An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of
hydrogen at 127 oC, 2 bar and the other contains nitrogen at 27 oC, 4 bar. The gases are allowed to mix until an equilibrium state is attained. Assume the ideal gas model with constant specific heats, determine
a. The final temperature, in oC.
b. The final pressure, in bar.
c. The amount of entropy produced, in KJ/K.
a. The final temperature, in oC.
b. The final pressure, in bar.
c. The amount of entropy produced, in KJ/K.
1 answer:
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Answer:
v=82 m/s
s=116m
Explanation:
a=20t
using condition given at t=0
c=-8
now equation becomes
v=10t²-8
v at t= 3s v=82 m/s
again
now using condition given s=50 at t=0
b=50
now equation becomes
calculating s at t=3s
s=116m
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e
r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m