Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1

For pipe 2

Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV




head loss (h)

Now putting the all values

So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.
Answer:
a) 
b) attached below
c) type zero system
d) k > 
e) The gain K increases above % error as the steady state speed increases
Explanation:
Given data:
Motor voltage = 12 v
steady state speed = 200 rad/s
time taken to reach 63.2% = 1.2 seconds
<u>a) The transfer function of the motor from voltage to speed</u>
let ;
be the transfer function of a motor
when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec
hence the transfer function of the motor from voltage to speed
= 
<u>b) draw the block diagram of the system with plant controller and the feedback path </u>
attached below is the remaining part of the detailed solution
c) The system is a type-zero system because the pole at the origin is zero
d) ) k > 
Answer:
COP of the heat pump is 3.013
OP of the cycle is 1.124
Explanation:
W = Q₂ - Q₁
Given
a)
Q₂ = Q₁ + W
= 15 + 7.45
= 22.45 kw
COP = Q₂ / W = 22.45 / 7.45 = 3.013
b)
Q₂ = 15 x 1.055 = 15.825 kw
therefore,
Q₁ = Q₂ - W
Q₁ = 15.825 - 7.45 = 8.375
∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124
Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation

under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that

hence , actual velocity will be equal to 8.39 m/s