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JulijaS [17]
2 years ago
8

Which statement is not correct regarding the deformation of a circular shaft in torsion? a. Cross sections remain flat. b. Longi

tudinal lines remain straight. c. Circular sections remain circular. d. Radial lines on the sections remain straight.
Engineering
1 answer:
sveta [45]2 years ago
5 0

Answer:

B

Explanation:

Torsion is application of torque to a shaft to turn it about its longitudinal axis. When torque is applied to a shaft the circle remains unchanged in a circular state, its cross section does not warp but remains flat with a straight radial lines but its longitudinal lines changes into an helix intersecting the circular shaft

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The total floor area of a building, including below-grade space but excluding unenclosed areas, measured from the exterior of th
alex41 [277]

Answer:

Gross building area

Explanation:

The Gross building area refers to the entire area of a building covering all the floors. The measurement is expressed in square feet. The Gross building area also includes basements, penthouses, and mezzanines. It is calculated by estimating the exterior dimension of the building. Storage rooms, laundries, staircases are also a part of the gross building area.

6 0
2 years ago
air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying
notsponge [240]

Answer:

a) 251.31 m/s and 55.29 m/s

b) The mass flow rate is 0.0396 kg/s

c) The rate of entropy production is 0.0144 kW/K

Explanation:

a) The steady state is:

mi = mo

\frac{A_{i}V_{i}  }{v_{i} } =\frac{A_{o}V_{o}  }{v_{o} } \\V_{i} =V_{o}(\frac{v_{i}}{v_{o}} )=V_{o}(\frac{T_{i}P_{o} }{T_{o}P_{i} } )=V_{o}(\frac{330*120}{300*600}) =0.22V_{o}

The energy balance is:

h_{i}+\frac{V_{i}^{2} }{2} =h_{2}+\frac{V_{o}^{2} }{2} \\h_{i}-h_{o}+(\frac{V_{i}^{2}-V_{o}^{2}  }{2} )=0\\V_{o}=\sqrt{\frac{2(h_{i}-h_{o})}{0.9516} } =\sqrt{\frac{2*(330.24-300.19)x10^{3} }{0.9516} } =251.31m/s

Vi = 0.22 * 251.31 = 55.29 m/s

b) The mass flow rate is:

m=\frac{A_{o}V_{o}}{v_{o}} =\frac{\pi d^{2}V_{o}P_{o} }{4RT_{o}} =\frac{\pi *(0.012^{2})*251.31*120x10^{3}  }{4*287*300} =0.0396kg/s

c) The entropy produced is equal to:

\frac{Q}{T} +S_{gen} =m(s_{2} -s_{1} )\\0+S_{gen} =m(s_{2} -s_{1} )\\S_{gen} =m(s_{2} -s_{1} )\\S_{gen}=0.0396*(c_{p} ln\frac{T_{o}}{T_{i}} -Rln\frac{P_{o}}{P_{i}} )=0.0396*(1.004ln\frac{300}{330} -0.287ln\frac{120}{600} )=0.0144kW/K

3 0
2 years ago
An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30
sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

F = LwY_{avg}

where;

w is the width of the annealed copper

Y_{avg} is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm

Now, determine the average true stress, Y_{avg}, for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

F = LwY_{avg}

F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F =   1.59 \ MN

The power required in this operation is given by;

P = \frac{2\pi FLN}{60}\\\\P =  \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW

5 0
2 years ago
Which of the following is NOT true about hydraulic valves? A. Directional control valves determine the path of a fluid in a give
Lelechka [254]

Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path

Explanation:

The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction

5 0
2 years ago
A person puts a few apples into the freezer at -15oC to cool them quickly for guests who are about to arrive. Initially, the app
frosja888 [35]

Answer:

Temperature at center of apples = 11.2⁰C

Temperature at surface of apples = 2.7⁰C

Amount of Heat transferred = 17.2kJ

Explanation:

The properties of apple are given as:

k = 0.418 W/m.°C

ρ = 840 kg/m³

Cр = 3.81 kJ/kg.°C

α = 1.3*10 ⁻⁷ m²/s

h = 8 W/m².°C

d = 0.09m

r = 0.045m

t = 1 hour = 3600s

<h2>Solution</h2>

Biot number is given as:

Bi = \frac{hr}{k}= \frac{8\cdot0.045}{0.418}=0.861

The constants λ₁ and A₁ corresponding to Biot number (from the table) are:

λ₁ = 1.476

A₁ = 1.239

Fourier Number is:

T = \frac{a\cdot{t}}{r^2} = \frac{(1.3\cdot10^{-7})(3600)}{0.045^2}= 0.231> 0.2

As Fourier Number > 0.2 , one term approximates solutions are applicable

The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.

8 0
2 years ago
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