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kolezko [41]
2 years ago
11

Water from an upper tank is drained into a lower tank through a 5 cm diameter iron pipe with roughness 2 mm. The entrance to the

pipe has minor loss coefficient 0.4 and the exit has minor loss coefficient of 1, both referenced to the velocity in the pipe. The water level of the upper tank is 4 m above the level of the lower tank, and the pipe is 5 m long. You will find the drainage volumetric flow rate. a) What is the relative roughness
Engineering
1 answer:
nignag [31]2 years ago
5 0

Answer:

Relative roughness = 0.04

Explanation:

Given that:

Diameter = 5 cm

roughness = 2 mm

At inlet:

Minor coefficient loss k_{L1} = 0.4

At exit:

Minor coefficient loss k_{L2} = 1

Height h = 4m

Length = 5 m

To find the relative roughness:

Relative roughness is a term that is used to describe the set of irregularities that exist inside commercial pipes that transport fluids. The relative roughness can be evaluated by knowing the diameter of the pipe made with the absolute roughness in question. If we denote the absolute roughness as e and the diameter as D, the relative roughness is expressed as:

e_r = \dfrac{e}{D}

e_r = \dfrac{0.2 }{5}

\mathbf{e_r = 0.04}

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A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the
raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

E = G \times H

Where G represents complex rated power and H is the inertia constant of turbo-generator.

E = 100 \times 8 \\\\E = 800 \: MJ

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}  $

Where M is given by

$ M = \frac{E}{180 \times f} $

$ M = \frac{800}{180 \times 50} $

M = 0.0889 \: MJ \cdot s/ elec \: \: deg

So, the rotor acceleration is

$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$  30 = 0.0889 \frac{d^2 \delta}{dt^2}  $

$   \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}  $

$   \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$ \Delta  \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $

Where t is given by

1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2  \\\\t = 0.2 \: sec

So,

$ \Delta  \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $

$ \Delta  \delta = 6.75 \: elec \: deg

The change in torque in rpm/s is given by

$ \Delta  \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ  }   $

$ \Delta  \delta =28.12 \: \: rpm/s $

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$ Rotor \: speed = \frac{120 \cdot f}{P}  + (\Delta  \delta)\cdot t  $

Where P is the number of poles of the turbo-generator.

$ Rotor \: speed = \frac{120 \cdot 50}{4}  + (28.12)\cdot 0.2  $

$ Rotor \: speed = 1500  + 5.62  $

$ Rotor \: speed = 1505.62 \:\: rpm

4 0
3 years ago
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