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Sloan [31]
3 years ago
8

Two technicians are discussing cylinder- testing. Technician A says that when testing the power level of a specific cylinder usi

ng a Power balance test, it's best to ground it's secondart current. Technician B says that it's best to disconnect one spark plugs wire at a time to test cylinder power levels. Who is correct?​
Engineering
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

The answer is "Both Technician A and Technician B".

Explanation:

The cylinder Testing is intended to assess locomotive inconsistency in CNS rodents, for example, whenever the animal moves within a transparent plastic tube, its preliminary activity is registered as it rises against the stadium wall.

In the given question both technicians are correct because both are reliable ways to check cylinders and the influence of the belief if every pathway has many more advantages than each other.

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A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal
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a)  the temperature to which the pin must be cooled for assembly is T_2 = -101.89^ \ ^0}C

b) the radial pressure at room temperature after assembly is P_f = 62.8 \ MPa

c) the  safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion \alpha = 12.3*10^{-6} \  ^0 C

Yield strength \sigma_y = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature T_1 = 20°C

outer diameter of the collar D_o = 95 \ mm

inner diameter of the collarD_i = 60 \ mm

pin diameter D_p = 60.03 \ mm

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

(D_i - c )-D_p = \alpha * D_p(T_2-T_1)

(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}T_2  \ \ - \ \ 0.01476738

-0.09 +  0.01476738 = 7.38369*10^{-4}T_2

−0.07523262 =7.38369*10^{-4}T_2

T_2 = \frac{-0.07523262}{7.38369*10^{-4}}

T_2 = -101.89^ \ ^0}C

b)

To determine the radial pressure at room temperature after assembly ;we have:

P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\  P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2}  \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa

c)  the safety factor of the resulting assembly is calculated as:

safety factor =  \frac{Yield \ strength }{walking \ stress}

safety factor =  \frac{400}{62.8}

safety factor = 6.4

Thus, the  safety factor in the resulting assembly = 6.4

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