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) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
DIAMETER = 9.797 m
POWER =
Explanation:
Given data:
circular windmill diamter D1 = 8m
v1 = 12 m/s
wind speed = 8 m/s
we know that specific volume is given as
where v is specific volume of air
considering air pressure is 100 kPa and temperature 20 degree celcius
v = 0.8409 m^3/ kg
from continuity equation
mass flow rate is given as
the power produced