hooooooooooooooooooooooooooooooooooooooooooooooooooooooe
Answer:
What's your question?
Explanation:
I'm not able to understand it....
Sorry.
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.
Answer: $17,206.13
Explanation:
Hi, to answer this question we have to apply the next formula:
Annual electricity cost = (P x 0.746 x Ckwh x h) /η
P = compressor power = 78 hp
0.746 kw/hp= constant (conversion to kw)
Ckwh = Cost per kilowatt hour = $0.11/kWh
h = operating hours per year = 2500 h
η = efficiency = 93% = 0.93 (decimal form)
Replacing with the values given :
C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13
