Answer: It does make sense, because I've been involved in these careers and have a long family line of them. And other questions?
Explanation:
Answer:
Qx = 9.10 m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 )²× 9 × sin18 × cos18
Qd = 94.305 × m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × × π × 85 × ( 9 )³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × - 85.2 ×
Qx = 9.10 m³/s
Answer:
a) 280MPa
b) -100MPa
c) -0.35
d) 380 MPa
Explanation:
GIVEN DATA:
mean stress
stress amplitude
a)
--------------1
-----------2
solving 1 and 2 equation we get
b)
c)
stress ratio
d)magnitude of stress range
= 280 -(-100) = 380 MPa
Answer:
The strength coefficient is K = 591.87 MPa
Explanation:
We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by
where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.
Solving for strength coefficient
From the strain hardening equation we can solve for K
And we can replace values
Thus we get that the strength coefficient is K = 591.87 MPa
Answer:
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