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9966 [12]
3 years ago
8

You just purchased a 400-L rigid tank for a client who works in the gas industry. The tank is delivered pre-filled with 3 kg of

air at a temperature of 25C. Assuming an atmospheric pressure of 98 kPa, determine the pressure reading if you connected a pressure gage to the tank.
Engineering
1 answer:
solniwko [45]3 years ago
6 0

Answer:

the pressure reading when connected a pressure gauge is 543.44 kPa

Explanation:

Given data

tank volume (V) = 400 L i.e 0.4 m³

temperature (T) =  25°C  i.e. 25°C + 273 = 298 K

air mass (m)  = 3 kg

atmospheric pressure  = 98 kPa

To find out

pressure reading

Solution

we have find out pressure reading by gauge pressure

i.e. gauge pressure = absolute pressure - atmospheric pressure

first we find absolute pressure (p) by the ideal gas condition

i.e pV = mRT

p = mRT / V

p = ( 3 × 0.287 × 298 ) / 0.4

p = 641.44 kPa

so

gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 641.44 - 98

gauge pressure = 543.44 kPa

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The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0
3 years ago
Question 4 (1 point) Sophia is working in her garden. She made two trips with her wagon along the same path to haul supplies. Th
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4 0
3 years ago
A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is 8.621 \times 10^{-4} in/in

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

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

Here

  • M is the moment which is to be calculated
  • I is the moment of inertia given as

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

Here

  • b is the breath given as 0.75"
  • d is the depth which is given as 8"

                     I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4

  • y is given as

                     y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\

  • Force is 50 ksi

\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

So the strain is 8.621 \times 10^{-4} in/in

Part c:

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

The plastic moment is 600 ksi.

3 0
3 years ago
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