1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Pavlova-9 [17]
2 years ago
11

Air enters the compressor of an ideal air-standard Braytoncycle at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s.The tur

bine inlet temperature is 1400 K. For compressorpressure ratios of 6, 8, and 12, determine(a)the thermal efficiency of the cycle.(b)the back-workratio.(c)the netpower developed, in kW.

Physics
2 answers:
bazaltina [42]2 years ago
5 0

Answer / Explanation:

First, we start solving the question by interpreting and representing it in a flow chart diagram.

The illustration have been  attached in a image below:

Now, if we reference the flow chart diagram attached below, we can see that:

from table A-22 in the flow chart diagram,

h1 = 300.19 kJ/kg

pr1 = 1.386 at T1 = 300 K

and we also note that:

Process 1-2 is isentropy, we have:

p2 / p1 = pr2 / pr1

= pr2 = p2 / p1

=  (1.386)(10)

= 13.86

Now, if we recall the table for the ideal gas property of air, which has been also attached below:

We will now interpolate the table:

On interpolation, we obtain:

h2 = 579.9 kJ/kg

and From Table A-22, we discover that:

h3 = 1515.42 kJ/kg and

pr3 = 450.5 at T3 = 1400 K

Process 3-4 is isentropy, we have:

p4/p3 = pr4/pr3

=pr4 = pr3. p4/p3

= (450.5)(0.1)

= 45.05

Now, going ahead to Interpolating  Table A22, we obtain:

h4 = 808.5 kJ/kg

So on discovering the above values, we go ahead to solving:

(a) The thermal efficiency of the cycle

η = (Wt / m) - (Wc / m) / Qm / m

= (h3 - h4) - (h2 - h1) / h3 - h2

η = (1515.4 - 808.5) - (579.9  - 300.19) / 1515.4  - 579.9

Solving further, we arrive at:

η = 0.457

(b) The back work ratio

= Wc / Wt = h2 - h1 / h3 - h4

= 579.9 - 300.19 / 1515.4 - 808.5

= 0.396

(c) The net power developed, in kW

Wcycle  = m [(h3 − h4) − (h2 − h1)]

Where, the air mass flow rate is given by:

m = (AV)₁ / V₁

= (AV)₁ p₁ / RT₁

m = (5.0 m ³ /s) [ 10⁵ N/m² / 8.314 kj ÷ 28.97kg.k] (300k) / 1kj /10³N.M/

= 5.807 kg/s

The power developed is then:

Wcycle   = (5.807 kg/s)[(1515.4 − 808.5) − (579.9 − 300.19)] kJ/kg

Wcycle  = 2481 kW

Harrizon [31]2 years ago
3 0

Answer:

Explanation:

Given that

Air Inlet Pressure, P1 = 100 KPa

Air Inlet temperature, T1 = 300 K  

Volume flow rate, Q = 5 m³/s

Turbine inlet temperature, T₃ = 1400 K

Compressor pressure ratio, r = 6, 8, 12

Heat capacity ratio or air = 1.4

γ= 1.4

Specific heat constant pressure of air, cp = 1.005 KJ/kg.k

At r = 6,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 6 ^ (1.4 - 1)/1.4

1400/T4 = 6 ^ (1.4 - 1)/1.4

T₂ = 1.67 × 300

= 500 K

T₄ = 1400/1.67

= 839.07 K

a)

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((839.07 - 300)/(1400 - 500))

= 0.40

= 40%

B.

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((500 - 300)/(1400 - 839.07))

= 0.36

C.

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 500 - 839.07 + 300)

= 1.005 × 360.93

= 362.74 kJ/kg

Net heat, Qnet = 362.74 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P  = ρ R T

By putting the values

P  = ρ R T

Inputting values,

100  = ρ x 0.287 x 300

ρ = 1.16  kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 362.74

= 2103.9 kW.

At r = 8,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 8 ^ (1.4 - 1)/1.4

1400/T4 = 8 ^ (1.4 - 1)/1.4

T₂ = 1.81 × 300

= 543.4 K

T₄ = 1400/1.81

= 772.9 K

a)

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((772.9 - 300)/(1400 - 543.4))

= 0.448

= 45%

B.

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((543.4 - 300)/(1400 - 772.9))

= 0.39

C.

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 543.4 - 772.9 + 300)

= 1.005 × 383.7

= 385.62 kJ/kg

Net heat, Qnet = 385.62 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P  = ρ R T

By putting the values

P  = ρ R T

Inputting values,

100  = ρ x 0.287 x 300

ρ = 1.16  kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 385.62

= 2236.59 kW.

At r = 12,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 12 ^ (1.4 - 1)/1.4

1400/T4 = 12 ^ (1.4 - 1)/1.4

T₂ = 2.03 × 300

= 610 K

T₄ = 1400/2.03

= 688.32 K

a)

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((688.32 - 300)/(1400 - 610))

= 0.509

= 51%

B.

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((610 - 300)/(1400 - 688.32))

= 0.44

C.

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 610 - 688.32 + 300)

= 1.005 × 401.68

= 403.7 kJ/kg

Net heat, Qnet = 403.7 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P  = ρ R T

By putting the values

P  = ρ R T

Inputting values,

100  = ρ x 0.287 x 300

ρ = 1.16  kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 403.7

= 2341.39 kW.

You might be interested in
Please hurry! 30 points
Cloud [144]

Answer:

50.000

Explanation:

4 0
2 years ago
What was the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
gregori [183]

Answer:

v = 7121.3 m/s

Explanation:

As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed

so here we will have

\frac{mv^2}{r} = \frac{GMm}{r^2}

here we know that

r = orbital radius = 6370 km + 1482 km

r = 7.852 \times 10^6 m

also we know that

M = 5.97 \times 10^{24} kg

now we will have

v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}

v^2 = 5.07 \times 10^7

v = 7121.3 m/s

3 0
3 years ago
Read 2 more answers
Seesaw unbalanced force explain
wolverine [178]
Like a seesaw, it shows that the forces aren’t equal because if it was the seesaw would stay put
3 0
2 years ago
My kitty just past her name was winter :c
Yuliya22 [10]
Oh I’m so sorry rip winter
7 0
3 years ago
Read 2 more answers
Which of the following is an example of the Doppler effect?
Oliga [24]

b is your answers in this thread

3 0
2 years ago
Other questions:
  • The molecule that traps the sun's energy is ?.
    13·1 answer
  • A woman falls to the ground while wearing a parachute. The air resistance of a parachute is 500N. If the woman falls at a consta
    10·1 answer
  • Which equation shows a conservation of mass?
    7·1 answer
  • What are the similarities and differences between light and sound waves?
    15·2 answers
  • Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. Fo
    14·1 answer
  • Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some v
    15·1 answer
  • An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency doe
    9·1 answer
  • 1. To get to school, a girl walks 1 km North in 15 minutes. She
    7·1 answer
  • In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background sign
    6·1 answer
  • explain why using a parabolic mirror for a car headlight throws much more light on the highway than a flat mirror.
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!