The U.S. Coast Guard requires certain equipment be carried onboard a vessel based on the vessel's length. How must an operator m
1 answer:
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Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:
We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
Substituting figures in the expression above, we have:
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection
Answer:
Head loss is 1.64
Explanation:
Given data:
Length (L) = 200 m
Discharge (Q) = 0.16 m3/s
According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m
We know,
where, f = Darcy friction factor
V = flow velocity
g = acceleration due to gravity
We know, flow rate Q = A x V
solving for V
obtained Darcy friction factor
calculate Reynold number (Re) ,
where, = density of water
= Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2
so reynold number is
= 6.4 x 10^5
For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm
Relative roughness (e/D) = 0.0015 / 318 = 0.00005
from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126
Therefore head loss is
HL = 1.64 m