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lyudmila [28]
3 years ago
9

For a column that is pinned at both ends, the critical buckling load can be calculated as, Pcr = π2 E I /L^2 where E is Young's

modulus, I is the moment of inertia of the cross sectional area, and L is the length of the column. If E = 200 × 109 N/m^2, I = 21 × 106 mm^4, and L = 12 ft what is the critical buckling load in kip?
Engineering
1 answer:
gulaghasi [49]3 years ago
6 0

When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area

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Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c
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Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress  acting on the bottom plate = 157.5 N/m²

(c) The velocity along the centerline of the channel = 0.93 m/s

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

  = 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

    = (1.8225*10⁻⁷/1.14) * (35000)

    = 1.60*10⁻⁷ * 35000

   = 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ  = h*(Δp/L)

    = 0.0045* 35000

    = 157.5 N/m²

(c) Calculating the velocity along the centre of the channel using the formula;

u(max) = h²/2μ)* (Δp/L)

   = (0.0045²/2*0.38) * 35000

   =2.664*10⁻⁵ *35000

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3 years ago
Describe how you would control employee exposure to excessive noise in a mining environment
Alexandra [31]

Answer:

1. Buy Quiet – select and purchase low-noise tools and machinery

2. Maintain tools and equipment routinely (such 3. as lubricate gears)

3. Reduce vibration where possible

4. Isolate the noise source in an insulated room or enclosure

5. Place a barrier between the noise source and the employee

6. Isolate the employee from the source in a room or booth (such as sound wall or window

Explanation:

Hope my answer will help u.

7 0
1 year ago
Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool
MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T  = 43.477° C

Explanation:

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mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

<u>Determine temperature of the cooled water exiting the cooling tower</u>

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

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<u>First step : calculate the value of Q </u>

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) +  43.13 (18.26089)

Q =  95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

<u>Hence the temperature of the cooled water can be calculated using the equation below</u>

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000.    ∴ T  = 43.477° C

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Explanation:

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Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

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