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Mila [183]
3 years ago
15

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

Engineering
1 answer:
Radda [10]3 years ago
3 0

Answer:

LMC wall thickness= 5.05

Explanation:

Given:

Minimum inner diameter = 25 - 0.05 = 24.95

Maximum outer diameter = 35 + 0.05 = 35.05

Find:

LMC wall thickness

Computation:

LMC wall thickness = (maximum outer diameter - minimum inner diameter) / 2

LMC wall thickness = (35.05 - 24.95) / 2

LMC wall thickness= 5.05

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Answer:

22

Explanation:

The question talks about a given program to which you input a set of numbers. Once you input number 10, the program stops receiving inputs and adds the numbers you previously inputed, except for the final 10 which only works as a stopping signal. So, imagine you are working on the program and input the numbers 8, 3, 11 and 10 consecutively. Once you input the 10, the program immediately adds the previous numbers: 8+3+11=22.

So the output should be 22.

4 0
4 years ago
What is the saturation pressure, Vf, and Vg for saturated water at a temperature of 287 °C?
Dvinal [7]

Answer:

Vf = specific volume of saturated liquid  = 0.0217158 ft^3/lb

Vg = specific volume of saturated steam = 0.430129 ft^3/lb

Explanation:

Given data:

water temperature is given as 287-degree Celcius

we have to find Vf and Vg

Vf = specific volume of saturated liquid  

Vg = specific volume of saturated steam

we know that from the saturated steam table we can find these value

therefore for temperature 287-degree Celcius

Vf = specific volume of saturated liquid  = 0.0217158 ft^3/lb

Vg = specific volume of saturated steam = 0.430129 ft^3/lb

6 0
3 years ago
On some engines, after torquing cylinder head fasteners, you
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Answer:

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7 0
3 years ago
While supporting load P, the maximum contraction permitted in a 4.75-m long pipe column is 7 mm. Given that E =180 GPa for the c
Alexeev081 [22]

Answer:

\mathbf{stress \ \sigma = 264.6 \ Mpa}

Explanation:

From the concept of Hooke's  Law,

E =\dfrac{ stress \ \sigma}{ strain \  \varepsilon}

where;

strain \ \varepsilon = \dfrac{change \ in \ dimension }{original \ dimension}

strain \ \varepsilon = \dfrac{7 \ mm }{4.75 \times 10^{3} \ mm}

strain \ \varepsilon =0.00147368

Recall:

E =\dfrac{ stress \ \sigma}{ strain \  \varepsilon}

stress \ \sigma = E \times { strain \  \varepsilon}

stress \ \sigma = 180 \times 10^{3} \ Mpa \times 0.00147

\mathbf{stress \ \sigma = 264.6 \ Mpa}

Thus, the stress in the pipe at the maximum allowable contraction = 264.6 Mpa

7 0
3 years ago
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