Answer:
The module is why it’s goin to work
Explanation:
The answer to your quesition is b
1. Define <em>Viscosity</em>
In physics, <em>Viscosity</em> refers to the level of resistance of a fluid to flow due to internal friction, in other words, viscosity is the result of the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow. For example, the honey is a fluid with high viscosity while the water has low viscosity.
What are the main differences between viscous and inviscid flows?
Viscous flows are flows that has a thick, sticky consistency between solid and liquid, contain and conduct heat, does not have a rest frame mass density and whose motion at a fixed point always remains constant. Inviscid flows, on the other hand, are flows characterized for having zero viscosity (it does not have a thick, sticky consistency), for not containing or conducting heat, for the lack of steady flow and for having a rest frame mass density
Furthermore, viscous flows are much more common than inviscid flows, while this latter is often considered an idealized model since helium is the only fluid that can become inviscid.
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
The shear plane angle and shear strain are 28.21° and 2.155 respectively.
Explanation:
(a)
Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.
Given:
Rake angle is 12°.
Chip thickness before cut is 0.32 mm.
Chip thickness is 0.65 mm.
Calculation:
Step1
Chip reduction ratio is calculated as follows:
![r=\frac{t}{t_{c}}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bt%7D%7Bt_%7Bc%7D%7D)
![r=\frac{0.32}{0.65}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B0.32%7D%7B0.65%7D)
r = 0.4923
Step2
Shear angle is calculated as follows:
![tan\phi=\frac{rcos\alpha}{1-rsin\alpha}](https://tex.z-dn.net/?f=tan%5Cphi%3D%5Cfrac%7Brcos%5Calpha%7D%7B1-rsin%5Calpha%7D)
Here,
is shear plane angle, r is chip reduction ratio and
is rake angle.
Substitute all the values in the above equation as follows:
![tan\phi=\frac{rcos\alpha}{1-rsin\alpha}](https://tex.z-dn.net/?f=tan%5Cphi%3D%5Cfrac%7Brcos%5Calpha%7D%7B1-rsin%5Calpha%7D)
![tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}](https://tex.z-dn.net/?f=tan%5Cphi%3D%5Cfrac%7B0.4923cos12%5E%7B%5Ccirc%7D%7D%7B1-0.4923sin12%5E%7B%5Ccirc%7D%7D)
![tan\phi=\frac{0.48155}{0.8976}](https://tex.z-dn.net/?f=tan%5Cphi%3D%5Cfrac%7B0.48155%7D%7B0.8976%7D)
![\phi=28.21^{\circ}](https://tex.z-dn.net/?f=%5Cphi%3D28.21%5E%7B%5Ccirc%7D)
Thus, the shear plane angle is 28.21°.
(b)
Step3
Shears train is calculated as follows:
![\gamma=cot\phi+tan(\phi-\alpha)](https://tex.z-dn.net/?f=%5Cgamma%3Dcot%5Cphi%2Btan%28%5Cphi-%5Calpha%29)
![\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})](https://tex.z-dn.net/?f=%5Cgamma%3Dcot28.21%5E%7B%5Ccirc%7D%2Btan%2828.21%5E%7B%5Ccirc%7D-12%5E%7B%5Ccirc%7D%29)
.
Thus, the shear strain rate is 2.155.