Examples – Propulsion Systems

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

Worked Example #1

A static thrust stand is being designed for the Air Force for testing their jet engines. The following conditions are expected for a typical test: intake air velocity = 200 m/s, exhaust gas velocity = 500 m/s, intake cross-section area = 1 m^2, intake static pressure = -22.5 kPa (= 78.5 kPa absolute), intake static temperature = 268K, exhaust static pressure = 0 kPa (101 kPa absolute). Estimate the thrust reaction force to design the test stand to withstand if it must have a structural safety factor of 3.

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

    \[ \rho_1 = \frac{p_1}{R T_1} = \frac{78.5 \times 10^3}{(287)(268)} = 1.0206~\mbox{kg m$^{-3}$} \]

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

    \[ \rho_1 A_1 V_1 = \rho_2 A_2 V_2 = \dot{m} = 204.12~\mbox{kg s$^{-1}$} \]

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

    \[ F = -p_1 A_1 + p_2 A_2 + \dot{m} ( V_2 - V_1 ) \]

and substituting values gives

    \begin{eqnarray*} F & = & (-78.5 \times 10^3)(1.0) + (101.0 \times 10^3)(1.0) + 204.12 (500.0-200.0) \nonumber \\ & = & 83,736~\mbox{N} \end{eqnarray*}

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, then 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 V_{\infty} of 530 mph; refer to the schematic below. The pressure ratio \delta = p/p_0 at this altitude is 0.2353 and the density ratio \sigma = \rho / \rho_0 is 0.3099. The inlet and exit areas of the turbojet engine are A_i =13 ft^2 and A_e = 8 ft^2, respectively. The fuel-to-air ratio by mass injected into the engine is 0.005. The jet exit velocity V_j is 1,510 ft s^{-1} and the pressure at the exit p_e is 450 lb ft^{-2}. Assume one dimensional flow.

  1.  Draw an appropriately annotated control volume to analyze this problem.
  2. Calculate the mass flow rate of air at the inlet to the turbojet.
  3. Calculate the thrust of the turbojet at this flight condition.
  4. Calculate the fuel consumption of the turbojet in units of lb hr^{-1}.
  5. Determine the thrust-specific fuel consumption, TSFC, of the turbojet.

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

2. The air density at the given altitude is

    \[ \rho_\infty = \sigma \, \rho_0 = 0.3099 \times 0.002378 = 0.000737~\mbox{slug ft$^{-3}$} \]

The corresponding ambient pressure of the air is

    \[ p_\infty = \delta \, p_0 = 0.2353 \times 2116.4 = 498.0~\mbox{lb ft$^{-2}$} \]

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

    \[ \dot{m}_{\rm air} = \rho_{\infty} \, A_i \, V_\infty \]

and substituting the known values gives

    \[ \dot{m}_{\rm air} = 0.000737 \times 13.0 \times 530.0 \times 1.46668 = 7.45~\mbox{slug ft$^{-3}$} \]

4. At the exit, the mass flow rate is

    \[ \dot{m}_e = \dot{m}_{\rm air} + \dot{m}_{\rm fuel} = 7.45 + 7.45 \times 0.005 = 7.487~\mbox{slug ft$^{-3}$} \]

noting that the mass flow rate of fuel \dot{m}_{\rm fuel} has been. accounted for. With the effects of the pressure on the inlet and outlet, then thrust generated at this flight condition is

    \[ T = \dot{m}_e V_j - \dot{m}_{\rm air} V_\infty + p_e A_e -p_\infty A_{\rm in} \]

Inserting the numerical values gives

    \[ T = 7.487 \times 1,510 - 7.45 \times 777.0 + 450.0 \times 10.0 - 498.0 \times 13.0 = 3,542~\mbox{lb} \]

5. The fuel flow rate is

    \[ \dot{W}_{\rm fuel} = \dot{m}_{\rm fuel} \, g = 7.45 \times 0.005 \times 3, 600 \times 32.17  =2,584~\mbox{lb~hr}^{-1} \]

5. The TSFC is

    \[ \mbox{TSFC} = \frac{\dot{W}_{\rm fuel}}{T} = \frac{2,584}{3,595} = 0.719~\mbox{lb lb$^{-1}$  hr$^{-1}$} \]

 

Worked Example #3

Consider the flow through a turbofan engine, i.e., an engine with a hot jet core of diameter D_j and a cold bypass fan of diameter D_f, 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 from the fan) in terms of the intake flow velocity V_{\infty} and the jet velocities from the hot core V_j and bypass fan V_f, 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

    \[ \dot{m}_{\rm fan} = \rho_{\infty} A_{\rm fan} V_{\infty} \]

where A_{\rm fan} = \pi (D_{\rm fan} - D_{\rm core})^2/4.

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

    \[ \dot{m}_{\rm core} = \rho_{\infty} A_{\rm core} V_{\infty} + \dot{m}_{\rm fuel} \]

where where A_{\rm core} = \pi D_{\rm core}^2/4 and the mass of fuel added to the engine has been taken into consideration.

The application of the principle of conservation of momentum gives for the fan part that

    \[ T_{\rm fan} = \dot{m}_{\rm fan} \left(V_{f}  - V_{\infty}\right) \]

and for the engine core part that

    \[ T_{\rm core} = \dot{m}_{\rm core} \left(V_{j} - V_{\infty} \right) \]

where the pressure effects have been neglected, as per the question.

Therefore, the total thrust is the sum of the parts from the fan and the core, i.e.,

    \[ T = T_{\rm fan} + T_{\rm core} \]

so that

    \[ T = \dot{m}_{\rm fan} \left(V_{f}  - V_{\infty} \right) + \dot{m}_{\rm core} \left(V_{j}  - V_{\infty} \right) \]

If we add in the values of the mass flow rates then this latter equation can be written as

    \[ T = \rho_{\infty} A_{\rm fan} V_{\infty}\left(V_{f}  - V_{\infty}\right) + \left(\rho_{\infty} A_{\rm core} V_{\infty} + \dot{m}_{\rm fuel} \right) \left(V_{j}  - V_{\infty}\right) \]

 

Worked Example #4

An aircraft flying at a true airspeed of 200 m/s is powered by a turbofan engine with a bypass ratio of 5:1. The mass flow rate into the engine core is 20 kg/s and the fuel consumption is 2 kg/s. The average velocity of the discharge flow velocity from the engine core is 410 m/s and from the bypass fan it is 270 m/s, both with respect to the engine. Determine the thrust and equivalent power developed by this turbofan engine. Neglect all pressure difference effects.

The mass flow of air into core of the engine is

    \[ \dot{m}_{\rm core} = 20~\mbox{kg/s} \]

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

    \[ \dot{m}_{\rm fan} = 5 \dot{m}_{\rm core} = 100~\mbox{kg/s} \]

The mass flow rate of fuel is

    \[ \dot{m}_{\rm fuel} = 2~\mbox{kg/s} \]

Therefore, the thrust from the engine core is

    \[ T_{\rm core} = (\dot{m}_{\rm core} + \dot{m}_{\rm fuel} ) V_{\rm core} - \dot{m}_{\rm core} V_{\infty} \]

and from the fan the thrust is

    \[ T_{\rm fan} = \dot{m}_{\rm fan} V_{\rm fan} - \dot{m}_{\rm fan} V_{\infty}  = \dot{m}_{\rm fan} \left( V_{\rm fan} - V_{\infty} \right) = 5 \dot{m}_{\rm core} \left( V_{\rm fan} - V_{\infty} \right) \]

The total thrust is then

    \[ T = T_{\rm core} + T_{\rm fan} \]

In this case we are given that \dot{m}_{\rm core} = 20 kg/s, \dot{m}_{\rm fuel} = 2 kg/s, and also that V_{\rm core} = 410 m/s, V_{\rm fan} = 270 m/s, and V_{\infty} = 200 m/s. Therefore, substituting the numbers gives

    \[ T_{\rm core} = (20.0 + 2.0 ) 410.0 - (20.0) 200 = 9,020 - 4,000 = 5.02~\mbox{kN} \]

and

    \[ T_{\rm fan} =  5 (20.0) (270.0 - 200.0) =  7.0~\mbox{kN} \]

