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Paha777 [63]
2 years ago
3

Due to loading, a line segment has length 2 m with constant normal strain 0.25. What is the original length of the line segment?

Due to loading, a line segment has length 2 with constant normal strain 0.25. What is the original length of the line segment? 1.60 m 1.50 m 1.75 m 2.67 m 2.50 m 2.25 m
Engineering
1 answer:
dezoksy [38]2 years ago
5 0

Answer: A

Original length = 1.60 m

Explanation: given that due to loading, a line segment has length 2 with constant normal strain 0.25

Strain is the ratio of extension to original length. That is,

Strain = e/L

If a line segment has length 2, that means:

e + L = 2

e = 2 - L

And given that the strain = 0.25

Substitute all the parameters into the formula

0.25 = ( 2 - L ) / L

Cross multiply

0.25L = 2 - L

Collect the like terms

0.25L + L = 2

1.25L = 2

L = 2/ 1.25

L = 1.6 m

Therefore, original length is 1.6 metres

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Answer:

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Now,

The velocity in the bore is given by:

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v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s

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The velocity head is given by:

\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553

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\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45

P_{tank} = 5.5\times \rho \times g

P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}

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Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20
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The total amount of daily heat transfer is 1382.38 M w.

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<u>Explanation:</u>

Given data,

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h_{0} = 250 w/ m^{2} k

Pipe length = 20 m

Inner diameter d_{1} = 6 cm, r_{1} = 3 cm

Outer diameter d_{2} = 8 cm, r_{2} = 4 cm

The thickness of insulation is 4 cm.

r_{3} = r_{2} + 4

= 4+4

r_{3} = 8 cm

h_{0} is the heat transfer coefficient of  convection inside, h_{i} is the heat transfer coefficient of  convection outside.

The heat transfer rate between ambient and steam is

q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}} watt

=  \begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned} watt

= \frac{190}{0.0003317+0.0000458+0.0110+0.0004976} watt

q = 15999.86 watt

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= 1382.387904 watt

= 1382.38 M w

The total amount of daily heat transfer is 1382.38 M w.

b) The temperature on the outside surface of the gypsum plaster insulation.

q = \frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}

15999.86   =\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}

T_{3} - 10 = 7.96

T_{3} = 17.96 ° C.

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2 years ago
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