Homework check for On/Off Design of a Fixed Area Turbojet

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SUMMARY

The discussion focuses on the performance analysis of an ideal fixed-area turbojet engine under both on-design and off-design conditions. Key parameters include a compression ratio (\pi_c) of 15, an inlet Mach number (M_o) of 0.8, and a total temperature (T_{t4}) of 2000 K for on-design conditions. For off-design conditions at M_o of 0.3 and T_{t4} reduced to 1500 K, thrust calculations yield 103,826 N for on-design and a mass flow rate of air (\dot{m_a}) of 457.7 kg/s for off-design, raising questions about the validity of the area calculation (A_o) of 0.00504 m².

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roldy
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Homework Statement


An ideal fixed-area turbojet is operated where \pi_c=15,M_o=0.8,T_o=260K, T_{t4}=2000 K, and P_o=20000 Pa. Mass flow rate of air processed by this engine at on-design is 100 kg/sec.

What will be the performance of this engine (thrust, fuel consumption) compared to the on-design conditions if it is flown at a Mach of 0.3 and at an altitude where temperature and pressure are 288K and 101325 Pa. Furthermore, the fuel throttle is set such that T_{t4}=1500 K at this off-design point. Assume that A_9 is varied to keep P_9=P_o.


Homework Equations


<br /> \tau_r=1+\frac{(\gamma-1)}{2}M_o^2<br />

<br /> \pi_r=\tau_r^{\frac{\gamma}{\gamma-1}}<br />

<br /> \tau_c=\pi_c^{\frac{\gamma-1}{\gamma}}<br />

<br /> \tau_\lambda=\frac{T_{t4}}{T_o}<br />

<br /> \pi_t=\tau_t^{\frac{\gamma}{\gamma-1}}<br />

<br /> \frac{T_{t3}}{T_o}=\tau_r{\tau_c}<br />

<br /> a_o=\sqrt{\gamma{R}{T_o}}<br />

<br /> U_o=a_o{M_o}<br />

<br /> U_9=a_o\left[\frac{2}{\gamma-1}{\tau_\lambda}{\tau_t}\left[1-\left({\pi_r}{\pi_d}{\pi_c}{\pi_b}{\pi_t}{\pi_N}\right)^{-\frac{(\gamma-1)}{\gamma}}\right]\right]^{1/2}<br />

<br /> \frac{THRUST}{\dot{m}}=U_9-U_o<br />


<br /> \dot{m_f}h=(\dot{m_a}+\dot{m_f})C_p{T_{t4}}-\dot{m_a}C_o{T_{t3}}<br />


The Attempt at a Solution


So the givens for on-design analysis:

\pi_c=15

M_o=0.8

T_o=260 K

T_{t4}=2000 K

P_o=20,000 Pa

\dot{m_a}=100 kg/s

h=4.5 *10^7

C_p=1004

Solving for each variable I get the following:

\tau_r=1.128

\pi_r=1.524

\tau_c=2.168

\tau_\lambda=7.69

\tau_t=.829

\pi_t=.519

Re-arranging \frac{T_{t3}}{T_o} to solve for T_{t3} I get 635.83 K

a_o=322.65 m/s

U_o=258.12 m/s

Assuming unknown \pi{&#039;s}=1, then

U_9=1296.38 m/s

Solving for THRUST I get 103,826 N

And finally

\dot{m_f}=3.185 kg/s

Now for the off-design:

M_o=0.3

T_o=288 K

T_{t4}=1500 K

P_o=101325 Pa

h=4.5 *10^7

C_p=1004

For off-design we keep \tau_t the same value for on-design, so
\tau_t=.829

\pi_t=.519

Solving for each variable I get the following:

\tau_r=1.018

\pi_r=1.064

/tau_c=1.875

\tau_\lambda=5.21

T_{t3}=549.72 K

a_o=339.58 m/s

U_o=101.87 m/s

U_9=957.36 m/s

And now I need to calculate thrust but I need to find \dot{m_a} first.




Also I found what the area is using A_o=\frac{\dot{m_a}}{\rho_o{R}{U_o}} which came out to be .00504 m^2. This seems like a small area for the turbojet. Did I mess something up?

If I use this area value to find \dot{m_f} for the off-design analysis using \rho=\frac{P_o}{RT_o}=1.23 at this new pressure and temperature I get the mass flow rate of air to be 457.7 kg/s. Does this value make since? To me it doesn't because of the slower velocity in the off-design vs. the higher velocity in the on-design Then again the density for the on-design case is much lower than the off-design (.268 kg/m^3 for on-design).

The mass flow rate of fuel is then 10.04 kg/s. I would be grateful if someone could look over this and see if I made an error in logic or calculations.
 
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