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16.50 Propulsion Systems Spring 2011 Quiz 2 May 18, 2011 Two hours, open book, open notes TRUE-FALSE QUESTIONS Justify your answer in no more than two lines. 4 points for correct answer and explanation 2-3 points for a correct answer with only partially correct explanation 1-2 points for an incorrect answer with some valid argument 0 for an incorrect answer with an incorrect explanation, or any answer with no explanation Statement ܶ௧ସ 1. For a turbojet, a high ߠ௧ ൌ ൗܶ gives a high thermodynamic efficiency ߟ௧ at any ௧ compression ratio ߨ . ܶ One can have a ௧ସൗܶ with very small pressure ratio in the cycle. This gives a Brayton ௧ cycle low efficiency. 2. The pressure ratio of the turbine does not change when the pilot changes the fuel/air ratio f. ߬௧ ൌ ቀ ళ ቁ ర మሺംషభሻ ംశభ True False √ √ is fixed by geometry (due to double choking of the flow). 3. If the throat area ܣof a turbojet decreases due to some obstruction, the compressor operating line moves closer to the stall line. The compressor operating line is ݉ ഥଶ ൌ ళ ర ߨ ඨ ଵିఛ , and ߬௧ ൌ ቀ ళ ቁ ംషభ ം గ ିଵ ర మሺംబభሻ ംశభ . If ܣ՛,߬௧ ՝,݉ ഥ ଶ ՝. 4. The bypass ratio ߙ of a turbofan engine is fixed by the geometry, and does not change with operating conditions. It does not vary with ߠ. 5. For a fixed compressor face Mach number ܯଶ , the cowl lip of a subsonic inlet would choke if its area ܣଵ were less than ݉ ഥ ଶ ሺܯଶ ሻܣଶ, where ݉ ഥ ଶ is the non-dimensional flow factor. ഥ ଶ మ మ ൌ బ భ, and since ܲ௧ଶ ൌ ܲ௧ , ܶ௧ଶ ൌ ܶ௧ , Equate flows when ܣଵ chokes: ݉ ඥ்మ √ √ √ ඥ்బ ݉ ഥ ଶ ܣଶ ൌ ܣଵ 6. The Euler equation is only valid for ideal, isentropic flow. It is a mechanical work balance, so it is not sensitive to non-idealities. 7. The stall line on a compressor map can be pre-determined by flow matching conditions even before the specific compressor has been selected. The working line is pre-determined, but the stall line depends on compressor details. 8. In a multi-stage turbine in which each stage has the same isentropic efficiency ߟ௦௧ , the overall turbine isentropic efficiency ߟ ் is greater than ߟ௦௧ . The inefficiency ͳ െ ߟ௦௧ of each stage means that extra heat is deposited in the flow by each stage, and the subsequent stages convert some of it to work by expansion. 