximately 75% of its rated speed.
Engineering
1 answer:
Alenkasestr [34]3 years ago
7 0

A <u>centrifugal switch</u> is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when the motor reaches approximately 75% of its rated speed.

Hope that helps!

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A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350 °C. The air is first expanded isothermally to 500 kP
Semmy [17]

Answer:

Isothermal  expansion W₁ =-37198.9 J

Polytropic Compression W₂ =-34872.82 J

Isobaric Compression W₃ =  -6974.566 J

The net work for the cycle = -79046.29 J

Explanation:

Mass of air = 0.15 kg = 150 g

Molar mass = 28.9647 g/mol

Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air

PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³

For isothermal expansion we have

P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³

Therefore work done

W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J

Stage 2

Compression polytropically we have

\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n  where P₃ = 2 MPa

Therefore V₃ = (\frac{1}{4} )^{\frac{1}{1.2} }*V_2  = 1.6904×10⁻² m³

Work = W₂ = \frac{P_2V_2-P_3V_3}{n-1} =  -34872.82 J

\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}     or T₃ = T_2*(\frac{P_3}{P_2})^\frac{n-1}{n} = 785.12 K

Isobaric compression we have  thus

Work done W₃ = P(V₁ -V₃) = -6974.566 J

Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J

7 0
3 years ago
Describe a gear train that would transform a counterclockwise input rotation to a counterclockwise output rotation where the dri
masya89 [10]

Answer:

For a gear train that would train that transform a counterclockwise input into a counterclockwise output such that the gear that is driven rotates three times when the driver rotates once, we have;

1) The number of gears in the gear train = 3 gears with an arrangement such that there is a gear in between the input and the output gear that rotates clockwise for the output gear to rotate counter clockwise

2) The speed ratio of the driven gear to the driver gear = 3

Therefore, we have;

Speed \ Ratio =\dfrac{Speed \ of \ Driven \ Gear}{Speed \ of \ Driver \ Gear} = \dfrac{The \ Number \ of \ Teeth \ of \ Driver \ Gear}{The \ Number \ of \ Teeth \ of \ Driven \ Gear}

Therefore, for a speed ratio of 3, the number of teeth of the driver gear, driving the output gear, must be 3 times, the number of teeth of the driven gear

Explanation:

3 0
3 years ago
How can input from multiple individuals improve design solutions for problems that occur because of a natural disaster, such as
Alla [95]

Answer:

Map and avoid high-risk zones.

Build hazard-resistant structures and houses.

Protect and develop hazard buffers (forests, reefs, etc.)

Develop culture of prevention and resilience.

Improve early warning and response systems.

Build institutions, and development policies and plans.

Explanation:

5 0
3 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
Hello , how are yall:))))
SVEN [57.7K]

Answer:

eh I'm good hbu?????????

6 0
2 years ago
Read 2 more answers
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