Answer:
23.3808 kW
20.7088 kW
Explanation:
ρ = Density of oil = 800 kg/m³
P₁ = Initial Pressure = 0.6 bar
P₂ = Final Pressure = 1.4 bar
Q = Volumetric flow rate = 0.2 m³/s
A₁ = Area of inlet = 0.06 m²
A₂ = Area of outlet = 0.03 m²
Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s
Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s
Height between inlet and outlet = z₂ - z₁ = 3m
Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

Work done by pump

∴ Power input to the pump 23.3808 kW
Now neglecting kinetic energy

Work done by pump

∴ Power input to the pump 20.7088 kW
Answer:
a) 0.489
b) 54.42 kg/s
c) 247.36 kW/s
Explanation:
Note that all the initial enthalpy and entropy values were gotten from the tables.
See the attachment for calculations
Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.
Explanation:
In an airplane equipped with fuel pumps, the auxiliary electric fuel pump is used in the event the engine-driven fuel pump fails.. hope this helped !
Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
<u>a) Calculate Natural frequency </u>
Wn = √k/m = 
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn = 
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs