Ask question

Ask question

All categories

3 years ago

8

A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that

can be used, knowing that the normal stress must not exceed 150 MPa and that the increase in length of the wire must not exceed 25 mm

1 answer:

sasho [114]3 years ago

3 0

Answer:

d=13.81 mm

Explanation:

Given that

P = 15 KN ,L = 50 m

E= 200 GPa

ΔL = 25 mm

σ = 150 MPa

Lets take d=Diameter

There are we have two criteria to find out the diameter of the wire

Case I :

According to Stress ,σ = 150 MPa

P = σ A

$A=\dfrac{P}{\sigma}$

$d=\sqrt{\dfrac{4P}{\pi \sigma}}$

By putting the values

$d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}$

d= 11.28 mm

Case II:

According to elongation ,ΔL = 25 mm

$\Delta L=\dfrac{PL}{AE}$

$A=\dfrac{PL}{E\Delta L}$

$A=\dfrac{4PL}{\pi E\Delta L}$

$d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}$

d=13.81 mm

Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.

You might be interested in

In any system, the _____ in the system remains constant, while the PE or KE may change.

vladimir1956 [14]

Answer:

The sum of PE and KE remains constant

Explanation:

6 0

3 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

3 0

3 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

7 0

3 years ago

The slope of the line on a speed-time graph tells the speed , true or false ?

KengaRu [80]

True I hope this helps you out

8 0

3 years ago

Read 2 more answers

Determining the pH of substances such as purple grape juice and catsup using test strips can be difficult. Why?

Charra [1.4K]

Answer:

Because they would naturally dye the test strips in the colors violet and red, regardless of their pH values

(would really appreciate the brainliest)

4 0

3 years ago