Answer:critical stress= 20.23 MPa
Explanation:
Since there was an internal crack, we will divide the length of the internal crack by 2
Length of internal crack, a = 0.7mm,
Half length = 0.7mm/2= 0.35mm changing to meters becomes
0.35/ 1000= 0.35 x 10 ^-3m
The formulae for critical stress is calculated using
σC = (2Eγs /πa) ¹/₂
σC = critical stress=?
Given
E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²
γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m
a= Half Length of crack=0.35 x 10 ^-3m
σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂
=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂
=(4.0920 x10 ^14)¹/₂
σC=20.23 x10^6 N/m² = 20.23 MPa
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:
The power input to the pump is 292KW
Explanation:
In this question, we are asked to calculate the power input to a pump neglecting potential and kinetic energy changes.
Please check attachment for complete solution and step by step explanation
Answer:
<h2>False </h2>
Explanation:
The noun form of organize is just adding letter r
Answer:
When an additional layer of insulation is applied to a cylindrical pipe or a spherical shell, the insulation layer works to increase its conduction resistance but at the same time, lowers the convection resistance of the surface.
Explanation:
Generally, when more insulation is added to a wall, the resulting effect is that heat transfer decreases. With increasing thickness of the insulation, the heat transfer rate becomes lesser. This is because the thermal resistance of the wall becomes more with the added insulation, wherein the heat transfer area and convection resistance aren't affected.
However, a different scenario occurs when an additional layer of insulation is applied to a cylindrical pipe or spherical shell. The conduction resistance increases, but the surface convection resistance decreases since the outer surface area for convection also increases and does not remain constant.
Overall, the heat transfer from the cylindrical pipe or spherical shell may increase or decrease, which depends on the effect that dominates.