Answer:
minimum length of a surface crack is 15.043 mm
Explanation:
given data
strain fracture toughness K = 78 MPa
tensile stress = 345 MPa
Y = 1.04
to find out
minimum length of a surface crack
solution
we find here length of critical interior flaw from formula that is
α =
....................1
put here value we get
α = 
α = 15.043 mm
so minimum length of a surface crack is 15.043 mm
Answer:
The answer to this question is 1273885.3 ∅
Explanation:
<em>The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.</em>
<em>Given that,</em>
<em>The distance = 50mm</em>
<em>The time t =10 seconds</em>
<em>The force F = 10kN</em>
<em>The piston diameter is = 100mm</em>
<em>The pressure = F/A</em>
<em> 10 * 10^3/Δ/Δ </em>
<em> P = 1273885.3503 pa</em>
<em>Then</em>
<em>Power = work/time = Force * distance /time</em>
<em> = 10 * 1000 * 0.050/10</em>
<em>which is =50 watt</em>
<em>Power =∅ΔP</em>
<em>50 = 1273885.3 ∅</em>
Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In (
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
Answer:
Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.
Explanation: lol