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There are many diferent materials available for seal faces . List the following seal face materials in order of hardness. i.e Hardest first, softest last. (a) 316 Stainless Steel (b)-Mild steel (c)- Reaction bonded Silicon carbide (d)- Tungsten carbide

Answers

Answer 1

Answer:

Answer:

Reaction bonded Silicon carbide: 2500-3500 HV

Tungsten carbide: 1800-2500 HV.

316 Stainless Steel: 152 HV

Mild steel: 130 HV

Explanation:

In order to list those seal face materials by hardness, we look up what are the values of hardness for each material in a hardness scale.

We are going to use Vickers scale, an indentation method of measuring hardness, it measures the deformation left in a sample by a constant compression load from an indenter (a diamond pyramid) with an adequate (to the material) force, as the result is independent from the test force.

1. Reaction bonded Silicon carbide: 2500-3500 HV

2. Tungsten carbide: 1800-2500 HV

3. 316 Stainless Steel: 152 HV

4. Mild steel: 130 HV

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or a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 164 mm2 without plastic deformation

Answers

Answer:

The maximum load (in N) that may be applied to a specimen with this cross sectional area is $F=43788 N$

Explanation:

We know that the stress at which plastic deformation begins is 267 MPa.

We are going to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:

$P=(F)/(A)$ (equation 1)

We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.

Before apply the equation, we need to convert the units of area in $m^(2)$ . So,

$A[m^(2)]=A[mm^(2) ]((1 m)/(1000 mm)) ^(2)\nA[m^(2)]=164 mm^(2)((1 m)/(1000 mm))^(2) \nA[m^(2)]=0.000164 m^(2)$

And then, from the equation 1,

$F=PA\nF=(267 MPa)(0.000164 m^(2))\nF=(267x10^(6) Pa)(0.000164 m^(2))\nF=(267x10^(6) N/m^(2))(0.000164 m^(2))\nF=43788 N$

A system with a mass of 8 kg, initially moving horizontally with a velocity of 40 m/s, experiences a constant horizontal force of 25 N opposing the direction of motion. As a result, the system comes to rest.
Determine the amount of energy transfer by work, in kJ, for this process and the total distance, in m, that the system travels.

Answers

Answer:

The kinetic energy is 6.4 kJ and the distance traveled by the system is 256 m.

Explanation:

Given the mass of the system $(m)$ is 8 kg.

And initially, it moves with a velocity $(v)$ 40 m/s.

Also, it experiences 25 N force $(f)$ which opposes its motion.

We need to find the kinetic energy and the distance traveled by the system $(d)$ before going to rest.

It will be the kinetic energy of 8 kg mass with 40 m/s velocity that is transferred to work.

$K.E=(1)/(2)mv^2\nK.E=(1)/(2)* 8* 40^2\nK.E= 6400\ J\nK.E=6.4\ kJ$

Since this system is opposed by 25 N force, work done by the force will be.

$W=f(d)$

And the kinetic energy transferred to work. We can equate them.

$f(d)=6400\n25(d)=6400\nd=(6400)/(25)=256\ m$

So, the system will travel 256 m.

Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)

Answers

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

$E(t)=E_o[1-a(T)/(T_m)]$

where,

E(t) is the modulus of elasticity at any temperature 'T'

$E_o$ is the modulus of elasticity at absolute zero.

$T_(m)$ is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

A 100-mm-diameter cast Iron shaft is acted upon by a 10 kN.m bending moment, a 8 kN.m torque, and a 150 kN axial force simultaneously. The ultimate tensile strength of the shaft material is 210 MPa, and the ultimate compressive strength of the shaft material is 750 MPa, determine the safety factor against failure for the above types of loading using a proper theory of failure.

Answers

Answer:

Are you smart You seem smart

Explanation:

im ur dad

Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

Answers

Answer:

Decrease to typical from utilizing lambda-decrease:

The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2

The of taking the terms is significant in lambda - math,

For the term, (λy, y×3)2, we can substitute the incentive to the capacity.

Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6

Presently the tem becomes, (λf λx f(f(fx)))6

The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.

Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.

In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.

As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine representative power requirements, consider nominal flight conditions for which the plane moves at 100 m/s in air that is at a temperature of -23 degree C. If the characteristic length of the airfoil is L = 2 m and wind tunnel measurements indicate an average friction coefficient of of C_f = 0.0025 for the nominal conditions, what is the average heat flux needed to maintain a surface temperature of T_s = 5 degree C?

Answers

Answer:

Average heat flux=3729.82 W/ $m^(2)$

Explanation:

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