- #1
roldy
- 206
- 2
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{'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.
