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velikii [3]
3 years ago
6

Tech A says that over time a grease seal can wear a groove in the sealing surface of the axle or shaft. Tech B says that grease

seals need to be replaced every time the bearing is removed. Who is correct?
Engineering
2 answers:
Natasha_Volkova [10]3 years ago
7 0

Answer:

Tech A is correct

Explanation:

Tech A is correct because overtime a grease seal can wear a groove in the sealing surface of the axle or shaft. Most times, worn out or damaged bearings can wear a groove into the axle shaft. The seal may also wear a groove into the axle shaft. Anytime this becomes the case, it's very important to replace the axle shaft assembly, bearings and seal.

Tech B is correct but to an extent, because there is a way the bearings will be removed carefully so replacements of grease seals won't be needed which makes tech B somehow wrong

maw [93]3 years ago
7 0

Answer:

Tech A is correct

Explanation:

Over time a grease seal in an axle or shaft, which is having continuous pinion seal failure due to leaks, can have worn a groove in the sealing surface.

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Please help me with this, picture.
Alenkasestr [34]
Maybe try 086 degrees
3 0
3 years ago
According to fire regulations in a town, the pressure drop in a commercial steel, horizontal pipe must not exceed 2.0 psi per 25
bonufazy [111]

Answer:

6.37 inch

Explanation:

Thinking process:

We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.

To determine the pressure drop in the pipe:

Using the Bernoulli equation for mass conservation:

\frac{P1}{\rho } + \frac{v_{2} }{2g} +z_{1}  = \frac{P2}{\rho } + \frac{v2^{2} }{2g} + z_{2} + f\frac{l}{D} \frac{v^{2} }{2g}

thus

\frac{P1-P2}{\rho }  = f\frac{l}{D} \frac{v^{2} }{2g}

The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.

Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F

from the tables

Re= 2.01 × 10⁵

Hence, f = 0.018

Therefore, pressure drop, (P1-P2)/p = 2.70 ft

This occurs at ae presure change of 1.17 psi

Correlating with the chart, we find that the diameter will be D= 0.513

                                                                                                      = <u>6.37 in Ans</u>

7 0
4 years ago
Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
Why does teachers grade things that are not due yet​
pav-90 [236]
I think because if you’ve already turned it in they might as well grade asap instead of waiting
4 0
3 years ago
Read 2 more answers
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem
Llana [10]

Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

7 0
4 years ago
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