Answer:
Tc = 424.85 K
Explanation:
Given that,
HEAT FLOW Q is
= 47123.88 w per unit length of rod
volumetric heat rate
= 424.85 K
A spherical seed of 1 cm diameter is buried at a depth of 1 cm inside soil (thermal conductivity of 1 Wm-1K-1) in a sufficiently
large planter. There is a 1 cm thick layer of mulch (thermal conductivity of 2 Wm-1K-1) on top of the soil. The planter has top surface dimensions of 10 cm by 10 cm and is exposed to 200 Wm-2 of heat. You find the top surface temperature of the mulch (Ts,1) to be uniform at 50˚C. What is the surface temperature of the seed (Ts,2) in ˚C
1 answer:
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Answer:
r=0.31
Ф=18.03°
Explanation:
Given that
Diameter of bar before cutting = 75 mm
Diameter of bar after cutting = 73 mm
Mean diameter of bar d= (75+73)/2=74 mm
Mean length of uncut chip = πd
Mean length of uncut chip = π x 74 =232.45 mm
So cutting ratio r
r=0.31
So the cutting ratio is 0.31.
As we know that shear angle given as
Now by putting the values
\
Ф=18.03°
So the shear angle is 18.03°.