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roldy
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I'm studying on variable area turbojet problems in the book and I can't seem to get past this problem.
Consider the performance of an ideal non-afterburning turbojet with flow at station 4(turbine entrance) and station 8 (nozzle throat) choked. A4 is fixed and A8 is varied in order to maintain constant compressor total pressure ratio ([tex]\pi_c[/tex]).
On-design conditions are as follows:
[tex]
\pi_{cR}=15, M_{oR}=2.0, \tau_{\lambda{R}}=7.0
[/tex]
Find the required ratio of nozzle throat area (off-design) to nozzle throat area (on-design) for the engine operating at the same flight Mach number (2.0) but at the off-design condition such that [tex]\tau_\lambda=6.0[/tex].
My work:
Using Mach number 2
On-design
[tex]
\tau_{rR}=0.8, \pi_{rR}=.458, \tau_{cR}=2.168, \tau_{tR}=.867, \pi_{tR}=.607
[/tex]
[tex]
\tau_{tR}=1-\frac{\tau_{rR}}{\tau_{\lambda{R}}}(\tau_{cR}-1)=1-\frac{.8}{7}(2.168-1)=.867
[/tex]Off-design
Same values as on-design because Mach number does not change.
[tex]
\tau_{r}=0.8, \pi_{r}=.458, \tau_{c}=2.168, \tau_{t}=.867, \pi_{t}=.607
[/tex]
Using given [tex]\lambda=6[/tex] for off-design:
[tex]
\tau_t=1-\frac{\tau_r}{\tau_\lambda}(\tau_c-1)=1-\frac{.8}{6}(2.168-1)=.844
[/tex]
Hence, the area ratio:
[tex]
\frac{A_8}{A_{8R}}=\frac{\tau_t^\frac{1}{2}}{\pi_T}\frac{\pi_{tR}}{\tau_{tR}^\frac{1}{2}}=.9866.
[/tex]
Is this correct? This seems a little too easy.
Consider the performance of an ideal non-afterburning turbojet with flow at station 4(turbine entrance) and station 8 (nozzle throat) choked. A4 is fixed and A8 is varied in order to maintain constant compressor total pressure ratio ([tex]\pi_c[/tex]).
On-design conditions are as follows:
[tex]
\pi_{cR}=15, M_{oR}=2.0, \tau_{\lambda{R}}=7.0
[/tex]
Find the required ratio of nozzle throat area (off-design) to nozzle throat area (on-design) for the engine operating at the same flight Mach number (2.0) but at the off-design condition such that [tex]\tau_\lambda=6.0[/tex].
My work:
Using Mach number 2
On-design
[tex]
\tau_{rR}=0.8, \pi_{rR}=.458, \tau_{cR}=2.168, \tau_{tR}=.867, \pi_{tR}=.607
[/tex]
[tex]
\tau_{tR}=1-\frac{\tau_{rR}}{\tau_{\lambda{R}}}(\tau_{cR}-1)=1-\frac{.8}{7}(2.168-1)=.867
[/tex]Off-design
Same values as on-design because Mach number does not change.
[tex]
\tau_{r}=0.8, \pi_{r}=.458, \tau_{c}=2.168, \tau_{t}=.867, \pi_{t}=.607
[/tex]
Using given [tex]\lambda=6[/tex] for off-design:
[tex]
\tau_t=1-\frac{\tau_r}{\tau_\lambda}(\tau_c-1)=1-\frac{.8}{6}(2.168-1)=.844
[/tex]
Hence, the area ratio:
[tex]
\frac{A_8}{A_{8R}}=\frac{\tau_t^\frac{1}{2}}{\pi_T}\frac{\pi_{tR}}{\tau_{tR}^\frac{1}{2}}=.9866.
[/tex]
Is this correct? This seems a little too easy.
Last edited: