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Anna71 [15]
3 years ago
14

The equation sigma = My/l is used in the mechanics of materials to determine normal stresses in beams. When this equation is exp

ressed in terms of SI base units. M is in newton-meters (N-m), y is in meters (m), and l is in meters to the fourth power (m^4). What are the SI units of sigma? If M = 2000 N-m v = 0.1 m, and l = 7 times 10^-5 m^4, what is the value of sigma in U.S. Customary base units?
Engineering
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Part a : The SI unit of σ is Pascal.

Part b : The pressure is 414.28 psi.

Explanation:

Part a

The equation is given as

\sigma =\frac{My}{l}

As per the dimensional analysis

M=[N-m]\\y=[m]\\l=[m^4]

So the equation becomes

\sigma =\frac{[N-m][m]}{[m^4]}\\\sigma =\frac{[N][m^2]}{[m^4]}\\\sigma =\frac{[N]}{[m^{4-2}]}\\\sigma =\frac{[N]}{[m^{2}]}\\

As the dimensions are of pressure so the SI unit of σ is Pascal.

Part b

\sigma =\frac{My}{l}\\\sigma =\frac{2000 \times 0.1}{7 \times 10^{-5}}\\\sigma =2857142.857  \, Pa

Pressure in US customary base units is given in psi so

1 Pa =1.45 \times 10^{-4}\\

So

\sigma =2857142.857  \times 1.45 \times 10^{-4} psi\\\sigma =414.28 psi

So the pressure is 414.28 psi.

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valina [46]

Answer:

a) the rate of heat transfer from the pipe to the air is 23.866 watts

b) YES, the rate of heat transfer changes to 3518.61 watt

Explanation:

Given that:

steam is saturated at 17.90 bar.

the pipe is stainless steel and has an outside diameter of 6.75 cm

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Bulk fluid temperature of 27°C

we know that

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

Outside diameter of pipe = 6.75 cm

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Cp of air = 1.005 kJ/Kgk

viscosity of air = 1.81 × 10⁻⁵ kg/(m.sec)

thermal conductivity of air = 2.624 × 10⁻⁵ kw/m.k

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hD/k = 0.028 (Re)^0.8 (Pr)^0.33

hD/k = 0.028 (Dvp / u)^0.8 (Cpu / k)^0.33

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([0.0675 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 0.0414 w/m².k

a)

Now to find the rate of heat transfer Q

Q = hAΔT

Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 23.866 watts

therefore the rate of heat transfer from the pipe to the air is 23.866 watts

b)

Now the flow direction changes to parallel flow, then

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

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so to find the rate of heat transfer Q

Q = hAΔT

Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 3518.61 watt

Therefore the rate of heat transfer changes to 3518.61 watt

4 0
3 years ago
A piston/cylinder contains 1.5 kg of water at 200 kPa, 150°C. It is now heated by a process in which pressure is linearly relate
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Answer:

final volume V2 = 0.71136 m³

work done in process W = -291.24 kJ

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Explanation:

given data

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temperature t1 = 150°C

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final temperature t2 = 350°C

solution

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V2 = 0.47424 × 1.5

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P(avg) = \frac{p1+p2}{2}    = \frac{200+600}{2} = 400 × 10³

put here value

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