Answer:
Explanation:
Answer:
Explanation:
Answer:
Explanation:
This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.
(a). here we are asked to determine the Temperature and Pressure.
Given that the properties of Air;
ha = 230.02 KJ/Kg
Ta = 230 K
Pra = 0.5477
From the energy balance equation for a diffuser;
ha + Va²/2 = h₁ + V₁²/2
h₁ = ha + Va²/2 (where V₁²/2 = 0)
h₁ = 230.02 + 220²/2 ˣ 1/10³
h₁ = 254.22 KJ/Kg
⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg
from this we have;
Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]
Pr₁ = 0.77759
therefore T₁ = 254.15K
P₁ = (Pr₁/Pra)Pa
= 0.77759/0.5477 ˣ 26
P₁ = 36.91 kPa
now we calculate Pr₂
Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349
⇒ now we obtain properties of air at
Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg
calculating the enthalpy of air at state 2
ηc = h₁ - h₂ / h₁ - h₂
0.85 = 254.22 - 505.387 / 254.22 - h₂
h₂ = 549.71 KJ/Kg
to obtain the properties of air at h₂ = 549.71 KJ/Kg
T₂ = 545.15 K
⇒ to calculate the pressure of air at state 2
P₂/P₁ = 11
P₂ = 11 ˣ 36.913
p₂ = 406.043 kPa
but pressure of air at state 3 is the same,
i.e. P₂ = P₃ = 406.043 kPa
P₃ = 406.043 kPa
To obtain the properties of air at
T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5
for cases of turbojet engine,
we have that work output from turbine = work input to the compressor
Wt = Wr
(h₃ - h₄) = (h₂ - h₁)
h₄ = h₃ - h₂ + h₁
= 1515.42 - 549.71 + 254.22
h₄ = 1219.93 kJ/Kg
properties of air at h₄ = 1219.93 kJ/Kg
T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]
T₄ = 1150.58 K
Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]
Pr₄ = 200.5636
Calculating the ideal enthalpy of the air at state 4;
Лr = h₃ - h₄ / h₃ - h₄*
0.9 = 1515.42 - 1219.93 / 1515.42 - h₄
h₄* = 1187.09 kJ/Kg
now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg
P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]
P₄* = 181.316
P₄ = (Pr₄/Pr₃)P₃ i.e. 3-4 isentropic process
P₄ = 181.316/450.5 * 406.043
P₄ = 163.42 kPa
For the 4-5 process;
Pr₅ = (P₅/P₄)Pr₄
Pr₅ = 26/163.42 * 200.56 = 31.9095
to obtain the properties of air at Pr₅ = 31.9095
h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]
h₅ = 734.09 KJ/Kg
T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]
T₅ = 719.32 K
(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;
QH = m(h₃-h₂)
QH = 25(1515.42 - 549.71)
QH = 24142.75 kW
(c). To calculate the velocity at the nozzle exit;
we apply steady energy equation of a flow to nozzle
h₄ + V₄²/2 = h₅ + V₅²/2
h₄ + 0 = h₅₅ + V₅²/2
1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2
therefore, V₅ = 985.74 m/s
cheers i hope this helps
