Answer:
0.234
Explanation:
True stress is ratio of instantaneous load acting on instantaneous cross-sectional area
σ = k × (ε)^n
σ = true stress
ε = true strain
k = strength coefficient
n = strain hardening exponent
ε = ( σ / k) ^1/n
take log of both side
log ε = ( log σ - log k)
n = ( log σ - log k) / log ε
n = (log 578 - log 860) / log 0.20 = 0.247
the new ε = ( 600 / 860)^( 1 / 0.247) = 0.234
Answer:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
thus
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>