Answer:
a. The final temperature is 69.8°C
b. The final pressure is 2.67bar
c. The amount of entropy produced is 0.38568kJ/K
Explanation:
Pls, check the attached files for detail step by step explanation as typing the solution here can not be explicit.
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
Answer:
The probable grain-coarsening mechanism is : Ideal grain growth mechanism
( -
= kt )
Explanation:
The plot attached below shows the time dependence of the growth of grain.
The probable grain-coarsening mechanism is : Ideal grain growth mechanism
the ideal growth follows this principle = = kt
d = final grain size
= initial grain size
k = constant ( temperature dependent )
t = 0
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) / 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2