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levacccp [35]
3 years ago
8

A cylindrical bar of metal having a diameter of 20.3 mm and a length of 205 mm is deformed elastically in tension with a force o

f 46400 N. Given that the elastic modulus and Poisson's ratio of the metal are 66.9 GPa and 0.32, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Engineering
1 answer:
Zarrin [17]3 years ago
5 0

Answer:

a) 0.4393 mm

b)  -0.141 mm

Explanation:

Cylindrical bar :  Diameter = 20.3 mm , Length = 205 mm

Force that deforms bar of metal = 46400 N

elastic modulus = 66.9 GPa

Poisson's ratio ( u ) = 0.32

A) Determine the amount by which this specimen will elongate in the direction of applied stress

dl = \frac{P*L}{A*E}

where P = 46400 N

          L = 205 mm

          A = \frac{\pi }{4} * 20.3^2 = 323.65

          E = 66.9 GPa

dl =  ( 46400 * 205 ) / ( 323.65 * 66.9 * 10^3 )  = 0.4393 mm

B) determine the change in diameter of the specimen

change in diameter( compressed due to elongation in length )

= - u * dl

 = - 0.32 * 0.4393

=  -0.141 mm

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Answer:

The answers to the question are

(1) Process 1 to 2

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(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

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Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

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Explanation:

(a) Volume compression ratio \frac{v_1}{v_2}  = 10

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

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Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

\frac{p_1}{p_2}  = (\frac{v_2}{v_1} )^n Therefore p₂ = p₁ ÷ (\frac{v_2}{v_1} )^n = (1 bar) ÷ (\frac{1}{10} )^{1.3} = 19.953 bar

Similarly for a polytropic process we have

\frac{T_1}{T_2}  = (\frac{v_2}{v_1} )^{n-1} or T₂ = T₁ ÷ (\frac{v_2}{v_1} )^{n-1} = \frac{310}{0.1^{0.3}} = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = \frac{8.3145}{28.9628} = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R =  1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = \frac{R(T_2-T_1)}{n-1} = \frac{0.287(618.531-310)}{1.3 - 1}= 295.16 kJ/kg

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W = 295.16 kJ/kg

Q = -73.79 kJ/kg

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W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = \frac{R(T_4-T_3)}{n-1} = Where T₄ is given by  \frac{T_4}{T_3}  = (\frac{v_3}{v_4} )^{n-1} or T₄ = T₃ ×0.1^{0.3}

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W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

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Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

\eta = 1-\frac{T_4-T_1}{T_3-T_2} =1-\frac{1102.611-310}{2200-618.531} = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

p_{m}  = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}  where r = compression ratio and r_p = \frac{p_3}{p_2}

However p₃ = \frac{p_2T_3}{T_2} =\frac{(19.953)(2200)}{618.531} =70.97 atm

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Answer:

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Explanation:

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outside radius r2 = 8 cm

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inside temperature t1 = 430°C

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to find out

Determine the heat loss per 1-m length of this insulation

solution

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3 years ago
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Answer:

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