Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are
Real power in 3 phase
Now the Phase Voltage is,
The current phase would be,
Rearranging,
Replacing,
Therefore the current per phase is 2.26kA
Answer:
Maximum allowable chip power is 0.35 W
Explanation:
This question is incomplete. The complete question is
A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at t[infinity] = 15°c. from reliability considerations, the chip temperature must not exceed t = 85°c. f the coolant is air and the corresponding convection 200 w/m2 k, what is the maximum allowable chip power?
<u>ANSWER:</u>
The heat transfer through convection, we have the equation:
q = hA(T - T∞)
where,
q = power transfer through convection = ?
h = convection coefficient = 200 W/m²K
A = Area of convection surface = (0.005 m)² = 0.000025 m²
T = Chip surface temperature = 85° C
T∞ = Fluid temperature = 15° C
Therefore,
q = (200 W/m².K)(0.000025 m²)(85° C - 15° C)
<u>q = 0.35 W</u>
Since, difference in temperature is same on both Celsius and kelvin scale. Therefore, Celsius is written as kelvin for difference and they shall be cancelled.
Answer:
(a) Increases
(b) Increases
(c) Increases
(d) Increases
(e) Decreases
Explanation:
The tensile modulus of a semi-crystalline polymer depends on the given factors as:
(a) Molecular Weight:
It increases with the increase in the molecular weight of the polymer.
(b) Degree of crystallinity:
Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.
(c) Deformation by drawing:
The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.
(d) Annealing of an undeformed material:
This also results in an increase in the tensile strength of the material.
(e) Annealing of a drawn material:
A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.
