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Dmitry_Shevchenko [17]
3 years ago
7

Carbon dioxide at 20°C flows in a pipe at a rate of 0.005 kg/s. Determine the minimum diameter required if the flow is laminar (

answer in m).
Engineering
1 answer:
Vesna [10]3 years ago
4 0

Answer:

the required diameter is 0.344 m

Explanation:

given data:

flow is laminar

flow of carbon dioxide Q = 0.005 Kg/s

for  flow to be laminar,  Reynold's number must be less than 2300 for pipe flow and it is given as

\frac{\rho VD}{\mu }

arrange above equation for diameter

\frac{\rho Q D}{\mu A }<2300

dynamic density of carbon dioxide = 1.47×10^{-5} Pa sec

density of carbon dioxide is 1.83 kg/m³

\frac{1.83\times 0.0056\times D}{1.47\times 10^{-5}\times \frac{\pi}{4} \times D^{2} }

\frac{1.83\times 0.0056}{1.47\times 10^{-5}\times \frac{\pi}{4} \times 2300}= D

D = 0.344 m

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The distance traveled by car A when they pass each other is 1071m

<h3>Calculations and Parameters</h3>

Given:

t=0

constant acceleration= 6 ft/s^2

speed= 80 ft/s^2

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= 2 * (\frac{27}{2})^{2}/2 + 27 * 32.9

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2 years ago
A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T th
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Answer:

Explanation:

Using equation of pure torsion

\frac{T}{I_{polar} }=\frac{t}{r}

where

T is the applied Torque

I_{polar} is polar moment of inertia of the shaft

t is the shear stress at a distance r from the center

r is distance from center

For a shaft with

D_{0} = Outer Diameter

D_{i} = Inner Diameter

I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}

Applying values in the above equation we get

I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\I_{polar}= 1.74 x 10^{-7} m^{4}

Thus from the equation of torsion we get

T=\frac{I_{polar} t}{r}

Applying values we get

T=\frac{1.74X10^{-7}X100X10^{6}  }{.021}

T =829.97Nm

7 0
3 years ago
At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates su
UkoKoshka [18]

Answer:

x=2.19in

Explanation:

This is the equation that relates the force and displacement of a spring

F=Kx

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F=mg=0.39*32.2=12.52Lbf

then we calculate the spring count in lbf / ft

K=F/x

K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft

Finally we calculate the displacement with the initial equation

X=F/k

x=12.52/68.4=0.18ft=2.19in

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4 years ago
A good rule of thumb is to design the horizontal stabilizer so that its area is about 1/6 to 1/8 of the area of the wing. If the
erik [133]

Answer:

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i dont know why but its right

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3 years ago
Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could
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Answer:

4265.04\ \text{m}

2.38\times 10^{10}\ \text{W}

Explanation:

PE = Energy of food = 500 cal = 500\times4184=2.092\times10^6\ \text{J}

m = Mass of object = 50 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Potential energy of food is given by

PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}

Nancy could raise the weight to a maximum height of 4265.04\ \text{m}.

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Energy of H_2 = \dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}

Power

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The power requirement is 2.38\times 10^{10}\ \text{W}.

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