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Ne4ueva [31]
1 year ago
10

the left rear brake drum is scored, but the right rear drum looks as good as new. technician a says the left-side drum should be

resurfaced and the right-side drum should be reinstalled without resurfacing. technician b says both sides should be remachined to a similar diameter. who is correct?
Engineering
1 answer:
Romashka-Z-Leto [24]1 year ago
7 0

Technician b is correct because the left rear brake drum is scored, but the right rear drum looks as good as new then both sides should be remachined to a similar diameter.

A brake drum is a revolving cylinder-shaped component that is friction-activated by a set of shoes or pads pressing outward against it. Drum brakes are brakes where shoes are pressed against the inner surface of the drum. Drum brakes are frequently used at the rear. Compared to disc brakes, these brakes have more parts and are more difficult to maintain, but they cost less to make and are simple to add an emergency braking mechanism. The front and rear brakes are standard equipment on all automobiles. Your car may have a "disc brake" system on all four wheels, a "drum brake" system on all four wheels, or a combination of the two. Typically, the front brakes of your car have a disc system, and the rear brakes have a drum system.

Learn more about brake drum here:

brainly.com/question/14937026

#SPJ4

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A cylindrical specimen of a metal alloy 45.8 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 378 MPa ca
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390.242 MPa

Explanation:

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2 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
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Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

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Explanation:

4 0
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Complete Question

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Explanation:

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