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disa
14 days ago
6

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t

o a final pressure of 4 bar, during which the pressure–volume relationship is pV1.1 = constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.
Engineering
1 answer:
Viktor [391]14 days ago
8 0
The work completed is -19.7 KJ, and the heat transferred amounts to 17.4 KJ. Given the temperature is 27°C and the volume is 0.2 m³, with pressure shifting from 1 bar to 4 bar, we recognize the equation pV¹°¹ is a constant. From the superheated propane table, the corresponding specific volume and internal energy values at initial conditions allow us to assess the work and heat transfer between state changes.
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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
Mrrafil [318]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Inertia 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) Calculate the rotor's stored energy at synchronous speed.

The stored energy is represented as

E = G \times H

Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.

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

(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.

The formula for rotor acceleration is given by

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

Where M is defined as

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

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

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

$ 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 derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.

The change in torque angle is expressed as

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

Where t is determined from

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

Consequently,

$ \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 provided by

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

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

The rotor speed in rpm at the culmination of this 10-cycle period is calculated as

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

Where P indicates the number of poles on 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
1 month ago
The Machine Shop has received an order to turn three alloy steel cylinders. Starting diameter = 250 mm and length = 625 mm. Feed
alex41 [359]

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

....................1

where D refers to diameter, L refers to length and f refers to feed while V represents speed \frac{\pi DL}{1000*f*V}

Thus, we derive

Tc = \frac{\pi (250)*625}{1000*0.3*V}

Tc = \frac{1636.25}{V}

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 × \frac{1636.25}{V}

According to the Taylor tool formula, cutting speed is expressed as

VT^{0.25} = 700

 × V × 8.37 = 700

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

5 0
1 month ago
Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe
choli [298]

Answer:

Flame-resistant clothing and aprons

Explanation:

Workers involved with welding are generally mandated to wear flame-resistant clothing and aprons to shield them from various hazards, including heat, flames, burns, and potential radiation. In the context of welding, this gear protects individuals from flying sparks that can ignite and cause fires. Hence, such clothing helps to prevent accidents in these situations.

7 0
1 month ago
An insulated box containing helium gas falls from a balloon 4.5 km above the earth's surface. calculate the temperature rise in
iogann1982 [368]
The increase in temperature of the helium gas is calculated to be 14.25 K. The helium is located in an insulated box that falls from a height of 4.5 km. As it descends, the potential energy is transformed into internal energy of the helium gas. The equation for temperature change can be expressed as: 10 x 4.5 = 3.15 x ΔT, yielding a temperature increase of ΔT = 14.25 K.
6 0
23 days ago
6.15. In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20
Viktor [391]

Explanation:

At a temperature of 33^{\circ} C and relative humidity of 86%, the humidity ratio stands at 0.0223 with a specific volume of 14.289.

At a temperature of 33^{\circ} C and relative humidity of 40%, the humidity ratio is 0.0066 while the specific volume is 13.535.

To determine the mass of air, the following formula can be used:

\begin{aligned}m _{1} &=\frac{ v }{ v }(1- w ) \\&=\frac{1 \times 10^{5}}{13.535}(1-0.0066) \\&=7339.49 lb / min \\v _{ a } &=\frac{ m _{1} v }{(1- w )} \\v _{ a } &=\frac{7339.49 \times 14.289}{(1-0.0223)} \\v _{ a } &=107266.0 ft ^{3} / min\end{aligned}

Now, we will calculate the volume

\begin{aligned}m _{ w } &=\frac{ v _{ a }}{ v _{ a }} w _{ a }-\frac{ v _{ i }}{ v _{ i }} w _{ i } \\&=\frac{107266.0}{14.289} \times 0.0223-\frac{100000}{13.535} \times 0.0066 \\&=118.64 lb / min\end{aligned}

The duration required to fill the cistern can be determined with the equation:

Time \(=\frac{\text { cistern volume }}{\text { removal water perminute volume }}\)

By substituting the values into the preceding formula, we find:

\(\frac{\left(15 \times 10^{3} L\right) \times\left(0.0353147 ft ^{3} / L \right)}{(118.641 b / min ) \times\left(\frac{1}{62.41 lb / ft ^{3}}\right)}\)\\\(=279.09\) minutes\\\(=4.65\) hours.

Thus, the hours necessary to fill the cistern amount to 4.65 hours.

3 0
1 month ago
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