Respuesta :
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.
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
