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timama [110]
3 years ago
7

Technician A states that a brake lathe is used to make a used brake rotor surface "like new". Technician B states that a brake l

athe can make any rotor reusable. Who is correct?
Engineering
1 answer:
nikitadnepr [17]3 years ago
7 0

Answer:

Both Technician A and B are correct.

Explanation: A brake lathe is a special tool used to improve or work on the surface of brake pads it helps to smoothen the surface.

Brake lathe has been found to be very effective in removing rusts in rotors and unevenness in the brake pad surfaces in order to ensure the efficiency and effectiveness of the brake system of a vehicle. Hence, a brake lathe helps to make brake rotor surface as smooth as possible.

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Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which
Rudik [331]

ANSWERS:

-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor x_{1} =1

The temperature T_{1} =0^0 F

Isothermal process  T=C

a)

-V_{2} =2V_{1} ( double)

b)

-V_{2} =.5V_{2} (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb

then the state exists in the supper heated region.

a) from standard data

-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF

at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg

at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg

assume linear interpolation

\frac{P_{x}-P_{2(b)}  }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)}  }{v_{x}-v_{y}  }

P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y}  }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2

b)

-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}

from standard data

-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}

then the state exist in the wet zone

-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )

x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%

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Answer:

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