Answer:
Zero 1 = -1
Zero 2 = -3
Pole 1 = 0
Pole 2 = -2
Pole 3 = -4
Pole 4 = -6
Gain = 4
Explanation:
For any given transfer function, the general form is given as
T.F = k [N(s)] ÷ [D(s)]
where k = gain of the transfer function
N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.
D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.
k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]
it is evident that
Gain = k = 4
N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)
= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)
The zeros are -1 and -3
D(s) = s⁴ + 12s³ + 44s² + 48s
= s(s³ + 12s² + 44s + 48)
= s(s + 2)(s + 4)(s + 6)
The roots are then, 0, -2, -4 and -6.
Hope this Helps!!!
Answer:
Qin = 1857 kJ
Explanation:
Given
m = 0.5 Kg
T₁ = 25°C = (25 + 273) K = 298 K
P₁ = 100 kPa
P₂ = 500 kPa
First, the temperature when the piston starts rising is determined from the ideal gas equations at the initial state and at that state:
T₂ = T₁*P₂/P₁
⇒ T₂ = 298 K*(500 kPa/100 kPa) = 1490 K
Until the piston starts rising no work is done so the heat transfer is the change in internal energy
Qin = ΔU = m*cv*(T₂-T₁)
⇒ Qin = 0.5*3.1156*(1490 - 298) kJ = 1857 kJ
Answer:
The preliminary work breakdown structure will be divided into two steps, the first is to draw the first level and the second is to draw the second level.
Explanation:
Please look at attachment.
Given:
Pressure, 
= 1300 kPa
Temperature, 
= 
= 100 kPa
velocity, v = 40 m/s
A = 1
Solution:
For air propertiess at
= 1300 kPa
=
= 793kJ/K
= 
and also at
= 100 kPa
= 401 KJ/K
= 
a) Mass flow rate is given by:
Now,
= 46.51 kg/s
b) for the power produced by turbine, 
= 18.231 MW
