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tatuchka [14]
3 years ago
10

In designing a fixed-incline self-acting thrust pad when the width of the pad is much larger than the length, it is of interest

to know the magnitude and location of the maximum pressure. The viscosity of the lubricant is 0.05 N-s/m2, the sliding velocity is 10 m/s, the pad length is 0.3 m, the minimum film thickness is 15 m, and the inlet film thickness is twice the outlet film thickness.
Engineering
1 answer:
Elina [12.6K]3 years ago
6 0

Answer:

T= 33.33 Kpa

Explanation:

T = μ (dv / dy)

T = μ (V/t)

T = (0.05  * 10 )/ (15 * 10^-6)

Answer is T = 33.33 Kpa.

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Determine the maximum volume in gallons​ [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3
Olin [163]

Answer:

V=1601gal

Explanation:

Hello! This problem is solved as follows,

First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.

This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.

P=Poil+Patm

P=total pressure or absolute pressure=26psi=179213.28Pa

Patm= the atmospheric pressure =101325Pa

Poil=pressure due to the weight of olive oil=0.86αgh

α=density of water=1000kg/m^3

g=gravity=9.81m/s^2

h= height that olive oil reaches

solving

P=Poil+Patm

P=Patm+0.86αgh

h=\frac{P-Patm}{0.86\alpha g } =\frac{179214.28-101325}{(0.86)(1000)(9.81)} \\h=9.23m[/tex]

Now we can use the equation that defines the volume of a cylinder.

V=V=\frac{\pi }{4} D^{2} h

D=3ft=0.9144m

h=9.23m

solving

V=\frac{\pi }{4} (0.9144)^{2} 9.23=6.06m^3

finally we use conversion factors to find the volume in gallons

V=6.06m^3\frac{1000L}{1m^3} \frac{1gal}{3.785L} =1601gal

3 0
4 years ago
Consider a single bacterial cell as a discrete particle with a diameter of 1x10-6m and a specific gravity of 1.01. Assuming lami
Alinara [238K]

Answer:

Sedimentation is not a good method.

Explanation:

We need to apply Stoke laws and assume that is valid here.

So,

V_s= 418(G-1)d^2(\frac{3T+70}{100})

Replacing the values,

V_s= (418)(1.01-1)(10^-3)^3*(\frac{3*20+70}{100})\\V_s=5.434*10^-6mm/s

Here then we calculate the time,

t_{req}=\frac{x}{v}

Where x= Distance, v= velocity

t_{req}=\frac{1foot}{5.434*10^{-6}} = \frac{304.8}{5.434*10^{-6}}\\t_{req}=649.2days

To calculate the surface required we need first to calculate the volume through the volume,

So,

Q=4MGD=4*10^6*0.135 (ft^3/day)

Then,

V_{req} = Q*t\\V_{req}=4*10^6*0.134*649.2\\V_{req}=34.79*10^7ft^3

Here we can calculate the surface

S_{req}= \frac{Volume}{Distance}=\frac{34.79*10^7}{1}\\S_{req}=7987.8 Acres

<em>So, the requeriment of Area of tank and settlement time is huge, it's not a practical method.</em>

6 0
3 years ago
Please help me<br><br> there is no woodworking or woodshop choice
goldenfox [79]

Answer:

carpenter helped to set the cutr height and depth

what is its a jigsaw

Explanation:I took the test  

7 0
2 years ago
At a ski resort, water at 40 °F is pumped through a 3-in.-diameter, 2001-ft-long steel pipe from a pond at an elevation of 4286
deff fn [24]

Answer:

P = 24.38 hp

Explanation:

Given data:

diameter of steel pipe = 3 inc

length of pipe = 2001 ft

elevation of pond = 4286 ft

elevation of snow making machine = 4620 ft

Rate of water = 0.26 ft^3/s

Applying bernouli equation on bioth side

\frac{P}{\rho} + \frac{v_1^2}{2g} + z_1 + h_p =\frac{P_2}{\rho} + \frac{v_2^2}{2g} + z_2 + \frac{flv^2}{2gD}.....1

where P_2 = 182 psi

P_1 = 0, v_1 = 0

V =V_2 = \frac{Q}{\frac{\pi}{4} D^2}= \frac{0.26}{\frac{\pi}{4} \times (3/12)^2} = 5.30 ft/s

calculation fro friction factor

from standard table we have

\frac{\epsilon}{D} = \frac{0.00015}{(3/12)} = 6\times 10^{-4}

Re =\frac{VD}[\nu} = \frac{5.30 \times (3/12)}{1.66 \times 10^{-5}} = 7.96 \times 10^4

so for calculated R and\frac{\epsilon}{D}, F value = 0.0212

from equation 1

h_p = \frac{P_2}{\rho} + Z_2 -Z_1 +(1 + f\frac{l}{D}) \frac{V^2}{2g}

h_p = \frac{182 \times 144/ lb /ft^2}{62.4} + 4620 - 4286 + (1 + 0.0212 \frac{2001}{(3/12)}) \frac{5.30}{2\times 32.2}

h_p = 826.76ft so that

P =  \rho Q h_p = 62.4 \times 0.26 \times 826.76 = 13413.38 ft lb/s = 24.38 hp

3 0
3 years ago
List principles that are most relevant to drop jump exercise and justify your answers that selected principles you
ycow [4]

The principles referred to in the question are called biomechanical principles. Some of them are:

  • Principles of Force; and
  • The principle of impulse direction.
<h3>What principle is most relevant to jumping?</h3>


The principle that is most relevant to jumping is called Angular Momentum.

<h3>What is Angular Momentum?</h3>

Simply defined, this refers to the velocity of the rotation of a body around an axis or a fixed point.

Other principles of biomechanics are the principles of:

  • linked segments.
  • impulse-causing momentum.
  • the stretch-shorten cycle.
  • summing joint forces.
  • continuity of joint forces.

See more about biomechanical principles at:
brainly.com/question/1087668

7 0
2 years ago
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