The product of two factors is 4,500. If one of the factors is 90, which is the other factor?
1 answer:
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This question is not complete, the complete question is;
A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured
Answer:
length of crack is 5.585 cm
we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures
Explanation:
Given the data in the question;
vessel thickness = 6 cm
fracture toughness k =
yield strength = 1250 MPa
hoop stress equal = 300 MPa
we know that, the relation between fracture toughness and crack length is expressed as;
k = (1.1)(2/π)(r√(πa))
where k is the fracture toughness, r is hoop stress and a is length of crack
so we rearrange to find length of crack
a = 1/π[( k / 1.1(r)(2/π)]²
a = 1/π[( kπ / 1.1(r)(2)]²
so we substitute
a = 1/π [( 88π / 1.1(300)(2/π)]²
a = 1/π[ 0.1754596 ]
a = 0.05585 m
a = 0.05585 × 100 cm
a = 5.585 cm
so, length of crack is 5.585 cm
we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures
This question is not complete, the complete question is;
The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005
Answer:
the length of the pipe is 11583 in or 965.25 ft
Explanation:
Given the data in the question;
Static pressure ratio; p1/p2 = 10
friction coefficient f = 0.005
diameter of pipe, D =4 inch
first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so
4fL / D = 57.915
we substitute
(4×0.005×L) / 4 = 57.915
0.005L = 57.915
L = 57.915 / 0.005
L = 11583 in
Therefore, the length of the pipe is 11583 in or 965.25 ft
