Answer:
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Explanation:
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
Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
Answer:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above