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Paraphin [41]
3 years ago
12

3. (9 points) A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 10

0kg. See the following two figures. For power screws, assume a major diameter d = 20mm, and a pitch of p = 2mm with double threads. The friction coefficient of steel parts is f = 0.1. Gravity constant is g = 10????/kg. (a) (7 points) If the purchased Cordless Electric Power Screwdrivers rotates at 300rpm, determine the power you need to raise to the basket board. (b) (2 points) Check if the self-locking condition of the power screw is satisfied or not.
Engineering
1 answer:
Bezzdna [24]3 years ago
3 0

Answer:

a) 49.95 watts

b) The self locking condition is satisfied

Explanation:

Given data

weight of the square-thread power screw ( w ) = 100 kg = 1000 N

diameter (d) = 20 mm ,

pitch (p) = 2 mm

friction coefficient of steel parts ( f ) = 0.1

Gravity constant ( g ) = 10 N/kg

Rotation of electric power screwdrivers = 300 rpm

A ) Determine the power needed to raise to the basket board

first we have to calculate T

T = Wtan (∝ + Ф ) * \frac{Dm}{2} ------------- equation 1

Dm = d - 0.5 ( 2) = 19mm

Tan ∝ = \frac{L}{\pi  Dm}  where L = 2*2 = 4

hence ∝ = 3.83⁰

given f = 0.1 , Tan Ф = 0.1.     hence Ф = 5.71⁰

insert all the values into equation 1

T = 1.59 Nm

Determine the power needed using this equation

\frac{2\pi NT }{60}   =  \frac{2\pi * 300 * 1.59}{60}

= 49.95 watts

B) checking if the self-locking condition of the power screw is satisfied

Ф > ∝  hence it is self locking condition is satisfied

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Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
3 years ago
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