Answer:
Given Data:
concentration of sewer Csewer = 1.2 g/L
converting into mg/L = Csewer = 1.2 g/L x 1000 mg/g = 1200 mg/L
flow rate of sewer Qsewer = 2000 L/min
concentration of sewer Cstream = 20 mg/L
flow rate of sewer Qstream = 2m3/s
converting Q into L/min = 2m3/s x 1000 x 60 = 120000 L/min
mass diagram is
Answer:
Elastic modulus of steel = 202.27 GPa
Explanation:
given data
long = 110 mm = 0.11 m
cross section 22 mm = 0.022 m
load = 89,000 N
elongation = 0.10 mm = 1 ×
m
solution
we know that Elastic modulus is express as
Elastic modulus =
................1
here stress is
Stress =
.................2
Area = (0.022)²
and
Strain =
.............3
so here put value in equation 1 we get
Elastic modulus =
Elastic modulus of steel = 202.27 ×
Pa
Elastic modulus of steel = 202.27 GPa
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Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.