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2 years ago

12

The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single cry

stal of Fe pulled in tension.

1 answer:

jasenka [17]2 years ago

7 0

Answer:

Answer is mentioned below.

Explanation:

Answer: Calculation of maximum yield strength for a single crystal of Fe pulled in tension is shown in the image attached .

value of max. Yield strength= 54MPa

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Derive an equation for the work of a mechanically reversible, isothermal compression of 1 mol of a gas from an initial pressure

Lyrx [107]

Answer:

The long derivation for work of a mechanically reversible, isothermal compression was done with detailed steps as shown in the attachment.

Explanation:

what is applied here is a long derivation from Work done in an isobaric process, the expression for the compressibility factor (Z) and the equation of state that was given. The requisite knowledge of Differentiation and Integration was used.

The detailed derivation from firs principle is as shown in the attachment.

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2 years ago

Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.

pantera1 [17]

Answer:

Eye.

Explanation:

Dust, dirt, and metal chips are most unpleasant to get in your eyes. Just experience it and you'll know what I mean.

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3 0

2 years ago

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

Viktor [21]

Given:

Pressure, $P_{1}$ = 1300 kPa

Temperature, $T_{1}$ = $500^{\circ}$

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

$v_{2}$ = $1.15\frac{m^{3}}{kg}$

a) Mass flow rate is given by:

$m' = \frac{Av}{v_{1}}$

Now,

$m = \frac{0.2\times 40}{0.172}$ = 46.51 kg/s

b) for the power produced by turbine, $P = m'(h_{1} - h_{2})$

$P = 46.51\times(793 - 401)$ = 18.231 MW

5 0

2 years ago

A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe ha

poizon [28]

Answer :

The force needed to move the cylinder is 25.6 N

<h2>Further explanation </h2>

Given that,

Length of the cylinder, l = 0.5 m

Outer diameter of the cylinder, d = 8 cm = 0.08 m

Outer radius of the cylinder, $r=0.04\ m$

Inside diameter of the pipe, d = 8.5 cm = 0.085 m

Inside radius of the pipe, $r=0.0425\ m$

Specific gravity of the oil, $\rho=0.92$

Density of oil, $d=\rho\times \rho_w$

Kinematic viscosity of the oil, $v=5.57\times 10^{-4}\ m^2/s$

Velocity of the cylinder, u = 1 m/s

We need to find the force needed to move the cylinder. Let the force is F.

Specific gravity is defined as the ratio of the density of the substance to the density of water.

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

$v=\dfrac{\mu}{d}$

Where, $d$ = density of oil

And $d=\rho\times \rho_w$ (density of oil = specific gravity × density of water )

$d=0.92\times 10^3\ kg/m^3$

So,

$\mu=v\times d$ ..............(1)

$\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3$

$\mu=0.512\ Pa-s$

The separation between the cylinder and pipe is given by :

$dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m$

$d_p\ and\ d_c$ are diameter of pipe and cylinder respectively.

The mathematical expression for the Newton's law of viscosity can be written as:

$\tau\propto\dfrac{du}{dy}$

$\tau=\mu\times \dfrac{du}{dy}$ ..........(2)

Where

$\tau$ = Shear stress, $\tau=\dfrac{F}{A}$ ............(3)

$\mu$ = viscosity

$\dfrac{du}{dy}$ = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :

$\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}$ ...............(4)

A is the area of the cylinder, $A=2\pi rl$

Equation (4) becomes :

$F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl$ ..............(5)

$A=\pi d\times l$

$A=\pi \times 0.08\ m\times 0.5\ m$

$A=0.125\ m^2$

Now, equation (5) becomes :

$F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl$

$F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2$

F = 25.6 N

<h2>Learn more </h2>

Kinematic viscosity : brainly.com/question/12947932

<h2>Keyword : </h2>

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.

7 0

2 years ago

A 10-mm drill rod was heat-treated and ground. The measured hardness He was found to be 300 Brinell. Estimate the endurance stre

lesya [120]

Answer:

The endurance strength for the rod is 434.6 MPa

Explanation:

Since the rod is used in rotating bending, we need to use Marin equation given by

$S=k_ak_bS'_e$

Here S stands for the endurance strength for rotating bending, $S'_e$ is the endurance strength, and $k_a \text{ and } k_b$ are the parameters for Marin surface modification factor.

Endurance strength.

We can start finding the endurance strength, from the directions we know that the hardness $H_e$ was found to be 300 Brinell, thus for such value we can find the ultimate tensile strength using

$S_{ut}=3.41H_e$

Replacing the hardness we get

$S_{ut}=3.41(300) MPa \\ S_{ut}=1023 MPa$

Now since the ultimate tensile strength has a value less than 1400 MPa, we can find the endurance strength using

$S'_e =0.5S_{ut}$

Replacing the tensile strength we get

$S'_e=0.5(1023) MPa \\ S'_e = 511.5 MPa$

Parameters for Marin surface modification factor.

From the directions we know that the drill rod has a ground surface finish, so then from tables we get

$a=1.58 \text{ and } b = -0.085$

Thus the surface factor will be

$k_a=a(S_{ut})^b$

Replacing values and the ultimate tensile strength

$k_a=(1.58)1023^{-0.085}\\k_a=0.8766$

Then we can find the rotating shaft factor, for a diameter of 10 mm, we can use the equation

$k_b=1.24d^{-0.107}$

Replacing the diameter we get

$k_b=1.24(10)^{-0.107}\\k_b=0.9692$

Estimating endurance strength for rotating shaft.

We can replace now all values we have found in Marin equation.

$S=k_ak_bS'_e$

$S=(0.8766)(0.9692)(511.5) MPa$

$S=434.6 MPa$

Thus the endurance strength for the rod is 434.6 MPa

3 0

2 years ago