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SSSSS [86.1K]
3 years ago
13

The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr

ic axial load F is applied at the end cap. Use p = 1402 kPa, F= 13 kN, t 18 mm and r =306 mm.
Required:
Obtain the state of plane stress in the x - y coordinate system
Engineering
1 answer:
Sedaia [141]3 years ago
3 0

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

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On the position time curve, if the slope of a tangent at a point is positive, that means:
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Answer:

  C. the object is moving forward

Explanation:

A positive slope means position is increasing when time is increasing. Generally, increasing position is "moving forward."

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3 years ago
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Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with
Scilla [17]

Answer:

DIAMETER  = 9.797 m

POWER = \dot W = 28.6 kW

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

v =\frac{RT}{P}

  where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

v =  \frac{0.287\times 293}{100}

v = 0.8409 m^3/ kg

from continuity equation

A_1 V_1 = A_2 V_2

\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2

D_2 = D_1 \sqrt{\frac{V_1}{V_2}}

D_2 = 8 \times \sqrt{\frac{12}{8}}

D_2 = 9.797 m

mass flow rate is given as

\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}

\dot m = 717.309 kg/s

the power produced \dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]

\dot W = 28.6 kW

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3 years ago
Before finishing and installing a shelved cabinet you just constructed, you need to check the
Greeley [361]

Answer:

Carpenter's square

Explanation:

The most common hand tool used to measure or set angles with its application extending to setting angles of roofs and rafters. Another name of a Carpenter's square is a framing square.

Other hand tools that are used to measure angles are;

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  • A Bevel that allows users to set any angle they like.
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7 0
2 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
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