# 55 Worked Examples: Propulsion Systems

These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

Assume that the inlet condition is identified as station 1 and the outlet as station 2. Assume also that the exit area is the same as the inlet area. The density of the air at the inlet can be found using

From continuity considerations (assume 1-dimensional, steady flow), then

Conservation of momentum (assume 1-dimensional flow) gives the force on the fluid as

and substituting values gives

The residual force on the fluid for the conditions stated must be 83,736 N in the direction of fluid flow, and the reaction force on the test stand will be in the opposite direction. Because there is a structural factor of safety requirement of 3, the force to which the stand should be designed is about 251 kN.

Worked Example #2

Consider a turbojet-powered airplane flying at an altitude of 35,000 ft with a true airspeed of 530 mph; refer to the schematic below. At this altitude, the pressure ratio is 0.2353, and the density ratio is 0.3099. The inlet and exit areas of the turbojet engine are =13 ft and = 8 ft, respectively. The fuel-to-air ratio by mass injected into the engine is 0.005. The jet exit velocity is 1,510 ft s and the pressure at the exit is 450 lb ft. Assume one-dimensional flow.

- Draw an appropriately annotated control volume to analyze this problem.
- Calculate the mass flow rate of air at the inlet to the turbojet.
- Calculate the thrust of the turbojet at this flight condition.
- Calculate the fuel consumption of the turbojet in units of lb hr.
- Determine the turbojet’s thrust-specific fuel consumption, TSFC.

1. The control volume is shown in the figure below.

2. The air density at the given altitude is

The corresponding ambient pressure of the air is

3. The air mass flow rate at the inlet of the turbojet engine is

and substituting the known values gives

At the exit, the mass flow rate is

noting that the mass flow rate of fuel has been. accounted for. With the effects of the pressure on the inlet and outlet, the thrust generated at this flight condition is

Inserting the numerical values gives

4. The fuel flow rate is

5. The TSFC is

Worked Example #3

Consider the flow through a turbofan engine, i.e., an engine with a hot jet core of diameter and a cold bypass fan of diameter , as shown in the figure below. Using the flow conservation principles applied to the control volumes, show (using appropriate reduced forms of the conservation equations) how to calculate the net thrust from the engine (i.e., the combined thrust from the core and the fan) in terms of the intake flow velocity and the jet velocities from the hot core and bypass fan , respectively. Write your result in terms of the bypass ratio and comment on your result. State any assumptions that you use. Hints: 1. Consider also the mass of fuel introduced into the engine. 2. Neglect all pressure difference effects.

Assume one-dimensional flow throughout the problem. The mass flow through the fan (outer, cold section) is

The mass flow through the engine core (inner, hot section) is

Because is much smaller than the mass flow of air into the engine, then another assumption is that can be neglected.

Therefore, the application of conservation of momentum gives us that

and

where the pressure effects have also been neglected. The bypass ratio (BPR) is given by

Therefore, the total thrust is

or in terms of the BPR

Notice that the higher the BPR, the higher the thrust from the fan stage of the engine. This is also a more efficient way of producing thrust because for a turbofan, the jet velocity for the fan stage, , is less than that of the core, , i.e.,

and

Therefore, the higher the mass flow rate and the lower the jet velocity, the more efficient the turbojet engine will produce thrust.

Worked Example #4

The mass flow of air into the core of the engine is

and with a bypass ratio of 5:1 then, the mass flow into the fan will be

The mass flow rate of fuel is

Therefore, the thrust from the engine core is

and from the fan, the thrust is

The total thrust is then

In this case we are given that = 20 kg/s, = 2 kg/s, and also that = 410 m/s, = 270 m/s, and = 200 m/s. Therefore, substituting the numbers gives

and

The total thrust is then

The equivalent net power that is developed is

Worked Example #5

(a) What are the values of the advance ratio and corresponding airspeed?

(b) What are the values of the propeller thrust coefficient and the corresponding thrust produced by the propeller?

(c) What are the values of the propeller power coefficient, and the corresponding shaft torque and power required to spin the propeller?

(d) What are the values of helical tip speed and helical Mach number of the propeller? Comment on your results. Note: The helical tip speed is the vector sum of the rotational speed at the tip of the propeller and the free-stream (airspeed).

(a) At the peak efficiency the values of the the advance ratio can be read off the first chart. We can easily do this to two decimal places; for better accuracy the chart can be digitized. We are also given information about the specific propeller, which is fairly small would likely be for a general aviation aircraft, so in each case we can calculate the corresponding airspeed for a given value of , i.e.,

so

It is best to use a table to show the results, i.e.,

Blade pitch () | (ft/s) | ||

15 | 0.82 | 0.65 | 151.7 |

20 | 0.85 | 0.82 | 191.3 |

25 | 0.87 | 1.04 | 242.7 |

30 | 0.87 | 1.25 | 292.7 |

35 | 0.86 | 1.45 | 338.3 |

40 | 0.86 | 1.70 | 398.7 |

45 | 0.84 | 1.95 | 455.0 |

so the corresponding thrust (in units of force) from the propeller is

We are told that the propeller operates at the equivalent of 8,000 ft ISA density altitude, so according to the ISA equations then the density at this altitude is 0.001869 slugs ft. Inserting the information gives

Again, it is best to use a table to show the results, i.e.,

Blade pitch () | |||

15 | 0.65 | 0.025 | 124.7 |

20 | 0.82 | 0.038 | 189.5 |

25 | 1.04 | 0.040 | 199.4 |

30 | 1.25 | 0.047 | 234.4 |

35 | 1.45 | 0.052 | 259.3 |

40 | 1.70 | 0.060 | 299.2 |

45 | 1.95 | 0.072 | 359.0 |

so the corresponding power needed to drive the propeller is

Inserting the known information gives

where we have converted to horsepower (hp) by dividing the result in ft-lb s by 550. Again, it is best to use a table to show the results, i.e.,

Blade pitch () | (hp) | ||

15 | 0.65 | 0.022 | 46.5 |

20 | 0.82 | 0.035 | 74.0 |

25 | 1.04 | 0.048 | 101.3 |

30 | 1.25 | 0.065 | 137.5 |

35 | 1.45 | 0.09 | 190.3 |

40 | 1.70 | 0.12 | 253.8 |

45 | 1.95 | 0.17 | 359.7 |

(d) Let be the helical tip velocity based on the vector sum of the rotational and airspeed components. The rotational tip speed of a propeller of diameter is

where and so the helical tip speed of the propeller is

where is the forward airspeed. The helical Mach number will be

where is the local speed of sound at the conditions at which the propeller operates. At 8,000 ft ISA density altitude, 1085.3 ft/s.

As a final table to show the results, then the helical tip speed and Mach number are:

Blade pitch () | (ft/s) | (ft/s) | ||

15 | 0.65 | 151.7 | 748.6 | 0.69 |

20 | 0.82 | 191.3 | 757.6 | 0.70 |

25 | 1.04 | 242.7 | 772.2 | 0.71 |

30 | 1.25 | 292.7 | 788.9 | 0.73 |

35 | 1.45 | 338.3 | 807.3 | 0.74 |

40 | 1.70 | 398.7 | 833.5 | 0.77 |

45 | 1.95 | 455.0 | 862.8 | 0.79 |

Worked Example #6

The equivalent exhaust velocity is given in terms of the specific impulse, i.e.,

Therefore, the thrust produced will be