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blsea [12.9K]
3 years ago
9

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and gauge pressure of 300 kPa. The gas is h

eated, and the gauge pressure at the final state is 367 kPa. Determine the final temperature, in °C. The local atmospheric pressure is 1 atm.
Engineering
1 answer:
Sergio [31]3 years ago
3 0

Answer:

the final temperature is 77.1 °C

Explanation:

Given the data in the question;

Initial temperature; T₁ = 27°C = ( 27 + 273)K = 300 K

Initial absolute pressure P₁ = 300 kPa = ( 300 + 101.325 )kPa = 401.325 kPa

Final absolute pressure P₂ = 367 kPa = ( 367 + 101.325 )kPa = 468.325 kPa

Now, to calculate the final temperature, we use the ideal gas equation;

P₁V/T₁ = P₂V/T₂

but it is mentioned that the rigid tank is closed,

so the volume is the same both before and after.

Change in volume = 0

hence;

P₁/T₁ = P₂/T₂

we substitute

401.325 kPa / 300 K = 468.325 kPa / T₂

T₂ × 401.325 kPa  = 300 K × 468.325 kPa

T₂ = [ 300 K × 468.325 kPa ] / 401.325 kPa

T₂ = 140497.5 K / 401.325

T₂ =  350.08 K

T₂ = ( 350.08 - 273 ) °C

T₂ = 77.1 °C

Therefore, the final temperature is 77.1 °C

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Explanation:

a) The total volume equals the sum of the volumes.

500 = x + y

The total octane amount equals the sum of the octane amounts.

89(500) = 87x + 92y

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As x increases, y decreases.

c) Use substitution or elimination to solve the system of equations.

44500 = 87x + 92(500−x)

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6 0
3 years ago
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
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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

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(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  $

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Downstream, we have half the volumetric flow and a smaller area.

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