Answer:
The critical length of surface flaw = 6.176 mm
Explanation:
Given data-
Plane strain fracture toughness Kc = 29.6 MPa-m1/2
Yield Strength = 545 MPa
Design stress. =0.3 × yield strength
= 0.3 × 545
= 163.5 MPa
Dimensionless parameter. Y = 1.3
The critical length of surface flaw is given by
= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2
Now putting values in above equation we get,
= 1/3.14( 29.6 / 1.3 × 163.5)^2
=6.176 × 10^-3 m
=6.176 mm
Answer:
450,000m = 450km = 4.5E5
32,600,000W = 32.6MW = 3.26E7
59,700,000,000cal = 59.7Gcal = 5.97E10
0.000000083s = 83ns = 8.3E-8
35,000Ω = 35kΩ = 3.5E4
Explanation:
Giga = 1,000,000,000
Mega = 1,000,000
kilo = 1,000
unit = 1
deci = .1
centi = .01
milli = .001
micro = .000001
nano = .0000000001
pico = .000000000001
You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.
Answer:
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.
Step-by-step solution:
Step 1 of 5
Given data:-
The velocity of water is .
The water flow rate is.
Answer:
a) 
b)
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below
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