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nikklg [1K]
3 years ago
8

If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab

ove
Engineering
1 answer:
12345 [234]3 years ago
5 0

Answer:

Explanation:

load = 4500lb                   lift height= 30 ft

time =15 s

velocity=\frac{30}{15} ft/s

velocity=2 ft/s

power = force\times velocity

power={4500}\times2

power= 9000 lb ft/s

1 hp= 550 lb ft/s

power= \frac{9000}{550} =16.36 hp

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3 years ago
Read 2 more answers
Fluid Dynamics: How do I find gauge pressure of an air current at various points around a cylinder?A freestream air current of v
DaniilM [7]

Answer:

The answer is as given in the explanation.

Explanation:

The 1st thing to notice is the assumptions required. Thus as the diameter of the cylinder and the wind tunnel are given such that the difference is of the orders of the magnitude thus the assumptions as given below are validated.

  1. Flow is entirely laminar, there's no boundary layer release.
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By D'alembert's paradox, "The net pressure drag exerted on a circular cylinder that moves in an inviscid fluid of large extent is identically zero".Just in the surface of the cylinder, the velocity profile can be given in the next equation:

V=2Usin\theta

And the pressure P on the surface of cylinder is given by Bernoulli's equation along the streamline through that point:

P=P_{_{\infty }}+\frac{1}{2}\rho U^{2}(1-4sin^{2}\Theta ))

where P_∞ is  Pressure at stagnation point, U is the velocity given, ρ is the density of the fluid (in this case air) and θ is the angle measured from the center of cylinder to the adjacent point where your pressure point will be determine.

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3 years ago
Name safety hazards that should be included in the design of a school
dem82 [27]

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Explanation:

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3 years ago
Determine the flow velocities at the inlet and exit sections of an
VikaD [51]

Answer:

The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

Explanation:

Using Bernoulli's equation

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²

P₁ - P₂ + ρgh₁ -  ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

ΔP + ρgΔh = 1/2ρ(v₂² - v₁²)  (1)

where ΔP = pressure difference = 12.35 kPa = 12350 Pa

Δh = height difference = 1.35 m

From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.

Substituting v₁ into (1), we have

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂²v₂/d₁² )²)

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂/d₁)⁴v₂²)

v₂ = √[2(ΔP + ρgΔh)/ρ(1 - (d₂/d₁)⁴)}

substituting the values of the variables, we have

v₂ = √[2(12350 Pa + 1000 kg/m³ × 9.8 m/s² × 1.35 m)/(1000 kg/m³ (1 - (0.44 m/0.95 m)⁴))}

= √[2(12350 Pa + 13230 Pa)/(1000 kg/m³ × 0.954)]

= √[2(25580 Pa)/954 kg/m³]

= √[51160 Pa/954 kg/m³]

= √53.627

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v₁ = d₂²v₂/d₁²

  = (0.44 m/0.95 m)² × 7.32 m/s

  = (0.954)² × 7.32 m/s

  = 6.66 m/s

The volume flow rate Q = A₁v₁

= πd₁²v₁/4

= π(0.95 m)² × 6.66 m/s ÷ 4

= 18.883 m³/s ÷ 4

= 4.72 m³/s

So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

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3 years ago
Air is compressed from 100 kPa, 300 K to 1200 kPa in a two-stage compressor with intercooling between the stages. The intercoole
jenyasd209 [6]

Answer:

410.7 K

Explanation:

Given data:

P1 ( initial air pressure ) = 100 kPa

Pi ( intercooler pressure ) = 400 kPa

P2( compressed air pressure )  = 1200 kPa

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Ti = 300 k

<u>Calculate for the temperature at the exit of the second stage ( T2 )</u>

we can calculate this using the relation below

T2 / Ti = ( P2/Pi )^(r-1/r)

∴ T2 = 300 ( 1200 / 400 )^( 1.4 - 1 / 1.4 )

         = 300 ( 3 )^0.286

         = 300 * 1.369  = 410.7 K

8 0
2 years ago
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