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Ilya [14]
3 years ago
8

A steel cylindrical pressure vessel has an inside diameter of 2m and a wall thickness of 2cm. If the steel has a yield stress Y

= 425 MPa, calculate the tank pressure that would just cause yielding of the tank material:
Engineering
1 answer:
Feliz [49]3 years ago
4 0

Answer:

The pressure that will cause yielding in the tank in the longitudinal direction is 1.7 Mpa

Explanation:

We are dealing with a thin-walled pressure vessel here.

the formula for calculating the hoops stress of this pressure vessel is \sigma =P d/4t

\sigma Y = longitudinal yield stress = 425  \times 10^{6} Pa

Thickness = 2/1000 =0.002m

Outside diameter = inside siameter + thickness =  2m + 0.002m= 2.002m

Dm = Mean diameter = Outside diameter - thickness = 2.002 - 0.002 = 2m

P= 425 \times 10^{6} \times 4 \times 0.002 /2=1.7 Mpa

You might be interested in
). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially
Delicious77 [7]

Answer:

\frac{e'_z}{e_z} = 0.87142

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            \frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ]  }{-\frac{s_z}{E}}  \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142... Answer

5 0
3 years ago
‏What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Virty [35]

Answer:

E = 2940 J

Explanation:

It is given that,

Mass, m = 12 kg

Position at which the object is placed, h = 25 m

We need to find the potential energy of the mass. It is given by the formula as follows :

E = mgh

g is acceleration due to gravity

E=12\times 9.8\times 25\\\\E=2940\ J

So, the potential energy of the mass is 2940 J.

3 0
3 years ago
Water "bubbles up" h2 = 9 in. above the exit of the vertical pipe attached to three horizontal pipe segments. The total length o
8_murik_8 [283]

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

<h3>What is Bernoulli's equation?</h3>

The Bernoulli's equation for incompressible fluid can be given as,

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2

Here, ρ is density of fluid, g is acceleration due to gravity, (P) is the pressure v is velocity, h is height of elevation and subscript (1 and 2) is used for point 1 and 2.

Water "bubbles up" h2 = 9 in.  Above the exit of the vertical pipe attached to three horizontal pipe segments.

The total length of the 1.75-in.- diameter galvanized iron pipe between point (1) and the exit is 21 inches. h3 = 18 in.

Reynolds number is,

R_e=\dfrac{V_3D}{v}\\R_e=\dfrac{4.01\dfrac{0.75}{12}}{1.21\times10^{-15}}\\R_e=2.07\times10^4

The formula for the ratio of pressure to radius, when pressure and velocity at point 2 is zero can be given as,

\dfrac{p_1}{r}=z_2\left(f\dfrac{l}{d}+\sum k_L\right)\dfrac{v^2}{2g}-\dfrac{v_1^2}{2g}

Here, f=0.039 and ∑K(L)=4.5. Put the values,

\dfrac{p_1}{r}=\dfrac{7}{12}\left(0.03\dfrac{21}{0.75}+4.5-1\right)\dfrac{4.01^2}{2\times32.2}\\\dfrac{p_1}{r}=1.73\rm\; ft

Put the value of r we get,

p_1=62.4\times1.73\\p_1=108\rm\; ib/ft^2\\p_1=0.750\rm\; psi

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

Learn more about the Bernoulli's equation here;

brainly.com/question/7463690

#SPJ1

5 0
2 years ago
Before proceeding into an intersection at a green signal, you should look out for
Lyrx [107]

Answer:

vehicles follower you preparing to turn , vehicles in the intersection waiting to turn right. c

Explanation:

5 0
4 years ago
Read 2 more answers
The number of the size of the cup is determined by ____________ diameter of the cup.
nadya68 [22]

Answer:

its determined by the size of inside the cup

6 0
3 years ago
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