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:
a. V = 109.64 × 10⁵ ft/min
b. Mw = 654519.54 kg/hr
Explanation:
Given Parameters
mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s
inlet temperature of water, T1 = 84 F = 28.89 C
outlet temperature of water, T2 = 68 F = 20 C
specific heat capacity of water, c = 4.18kJ/kgK
rate of heat remover from water, Qw is given by
Qw = 6607.33[28.89 - 20] * 4.18
Qw = 245529.545kw
For air, inlet condition
DBT = 70 F hi = 43.43 kJ/kg
WBT = 60 F wi = 0.00874 kJ/kg
u1 = 0.8445 m/kg
oulet condition,
DBT = 70 F RH = 100.1
h1 = 83.504kJ/kg
Wo = 0.222kJ/kg
check the attached file for complete solution
