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Svetllana [295]
3 years ago
14

A shaft 70 mm in diameter is being pushed at a speed of 400 mm/s through a bearing sleeve 70.2 mm in diameter and 250 mm long. T

he clearance, assumed uniform, is filled with oil with kinematic viscosity v = 0.005 m²/s and density p= 900 kg/m². Find the force exerted by the oil on the shaft.
Engineering
1 answer:
Allushta [10]3 years ago
6 0

Answer:

F=989.6 N

Explanation:

Given that

Diameter of shaft = 70 mm

Diameter of bearing sleeve =70.2 mm

So clearance h=0.1 mm

Speed V= 400 mm/s

Length of shaft = 250 mm

\nu =0.005\ \frac{m^2}{s}\ , \rho=900\ \frac{kg}{m^3}

We know that

\mu =\rho\times\nu

μ= 900 x 0.005 Pa-s

μ= 4.5 Pa-s

As we know that

From Newton's law of viscosity ,the shear stress given as follows

\tau =\mu \dfrac{dU}{dy}

We also know that

Force = shear stress x area

Now by putting the values

\tau =4.5\times \dfrac{400}{0.1}

\tau=18,000 Pa

So force

F= 18,000 x π x 0.07 x 0.25

F=988.6 N

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53.05 m/s

Explanation:

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A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure
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Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

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taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

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Being that we have same thermal resistance in the first and second plane,

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R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

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R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

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from equation (3),

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q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

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