1 √ √ √ 9. The nitrogen oxides produced in the primary zone of a jet engine burner are largely destroyed by the cooler secondary air that is injected downstream. The destruction reactions are very slow. 10. A quadrupole made up of four monopoles emits less acoustic power than would each of the monopoles separately. There are partial cancellations between positive and negative monopoles. √ √ PROBLEM 1 (30 points) The design of a certain turbofan engine is such that the turbine inlet temperature at takeoff on a standard day ሺܶͲ ൌ ʹͺͺܭǡ ܲͲ ൌ ͳܽ ݉ݐis ͳͷͲܭ, and the compressor-face Mach number is ܯଶ ൌ ͲǤͷǤ The compressor is designed to provide maximum thrust at that condition. A set of such engines provides the required thrust (including margin) for takeoff of a passenger jet plane. Consider now a “hot day” situation ሺܶ ൌ ͵Ͳͷܭǡ ܲ ൌ ͳܽ݉ݐሻ for the same plane, with the same load and at the same take-off Mach number. How will the following quantities change from their design values?: x Thrust ܨ x Normalized thrust ߮ ൌ x Normalized peak temperature ߠ ൌ x Peak temperature ܶ௧ସ Normalized flow rate ݉ ഥଶ Flow rate ݉ሶ Fuel flow rate ݉ሶ Compressor pressure ratio ߨ x x x x ி మ మ ܶ௧ସ ൗܶ ௧ Same plane weight W, same ܯ , gives same lift L, and assuming same aerodynamic L/D, same drag D. Therefore, same thrust. ܨᇱ ൌ ܨ For the same ܯ and ܲ , same ܲ௧ , so same normalized thrust. ߮ᇱ ൌ ܨԢ ᇱ ܲ௧ ʹܣ ൌ߮ൌ ܨ ܲͲܣ Ͳݐ Now, from lesson 18b, ߮ ൌ ߮ሺߠǡ Ͳܯሻ, hence same ߠ: ߠᇱ ൌ ᇱ ܶ௧ସ ܶݐͶ ᇱ ൌߠൌ ܶ௧ ܶͲݐ 2 ᇲ ்బ Since ܯᇱ ൌ Ͳܯ, ்బ ்ᇲ ଷହ బ ଶ଼଼ ൌ ்బ ൌ ்ᇲ ்ᇲ ൌ ͳǤͲͷͻ. Therefore, ்ర ൌ ்బ ൌ ͳǤͲͷͻ. ర బ ᇱ ܶ௧ସ ൌ ͳǤͲͷͻ ൈ ͳͷͲ ൌ ͳͶ ܭ ഥ ଶ , the compressor ratios ߨ , ߬ , and the ratio The normalized flow rate ݉ ்బ depend exclusively on ߠ, so ߨᇱ ൌ ߨ ݉ ഥ ଶᇱ ൌ ݉ ഥ ଶ Now, ݉ሶ ൌ ݉ ഥ ଶȞ బ మ ඥோ்బ , and so ݉ሶԢ ܶ௧ ͳ ൌඨ ᇱ ൌ ൌ ͲǤͻʹ ܶ௧ ξͳǤͲͷͻ ݉ሶ Also, ݂ ൌ ሶ ሶ ᇲ , so ሶ ൌ ሶ ሶ ᇱ ᇱ ሶ ்ᇲ ்ᇲ ் ൌ ்బ ට்బᇲ ൌ ට்బ బ బ బ ݉ሶᇱ ൌ ξͳǤͲͷͻ ൌ ͳǤͲʹͻ ݉ሶ ݂ PROBLEM 2 (30 points) In designing one of the identical stages of a compressor, we wish to maximize the stage temperature rise οܶ ݐ, so as to minimize the number of stages, while limiting the stage loading to avoid excessive losses. Assume a 50% reaction design, with the axial velocity w determined by a compressor-face Mach number ܯଶ ൌ ͲǤͷǡ and an inlet total temperature ܶ௧ ൌ ʹͷͲܭ. a) Show from the Euler equation that high οܶ௧ per stage is favored by high wheel spin ߱ݎҧ and low stator exit angle ߚଵ (or, for this design, ߚଶᇱ ൌ ߚଵ ). Assume the wheel speed is as high as allowed by hoop stress limitations on the rim (assumed to be self-sustaining, namely, the blade centrifugal pull is compensated by the disk tension). The rim material is a Titanium alloy with ݎ working stress ߪ ൌ ൈ ͳͲ଼ ܲܽ, and density ߩ ൌ ͶͷͲͲ݇݃Ȁ݉ଷ . Take the blade ratio ுൗ ்ݎൌ ҧ ͲǤͺ, so that ݎൗ ݎு ൌ ͻȀͺ. Calculate ߱ݎ.