Answer:
screw is the answer of the question
the function is to provide sealed combustion so that the loss of gas is minimized
Answer:
special type
Explanation:
As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.
Answer:d
Explanation:
Given
Temperature![=200^{\circ}\approc 473 K](https://tex.z-dn.net/?f=%3D200%5E%7B%5Ccirc%7D%5Capproc%20473%20K)
Also ![\gamma for air=1.4](https://tex.z-dn.net/?f=%5Cgamma%20for%20air%3D1.4)
R=287 J/kg
Flow will be In-compressible when Mach no.<0.32
Mach no.![=\frac{V}{\sqrt{\gamma RT}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7BV%7D%7B%5Csqrt%7B%5Cgamma%20RT%7D%7D)
(a)![1000 km/h\approx 277.78 m/s](https://tex.z-dn.net/?f=1000%20km%2Fh%5Capprox%20277.78%20m%2Fs)
Mach no.![=\frac{277.78}{\sqrt{1.4\times 287\times 473}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B277.78%7D%7B%5Csqrt%7B1.4%5Ctimes%20287%5Ctimes%20473%7D%7D)
Mach no.=0.63
(b)![500 km/h\approx 138.89 m/s](https://tex.z-dn.net/?f=500%20km%2Fh%5Capprox%20138.89%20m%2Fs)
Mach no.![=\frac{138.89}{\sqrt{1.4\times 287\times 473}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B138.89%7D%7B%5Csqrt%7B1.4%5Ctimes%20287%5Ctimes%20473%7D%7D)
Mach no.=0.31
(c)![2000 km/h\approx 555.55 m/s](https://tex.z-dn.net/?f=2000%20km%2Fh%5Capprox%20555.55%20m%2Fs)
Mach no.![=\frac{555.55}{\sqrt{1.4\times 287\times 473}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B555.55%7D%7B%5Csqrt%7B1.4%5Ctimes%20287%5Ctimes%20473%7D%7D)
Mach no.=1.27
(d)![200 km/h\approx 55.55 m/s](https://tex.z-dn.net/?f=200%20km%2Fh%5Capprox%2055.55%20m%2Fs)
Mach no.![=\frac{55.55}{\sqrt{1.4\times 287\times 473}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B55.55%7D%7B%5Csqrt%7B1.4%5Ctimes%20287%5Ctimes%20473%7D%7D)
Mach no.=0.127
From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.
Complete Question
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
The elongation is ![=21.29mm](https://tex.z-dn.net/?f=%3D21.29mm)
Explanation:
In order to gain a good understanding of this solution let define some terms
True Stress
A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as
.
True Strain
A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as
.
The mathematical relation between stress to strain on the plastic region of deformation is
![\sigma _T =K\epsilon^n_T](https://tex.z-dn.net/?f=%5Csigma%20_T%20%3DK%5Cepsilon%5En_T)
Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows
![K = \frac{\sigma_T}{(\epsilon_T)^n}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5Csigma_T%7D%7B%28%5Cepsilon_T%29%5En%7D)
No substituting
from the question we have
![K = \frac{345}{(0.02)^{0.22}}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B345%7D%7B%280.02%29%5E%7B0.22%7D%7D)
![= 815.82MPa](https://tex.z-dn.net/?f=%3D%20815.82MPa)
Making
the subject from the equation above
![\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }](https://tex.z-dn.net/?f=%5Cepsilon_T%20%3D%20%28%5Cfrac%7B%5Csigma_T%7D%7BK%7D%20%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%7D)
![Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n](https://tex.z-dn.net/?f=Substituting%20%5C%20411MPa%20%5C%20for%20%5C%20%5Csigma_T%20%5C%20815.82MPa%20%5C%20for%20%5C%20K%20%20%5C%20and%20%20%5C%20%200.22%20%5C%20for%20%5C%20n)
![\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }](https://tex.z-dn.net/?f=%5Cepsilon_T%20%3D%20%28%5Cfrac%7B411MPa%7D%7B815.82MPa%7D%20%29%5E%7B%5Cfrac%7B1%7D%7B0.22%7D%20%7D)
![=0.0443](https://tex.z-dn.net/?f=%3D0.0443)
From the definition we mentioned instantaneous length and this can be obtained mathematically as follows
![l_i = l_o e^{\epsilon_T}](https://tex.z-dn.net/?f=l_i%20%3D%20l_o%20e%5E%7B%5Cepsilon_T%7D)
Where
is the instantaneous length
is the original length
![Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T](https://tex.z-dn.net/?f=Substituting%20%20%5C%20470mm%20%5C%20for%20%5C%20l_o%20%5C%20and%20%5C%200.0443%20%5C%20for%20%20%5C%20%5Cepsilon_T)
![l_i = 470 * e^{0.0443}](https://tex.z-dn.net/?f=l_i%20%3D%20470%20%2A%20e%5E%7B0.0443%7D)
![=491.28mm](https://tex.z-dn.net/?f=%3D491.28mm)
We can also obtain the elongated length mathematically as follows
![Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i](https://tex.z-dn.net/?f=Substituting%20%5C%20470mm%20%5C%20for%20l_o%20and%20%5C%20491.28%20%5C%20for%20%5C%20l_i)
![Elongated \ Length = 491.28 - 470](https://tex.z-dn.net/?f=Elongated%20%5C%20Length%20%3D%20491.28%20-%20470)
![=21.29mm](https://tex.z-dn.net/?f=%3D21.29mm)