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maksim [4K]
3 years ago
11

An empty metal can is heated to 908C and sealed. It is then placed in a room to cool to 208C. What is the pressure inside the ca

n upon cooling? Assume that can contains air under ideal conditions. ?
Engineering
1 answer:
Natali5045456 [20]3 years ago
3 0

The pressure inside the can upon cooling is 0.4 atm.

<u>Explanation:</u>

Given -

Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K

Final Temperature, T2 = 208°C = 208 + 273 K = 481 K

Pressure upon cooling, P2 = ?

Using Gay Lussac's law:

P1/T1 = P2/T2

P2 = P1 X T2 / T1

P2 = 1 atm X 481 / 1181

P2 = 0.4 atm

Therefore, the pressure inside the can upon cooling is 0.4 atm.

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A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
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Part a:

As per bending equation

\frac{M}{I}=\frac{F}{y}

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                         I=\frac{bd^3}{12}

Here

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\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in

The yield moment is 400 k.in.

Part b:

The strain is given as

Strain=\frac{Stress}{Elastic Modulus}

The stress at the station 2" down from the top is estimated by ratio of triangles as

                        F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi

Now the steel has the elastic modulus of E=29000 ksi

Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in

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For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi

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