ҧ ఊோ்బ Axial velocity: ݓൌ ܯଶ ඥߛܴܶଶ ൌ ܯଶ ට ംషభ మ ெమ మ ଵା 3 ͳǤͶ ൈ ʹͺ ൈ ʹͷͲ ൌ ͵ͲͻǤ͵݉Ȁݏ ݓൌ ͲǤͷඨ ͳ ͲǤʹ ൈ ͲǤͷଶ Wheel speed: or self-sustaining rim, ߪ ൌ ߩሺ߱ଶ ݎுଶ ሻ, and ݎҧ ൌ ͻȀͺሺ ܪݎሻ, so ߱ݎҧ ൌ ͻ ͻ ߪ ͻ ൈ ͳͲ଼ ߱ݎு ൌ ඨ ൌ ඨ ͺ ߩ ͺ ͶͷͲͲ ͺ ߱ݎҧ ൌ ͶͳͲǤͺ݉Ȁݏ ҧ ҧ െ ʹߚ ݓଵ ሻ Euler equation: ܿ οܶ௧ ൌ ߱ݎሺ߱ݎ Because ݒଵ ൌ ߚ ݓଵ ǡ ݒଶ ൌ ߱ݎҧ െ ߚ ݓଵ , so ݒଶ െ ݒଵ ൌ ߱ݎҧ െ ʹߚ ݓଵ Euler’s equation shows οܶ௧ increases with ߱ݎҧ and decreases with ߚଵ , as stated. b) Draw the velocity triangle and show that the flow turning angle ߜ (the angle between ܸଵ and ܸଶ , or between ܸଵᇱ and ܸଶᇱ) increases as ߚଵ decreases. Values of ߚଵ that are too small will therefore lead to excessive blade losses, and possibly to stall. Choose the smallest ߚଵ that keeps ߜ ͳͷι. Let us consider three values of ߚଵ : zero, a medium value, and the maximum that still makes ߱ݎҧ െ ʹߚ ݓଵ Ͳ: It is clear that as ߚଵ increases (left to right), ߜ decreases. So, the smallest ߚଵ angles give the highest blade loading (most flow turning). 4 If ߜ is prescribed, what is ߚଵ ? We have, from geometry, ߱ݎҧ െ ߚ ݓଵ ൰ െ ߚଵ ߜ ൌ ିଵ ൬ ݓ ௪ Define t = ߚଵ , ҧ ൌ ߶: ఠ ͳ ߜ ൌ ିଵ ൬ െ ݐ൰ െ ିଵ ݐ ߶ Take the tangent of both sides: ଵ ߜ ൌ థ െ ʹݐ ଵ ͳ ቀ െ ݐቁ ݐ థ This can be rearranged as a quadratic equation for t: ͳ ʹ ͳ ൰ െͳൌͲ ݐଶ െ ൬ ߶ ߜ ߶ ߜ With solution: ݐൌ ͳ ͳ ͳ ͳ േඨ ଶ ͳ ʹ߶ ߜ Ͷ߶ ߜ In our case, ߜ ൌ ͳͷι (using the negative square root, the positive makes ߱ݎҧ െ ʹߚ ݓଵ ൏ Ͳ). ͳ ͶͳͲǤͺ ͳ ͶͳͲǤͺ ଶ ඨ ൰ ݐൌ ߚଵ ൌ െ ൬ ͳ ʹ ൈ ͵ͲͻǤ͵ ͳͷι ʹ ൈ ͵ͲͻǤ͵ ଶ ͳͷι ߚଵ ൌ ʹͷǤͶͶι Note: This can be also found by direct trial-and-error on ͶͳͲǤͺ െ ߚͳ ൰ െ ߚͳ ߜ ൌ ିଵ ൬ ͵ͲͻǤ͵ c) With these choices, calculate the temperature rise οܶ௧ per stage. How many stages would be required to achieve an overall pressure ratio ߨ ൌ ʹͳ? For ߱ݎҧ ൌ ͶͳͲǤͺ݉Ȁݏ, ݓൌ ͵ͲͻǤ͵ ǡ ߚଵ ௦ οܶ௧ ൌ ംషభ ൌ ʹͷǤͶͶι. Euler gives: ͶͳͲǤͺሺͶͳͲǤͺ െ ʹ ൈ ͵ͲͻǤ͵ ൈ ͲǤͲͶͷሻ ͳͲͲͷ ሺοܶ௧ ሻ௦௧ ൌ ͶǤܭ భ If ߨ ൌ ʹͳ, ߬ ൌ ߨ ം ൌ ʹͳయǤఱ ൌ ʹǤ͵ͺͷ The number of stages is: ܰൌ ܶ௧ ሺ߬ െ ͳሻ ʹͷͲሺʹǤ͵ͺͷ െ ͳሻ ൌ ൌ Ǥʹͺ ͶǤ ሺοܶ௧ ሻ௦௧ Rounding up: ܰ ൌ ͺ 5 0,72SHQ&RXUVH:DUH KWWSRFZPLWHGX ,QWURGXFWLRQWR3URSXOVLRQ6\VWHPV 6SULQJ )RULQIRUPDWLRQDERXWFLWLQJWKHVHPDWHULDOVRURXU7HUPVRI8VHYLVLWKWWSRFZPLWHGXWHUPV