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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.

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