Question

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 turbine 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.

Answer 1

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

Answer 2

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.