Answer:
Free convection:
When heat transfer occurs due to density difference between fluid then this type of heat transfer is know as free convection.The velocity of fluid is zero or we can say that fluid is not moving.
Force convection:
When heat transfer occurs due to some external force then this type of heat transfer is know as force convection.The velocity of fluid is not zero or we can say that fluid is moving in force convection.
Heat transfer coefficient of force convection is high as compare to the natural convection.That is why heat force convection reach a steady-state faster than an object subjected to free-convection.
We know that convective heat transfer given as
q = h A ΔT
h=Heat transfer coefficient
A= Surface area
ΔT = Temperature difference
Explanation:
Specific cutting energy:
It the ratio of power required to cut the material to metal removal rate of material.If we take the force required to cut the material is F and velocity of cutting tool is V then cutting power will be the product of force and the cutting tool velocity.
Power P = F x V
Lets take the metal removal rate =MRR
Then the specific energy will be
If we consider that metal removal rate and cutting tool velocity is constant then when we increases the cutting force then specific energy will also increase.
Answer:
the minimum shaft diameter is 35.026 mm
the maximum shaft diameter is 35.042mm
Explanation:
Given data;
D-maximum = 35.020mm and d-minimum = 35.000mm
we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6
so From table, Selection of International Trade Grades metric series
the grade tolerance are;
ΔD = IT7(0.025 mm)
Δd = IT6(0.016 mm)
Also from Table "Fundamental Deviations for Shafts" metric series
Sf = 0.026
so
D-maximum
Dmax = d + Sf + Δd
we substitute
Dmax = 35 + 0.026 + 0.016
Dmax = 35.042 mm
therefore the maximum diameter of shaft is 35.042mm
d-minimum
Dmin = d + Sf
Dmin = 35 + 0.026
Dmin = 35.026 mm
therefore the minimum diameter of shaft is 35.026 mm
Answer:
The maximum length is 3.897×10^-5 mm
Explanation:
Extension = surface energy/elastic modulus
surface energy = 1.05 J/m^2
elastic modulus = 198 GPa = 198×10^9 Pa
Extension = 1.05/198×10^9 = 5.3×10^-12 m
Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4
Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm
