Question 16.1: The following data refer to a single shaft of a turboprop en......
The following data refer to a single shaft of a turboprop engine during ground run. The gas turbine is running at its design point.
The compressor delivery pressure and the ambient conditions are 1.01 bar and 288 K. The nondimensional flows are based on \dot m in kg/s, P in bar, and T in K. Neglect mechanical losses. The pressure losses in the combustion chamber are 3%. Assume also that the mass flow rate in both compressor and turbine are equal. Calculate the power consumed in the propeller when the turbine inlet temperature is 1200 K.
| Compressor characteristics | Turbine characteristics | ||||
| P_{02}/P_{01} | \dot{m}_1\sqrt{T_{01}}/P_{01} | \eta_{\text{C}} | \dot{m}_3\sqrt{T_{03}}/P_{03} | P_{03}/P_{04} | \eta_{\text{t}} |
| 4.0 | 350 | 0.76 | 130 | 4.2 | 0.88 |
| 4.5 | 330 | 0.8 | 144 | 4.6 | 0.9 |
| 5.0 | 310 | 0.84 | 144 | 5.0 | 0.91 |
Flow compatibility is expressed by the relation (16.5):
\frac{\dot{m}_3\sqrt{T_{03}}}{P_{03}} =\frac{\dot{m}_1 \sqrt{T_{01}}}{P_{01}} \frac{P_{01}}{P_{02}} \frac{P_{02}}{P_{03}} \frac{\dot{m}_3}{\dot{m}_1} \sqrt{\frac{T_{03}}{T_{01}} }
For equal mass flow rate through the compressor and turbine, 3% pressure loss in the combustion chamber, and equality of compressor inlet pressure and turbine outlet pressure (P_{01}=P_{04}), then
\frac{\dot{m}\sqrt{T_{03}}}{P_{03}} =\frac{\dot{m}\sqrt{T_{01}}}{P_{01}} \frac{P_{01}}{P_{03}} \sqrt{\frac{1200}{288} }=2.041\frac{\dot{m}\sqrt{T_{01}}}{P_{01}} \frac{P_{04}}{P_{03}} \quad \quad \quad (1)
The mass flow parameter is calculated from the Equation 1 for different turbine pressure ratios and tabulated along with the values given in the data above in the following table:
The two sets of turbine mass flow parameter are plotted in Figure 16.6.
From Figure 16.6, equilibrium between compressor and turbine occurs at P_{03}/P_{04}=4.64, \dot m_3 \sqrt{T_{03}}/P_{03 } = 144. Plotting the turbine efficiency on the same figure gives the corresponding turbine efficiency as \eta_{\text{t}} = 0.9012.
The corresponding compressor pressure ratio is
P_{02}/P_{01}=\frac{P_{03}/P_{04}}{0.97} =\frac{4.64}{0.97} =4.7835
The corresponding values of the compressor mass flow rate and efficiency are determined from Figure 16.7, which are \dot{m}_1 \sqrt{T_{01}}/P_{01}=319 \text{ and } \eta_{\text{C}}=0.82, respectively.
From Equation 16.2, the mass flow rate is
\dot{m}_1=\frac{\dot{m}_1\sqrt{T_{01}}}{P_{01}}\frac{P_{01}}{\sqrt{T_{01}}} =\frac{319 \times 1.01}{\sqrt{288}} =18.98 \text{ kg/s}
The available power is evaluated from Equation 16.7.
| P_{03}/P_{04} | \dot{m}_3 \sqrt{T_{03}}/P_{03}
From turbine map |
\dot{m}_3 \sqrt{T_{03}}/P_{03}
Equation 1 |
| 4.2 | 130 | 170.1 |
| 4.6 | 144 | 146.42 |
| 5.0 | 144 | 126.54 |
