What is the average distance (in terms of R) between the mobile on the fringe of the serving cell and the second and third tier
1 answer:
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Answer:
Flow velocity
50.48m/s
Pressure change at probe tip
1236.06Pa
Explanation:
Question is incomplete
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively
solution
In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe
please check attachment for complete solution and step by step explanation
Question in order:
A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?
Mirror Radius (mm) Bending Failure Stress (MPa)
0.603 225
0.203 368
0.162 442
Answer:
191 MPa
Explanation:
Failure stress of bending is Inversely proportional to the mirror radius
At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)
At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)
At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)
comparing case 1 and 2 using the above equation
Taking the natural logarithm of both side
ln(0.6114) = n ln(0.3366)
n₁ = ln(0.6114)/ln(0.3366)
n₁ = 0.452
comparing case 2 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.8326) = n ln(0.7980)
n₂ = ln(0.8326)/ln(0.7980)
n₂ = 0.821
comparing case 1 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.5090) = n ln(0.2687)
n₃ = ln(0.5090)/ln(0.2687)
n₃ = 0.514
average for n
n = 0.596
Hence to get bending stress x at mirror radius 0.796
stress x = 191 MPa