The total thrust is then

    \[ T = T_{\rm core} + T_{\rm fan} = 5.02 + 7.0 = 12.02~\mbox{kN} \]

The equivalent net power P that is developed is

    \[ P = T V_{\infty}  = 12.02 \times 200 = 2,404~\mbox{kW} \]

 

Worked Example #5

Refer to the attached propeller charts, which are a standard form of performance presentation for all propellers. Assume that an actual propeller has a diameter of 7 ft and it is a constant speed propeller with a rotational speed of 2,000 rpm. The propeller operates at the equivalent of 8,000 ft ISA density altitude.
For each blade pitch angle measured at 75% radius and at the point of maximum propulsive efficiency in each case then estimate the following:
(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 J, i.e.,

    \[ J = \frac{V_{\infty} }{n \, d} =\frac{V_{\infty} }{(\mbox{rpm}/60) \, d} \]

so

    \[ V_{\infty} = J \, (\mbox{rpm}/60) \, d = J (2,000/60) \times 7.0 = 233.34 \, J~\mbox{ft/s} \]

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

Blade pitch (^{\circ}) \eta J V_{\infty}  (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
(b) The propeller thrust coefficient can be read off the second chart for each value of the advance ratio as was identified in the previous part. The thrust coefficient for a propeller is defined as

    \[ C_T = \frac{T}{\rho n^2 d^4} \]

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

    \[ T = \rho n^2 d^4 C_T \]

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^{-3}. Inserting the information gives

    \[ T = \rho n^2 d^4 C_T = 0.001869 \times \left( \frac{2000}{60} \right)^2 \times 7.0^4 \, C_T = 4986.08 \, C_T \]

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

Blade pitch (^{\circ}) J C_T T
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
(c) The propeller power coefficient can be read off the third chart for each value of the advance ratio that was identified in the previous part. The power coefficient for a propeller is defined as

    \[ C_P = \frac{P}{\rho n^3 d^5} \]

so the corresponding power needed to drive the propeller is

    \[ P = \rho n^3 d^5 C_P \]

Inserting the known information gives

    \[ P = \rho n^3 d^5 C_P = 0.001869 \times \left( \frac{2000}{60} \right)^3 \times 7.0^5 \, C_P = 2115.31 \, C_P \]

where we have converted to horsepower (hp) by dividing the result in lb ft s^{-1} by 550. Again, it is best to use a table to show the results, i.e.,

Blade pitch (^{\circ}) J C_P P (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 V_{\rm hel} 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 d is

    \[ V_{\rm tip} = \Omega \left( \frac{d}{2} \right) \]

where \Omega = 2 \pi ( \mbox{rpm}/60 ) and so the helical tip speed of the propeller is

    \[ V_{\rm hel} = \sqrt{ V_{\rm tip}^2 + V_{\infty}^2} \]

where V_{\infty} is the forward airspeed. The helical Mach number M_{\rm hel} will be

    \[ M_{\rm hel} = \frac{V_{\rm hel} }{a} \]

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

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

Blade pitch (^{\circ}) J V_{\infty} (ft/s) V_{\rm hel} (ft/s) M_{\rm hel}
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
We notice that for an airspeed above 400 ft/s then the propeller blade tips would likely to begin to operate near or just beyond the critical Mach number (i.e., the onset of transonic flow), which for the thin tips of propeller blades is about 0.8. Under these conditions the propeller is likely to see some loss of propulsive efficiency.

 

Worked Example #6

A rocket motor is to be tested on a test stand. The burning the propellent occurs a steady rate of 150 kg/s, and the specific impulse of the propulsion system is 240 seconds. What thrust does the rocket engine develop?

The equivalent exhaust velocity V_{\rm eq} is given in terms of the specific impulse, i.e.,

    \[ V_{\rm eq} = I_{\rm sp} \, g = 240 \times 9.81 = 2,354.4~\mbox{m/s} \]

Therefore, the thrust produced T will be

    \[ T = \dot{m}_P \, V_{\rm eq} = 150 \times 2,354.4 = 353.16~\mbox{kN} \]

 

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