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Jlenok [28]
3 years ago
9

A thick spherical pressure vessel of inner radius 150 mm is subjected to maximum an internal pressure of 80 MPa. Calculate its w

all thickness based upon the (a) principal stress theory, and (b) total strain energy theory. Poisson's ratio = 0-30, yield strength 300 MPa.
Engineering
1 answer:
ivann1987 [24]3 years ago
5 0

Answer:

by principal stress theory

t = 20.226

by total strain theory

t = 20.36

Explanation:

given data

internal radius r_{1} = 150 mm

pressure p = 80 MPa

yield strength = 300 MPa

poisson's ratio = 0.3

a) by principal stress theory

thickness can be obtained as t

t  = r_{1}\left [ (\frac{\sigma _{y} +p}{\sigma _{y} - 0.5p})^{1/3}-1 \right ]

t = = 150\left [ (\frac{300 +80}{300-0.5*80})^{1/3}-1 \right ]

t = 20.226

b) by total strain theory

m =\frac{\sigma _{y}}{p}

m = \frac{300}{80} = 3.75

we know that

K = \frac{r_{2}}{r_{1} }

\frac{K^{3}+1}{K^{3}-1}= \frac{-2\mu +\sqrt{4\mu^{2}-2(1-\mu)(1-m^{^{2}}))}}{1-\mu}

\frac{K^{3}+1}{K^{3}-1}= \frac{-2*0.3 +\sqrt{4*0.3^{2}-2(1-0.3)(1-3.75^{^{2}}))}}{1-0.3}

\frac{K^{3}+1}{K^{3}-1}= 5.3

k = 1.13

1.13 = \frac{r_{2}}{150 }

r_{2} = 170.36 mm

t = r_{2}-r_{1}

t = 170.36 - 150

t = 20.36

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6 0
2 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

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z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

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3 0
3 years ago
A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63
Fudgin [204]

Answer:

The percentage volume change is -3.0%

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Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0
3 years ago
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