Answer:
Fatigue lifetimes will be ranked as B>A>C
Explanation:
Start by calculating the mean stress for all samples
σa= (σamax + σamin)/2 = (450 - 350)/2 = 50MPa
σb= (σbmax + σbmin)/2 = (400 - 300)/2 = 50MPa
σa= (σcmax + σcmin)/2 = (340 - 340)/2 = 0MPa
Now calculate the stress amplitudes of all three samples
σa= (σamax - σamin)/2 = (450 + 350)/2 = 400MPa
σb= (σbmax - σbmin)/2 = (400 + 300)/2 = 350MPa
σc= (σcmax - σcmin)/2 = (340 + 340)/2 = 340MPa
The mean stress of samples A and B is the same which is higher than sample C. Hence samples A and B will have a higher fatigue life than sample C. However, the higher stress amplitude means lower fatigue life, hence, sample B will have a higher fatigue life than sample A. So the order will be B> A> C.
Attached picture shows the justification using and S- N plot.
Answer:i need a picture not a link
Explanation:
Answer:
time need = 17.65 minutes
Explanation:
given data
water flows in = 22 liters/minute
water flows out = 5 liters/minute
volume of water needed for a bath = 0.30 m³ = 300 Liters
Density of water = 1000 kg/m³
solution
we get here time for need to fill bath that is express as
time need = .........................1
put here value and we will get
time need =
time need = 17.65 minutes
Answer:
The heat transfer is 29.75 kJ
Explanation:
The process is a polytropic expansion process
General polytropic expansion process is given by PV^n = constant
Comparing PV^n = constant with PV^1.2 = constant
n = 1.2
(V2/V1)^n = P1/P2
(V2/0.02)^1.2 = 8/2
V2/0.02 = 4^(1/1.2)
V2 = 0.02 × 3.2 = 0.064 m^3
W = (P2V2 - P1V1)/1-n
P1 = 8 bar = 8×100 = 800 kPa
P2 = 2 bar = 2×100 = 200 kPa
V1 = 0.02 m^3
V2 = 0.064 m^3
1 - n = 1 - 1.2 = -0.2
W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ
∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ
Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