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3 years ago

Where is the specific machine information concerning lubrication/maintenance found?

Engineering

1 answer:

artcher [175]3 years ago

8 0

The specific machine information concerning lubrication/maintenance found is Preventative Maintenance Manual.

<u>Explanation:</u>

Specific machine information concerning lubrication/maintenance is required for a long life term of the machine as well as for better availability and great profit. The specific machine information concerning lubrication/maintenance found is Preventative Maintenance Manual.

This manual helps the machines to have better maintenance and also prevents the danger that they might occur anytime.

This maintenance helps the equipment process day to day without any issues. Everything is kept under schedule, the past inspections are recorded, and every maintenance scheduled in future are followed under certain time and checks on the service reports.

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What are the 5 major forest types?

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Answer:

1. Equatorial Evergreen or Rainforest

2. Tropical forest

3. Mediterranean forest

4. Temperate broad-leaved forest

5. Warm temperate forest

Explanation:

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3 years ago

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Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe

Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe $\left ( D\right )=0.15m$

Velocity of water in pipe $\left ( V\right )=15cm/s$

We know viscosity of water is $\left (\mu\right )=8.90\times10^{-4}pa-s$

Pressure drop is given by hagen poiseuille equation

$\Delta P=\frac{128\mu \L Q}{\pi D^4}$

We have asked pressure Drop per unit length i.e.

$\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}$

Substituting Values

$\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}$

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3 years ago

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Umm the Water cycle sorry I’m trying

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3 years ago

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A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of

OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

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