Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:
![\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}](https://tex.z-dn.net/?f=%5Cepsilon%20_%7B1%2C2%7D%20%3D%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%20%5Cpm%20%5Csqrt%7B%28%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%29%5E2%20%2B%20%28%5Cfrac%7B%5Cgamma_xy%7D%7B2%7D%29%5E2%7D%20%5C%5C%5C%5C%5Cepsilon%20_%7B1%2C2%7D%20%3D%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%20%2B%200%7D%7B2%7D%20%20%5Cpm%20%5Csqrt%7B%28%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%2B%200%7D%7B2%7D%29%20%5E2%20%2B%20%28%5Cfrac%7B150%20%5Ctimes%2010%5E-6%7D%7B2%7D%29%5E2%7D%5C%5C%5C%5C%5Cepsilon%20_%7B1%2C2%7D%20%3D%20-150%20%5Ctimes%2010%5E%7B-6%7D%20%20%5Cpm%201.67%20%5Ctimes%2010%5E%7B-4%7D)
ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂
![tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5](https://tex.z-dn.net/?f=tan%202%20%5Ctheta_p%20%3D%20%5Cfrac%7B%5Cgamma_xy%7D%7B%5Cepsilon_x%20-%20%5Cepsilon_y%7D%20%5C%5C%5C%5Ctan%202%20%5Ctheta_p%20%3D%20%5Cfrac%7B150%20%5Ctimes%2010%5E%7B-6%7D%7D%7B-300%20%5Ctimes%2010%5E%7B-6%7D-%5C%200%7D%5C%5C%5C%5Ctan%202%20%5Ctheta_p%20%3D%20-0.5)
θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂
![\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\](https://tex.z-dn.net/?f=%5Cepsilon%20_%7Bx%27%20%7D%3D%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%20%2B%20%5Cfrac%7B%5Cepsilon_x%20-%5Cepsilon_y%7D%7B2%7D%20cos2%5Ctheta%20%20%2B%20%5Cfrac%7B%5Cgamma_xy%7D%7B2%7Dsin2%5Ctheta%20%5C%5C%5C%5C%5Cepsilon%20_%7Bx%27%7D%20%3D%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%2B%20%5C%200%7D%7B2%7D%20%20%2B%20%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%20-%5C%200%7D%7B2%7D%20cos%28-26.56%29%20%20%2B%20%5Cfrac%7B150%20%5Ctimes%2010%5E%7B-6%7D%7D%7B2%7Dsin%28-26.56%29%5C%5C%5C%5C)
=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain
![\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}](https://tex.z-dn.net/?f=%5Cgamma_%7Bmax%20%5C%20in%20%5C%20plane%7D%20%3D2%5Csqrt%7B%28%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%29%5E2%20%2B%20%28%5Cfrac%7B%5Cgamma_xy%7D%7B2%7D%29%5E2%7D%20%5C%5C%5C%5C%5Cgamma_%7Bmax%20%5C%20in%20%5C%20plane%7D%20%3D%202%5Csqrt%7B%28%5Cfrac%7B-300%20%2A10%5E%7B-6%7D%20%2B%200%7D%7B2%7D%20%29%5E2%20%2B%20%28%5Cfrac%7B150%20%2A10%5E%7B-6%7D%7D%7B2%7D%29%5E2%7D)
=3.335 *10^-4
![\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%20%3D%28%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%29)
ε(avg) =150 *10^-6
orientation of γmax
![tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}](https://tex.z-dn.net/?f=tan%202%20%5Ctheta_s%20%3D%20%5Cfrac%7B-%28%5Cepsilon_x%20-%20%5Cepsilon_y%29%7D%7B%5Cgamma_xy%7D%20%5C%5C%5C%5Ctan%202%20%5Ctheta_s%20%3D%20%5Cfrac%7B-%28-300%2A10%5E%7B-6%7D%20-%200%29%7D%7B150%2A10%5E%7B-6%7D%7D)
θ = 31.71 or -58.29
To determine the direction of γmax
![\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)](https://tex.z-dn.net/?f=%5Cgamma%20_%7Bx%27y%27%20%7D%3D%20%20-%20%5Cfrac%7B%5Cepsilon_x%20-%5Cepsilon_y%7D%7B2%7D%20sin2%5Ctheta%20%20%2B%20%5Cfrac%7B%5Cgamma_xy%7D%7B2%7Dcos2%5Ctheta%20%5C%5C%5C%5C%5Cgamma%20_%7Bx%27y%27%20%7D%3D%20%20-%20%5Cfrac%7B-300%2A10%5E%7B-6%7D%20-%20%5C%200%7D%7B2%7D%20sin%2863.42%29%20%20%2B%20%5Cfrac%7B150%2A10%5E%7B-6%7D%7D%7B2%7Dcos%2863.42%29)
= 1.67 *10^-4
Hopefully that helps you out and is this for history or science?
Explanation:
Yes Diesel engine have problem of knocking.
We know that knocking is phenomenon in which suddenly large amount of power generates this large amount of power will cause the failure of diesel engine.
Actually when one set of fuel inject inside the cylinder to burn with already compressed air (in general up to 10-15 bar) then this fuel does not burn complete and accumulate inside the cylinder.After that second set of fuel inject inside the cylinder then that one set of fuel burns with second set of fuel and produces large amount of sudden power for engine and causes the breaks in the crank or connecting rod of engine.it leads to damage the engine.
Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V =
V = Av = ![\frac{\pi}{4} * d^2 * v1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%20d%5E2%20%2A%20v1)
V = ![\frac{\pi}{4} * 0.28^2 * 5](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%200.28%5E2%20%2A%205)
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m =
= ![\frac{0.30787}{0.11418}](https://tex.z-dn.net/?f=%5Cfrac%7B0.30787%7D%7B0.11418%7D)
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 = ![\frac{v2}{A2}](https://tex.z-dn.net/?f=%5Cfrac%7Bv2%7D%7BA2%7D)
v2 = ![\frac{0.3705}{\frac{\pi}{4} * 0.28^2}](https://tex.z-dn.net/?f=%5Cfrac%7B0.3705%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%200.28%5E2%7D)
v2 = 6.017 m/s
Answer and Explanation:
The explanation is attached below