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

In summary, the engine would be less efficient at higher velocities, and the fuel throttle would need to be adjusted to keep the temperature the same at 1500K.f
  • #1
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Homework Statement


An ideal fixed-area turbojet is operated where [tex]\pi_c=15,M_o=0.8,T_o=260K, T_{t4}=2000 K[/tex], and [tex]P_o=20000 Pa[/tex]. 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 [tex]T_{t4}=1500 K[/tex] at this off-design point. Assume that [tex]A_9[/tex] is varied to keep [tex]P_9=P_o[/tex].


Homework Equations


[tex]
\tau_r=1+\frac{(\gamma-1)}{2}M_o^2
[/tex]

[tex]
\pi_r=\tau_r^{\frac{\gamma}{\gamma-1}}
[/tex]

[tex]
\tau_c=\pi_c^{\frac{\gamma-1}{\gamma}}
[/tex]

[tex]
\tau_\lambda=\frac{T_{t4}}{T_o}
[/tex]

[tex]
\pi_t=\tau_t^{\frac{\gamma}{\gamma-1}}
[/tex]

[tex]
\frac{T_{t3}}{T_o}=\tau_r{\tau_c}
[/tex]

[tex]
a_o=\sqrt{\gamma{R}{T_o}}
[/tex]

[tex]
U_o=a_o{M_o}
[/tex]

[tex]
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}
[/tex]

[tex]
\frac{THRUST}{\dot{m}}=U_9-U_o
[/tex]


[tex]
\dot{m_f}h=(\dot{m_a}+\dot{m_f})C_p{T_{t4}}-\dot{m_a}C_o{T_{t3}}
[/tex]


The Attempt at a Solution


So the givens for on-design analysis:

[tex]\pi_c=15[/tex]

[tex]M_o=0.8[/tex]

[tex]T_o=260 K[/tex]

[tex]T_{t4}=2000 K[/tex]

[tex]P_o=20,000 Pa[/tex]

[tex]\dot{m_a}=100 kg/s[/tex]

[tex]h=4.5 *10^7[/tex]

[tex]C_p=1004[/tex]

Solving for each variable I get the following:

[tex]\tau_r=1.128[/tex]

[tex]\pi_r=1.524[/tex]

[tex]\tau_c=2.168[/tex]

[tex]\tau_\lambda=7.69[/tex]

[tex]\tau_t=.829[/tex]

[tex]\pi_t=.519[/tex]

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

[tex]a_o=322.65 m/s[/tex]

[tex]U_o=258.12 m/s[/tex]

Assuming unknown [tex]\pi{'s}=1[/tex], then

[tex]U_9=1296.38 m/s[/tex]

Solving for THRUST I get 103,826 N

And finally

[tex]\dot{m_f}=3.185 kg/s[/tex]

Now for the off-design:

[tex]M_o=0.3[/tex]

[tex]T_o=288 K[/tex]

[tex]T_{t4}=1500 K[/tex]

[tex]P_o=101325 Pa[/tex]

[tex]h=4.5 *10^7[/tex]

[tex]C_p=1004[/tex]

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

[tex]\pi_t=.519[/tex]

Solving for each variable I get the following:

[tex]\tau_r=1.018[/tex]

[tex]\pi_r=1.064[/tex]

[tex]/tau_c=1.875[/tex]

[tex]\tau_\lambda=5.21[/tex]

[tex]T_{t3}=549.72 K[/tex]

[tex]a_o=339.58 m/s[/tex]

[tex]U_o=101.87 m/s[/tex]

[tex]U_9=957.36 m/s[/tex]

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




Also I found what the area is using [tex]A_o=\frac{\dot{m_a}}{\rho_o{R}{U_o}}[/tex] 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 [tex]\dot{m_f}[/tex] for the off-design analysis using [tex]\rho=\frac{P_o}{RT_o}=1.23[/tex] 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.
 
  • #2
Any takers on this?
 
  • #3
I think you've confused "Other Sciences" with Aeronautical Engineering? I'll try to have your question redirected to a more appropriate forum.
 
  • #4
Thank you.
 

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