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

  • Thread starter roldy
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
232
1
Any takers on this?
 
  • #3
chemisttree
Science Advisor
Homework Helper
Gold Member
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I think you've confused "Other Sciences" with Aeronautical Engineering? I'll try to have your question redirected to a more appropriate forum.
 
  • #4
232
1
Thank you.
 

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