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Sliva [168]
3 years ago
14

The flatbed truck carries a large section of circular pipe secured only by the two fixed blocks A and B of height h. The truck i

s in a left turn of radius rho. Determine the maximum speed for which the pipe will be restrained. Use the values rho = 60 m, h = 0.1 m, and R = 0.8 m.
Engineering
2 answers:
zalisa [80]3 years ago
6 0

Answer:

The maximum speed for which the pipe will be restrained is 18.05 m/s.

Explanation:

We note that the

Securing blocks have height, h = 0.1 m

The radius of travel, rho = 60 m

The radius of the circular pipe, R = 0.8 m

We note that the weight of the pipe is acting at the centroid of the pipe

Therefore, for the pipe to slip, it has to climb the wedge h

Taking moments about h, we have  

mg × R sin α = ma × R cos α  

a/g  =  Tan α

But a = \frac{V^{2} }{rho}

Therefore  \frac{V^{2} }{rho} = g tanα

Since height of the block, h = 0.1 m therefore,

R cos α = R - h

That is 0.8 cos α = 0.8 - 0.1 = 0.7

Therefore α = cos⁻¹ (0.7/0.8) = 28.96 °

From which V² = rho × g× tanα = 60 × 9.81 × tan 28.96

= 325.66 m²/s²

∴ V = √(325.66 m²/s²)  = 18.05 m/s

Maximum speed = 18.05 m/s.

garik1379 [7]3 years ago
6 0

Answer:

The maximum speed is 18.1 m/s

Explanation:

the angle made by the line of action of reaction forces and normal reaction with the center is equal to:

\alpha =cos^{-1}(\frac{R-h}{R})

If R=0.8 m and h=0.1 m, we have:

\alpha =cos^{-1}(\frac{0.8-0.1}{0.8})=29

the equilibrium of forces acting in y-direction is zero and we have the following:

Fy=0\\R_{B}^{2}cos\alpha  -mg=0

Clearing RB:

R_{B}=\frac{mg}{cos\alpha }

where RB is reaction force on the ring and m is the mass of circular ring

the equilibrium of forces acting in n-direction is equal to:

Fn=0\\R_{B}sin\alpha  =m\frac{v^{2} }{p}

where p is the radius and v is the speed. if RB=mg/cosα

v^{2}=pgtan\alpha

Replacing values:

v^{2}=60*9.8*tan29\\v=18.1 m/s

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The answer is "-121\  \frac{KJ}{Kg}".

Explanation:

Please find the correct question in the attachment file.

using formula:

\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \   or \ \  \frac{RT_1 -RT_2}{n-1}\\\\

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7 0
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At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t
nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

PGP = \rho *\frac{\pi r^2}{2} *V*e  \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

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