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Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT
ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system
For ideal gas
Δh = CpΔT
by putting the values
Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding