MAE 113, Summer Session 1, 2009

Transcription

1 HW #1 1., 1.7, 1.14,.3,.6 MAE 113, Summer Seion 1, 9 1. Develop the following analytical expreion for a turbojet engine: a) When m f << m o, P e =P a, and f inlet = f noz =, then the intalled thrut i given by: From equation 1.5, T = m o HV e -V L F = Im o +m fmv e -m V +HP e -P LA e when we apply m f << m o and P e =P a, we get F = m ov e -m V From equation 1.9, T =FH1-f inlet -f noz L but, f inlet = f noz =, o T=F. Thu, T = m o HV e -V L b) By the ame condition, how TSFC = T êm o + V h T h PR From equation 1., TSFC = V h P h T h PR and from equation 1.16, h P = V e êv +1 from part (a), T = m o HV e -V L T m o T m ov V e V =V e -V = V e V - 1 = T m ov + 1

2 MAE 113 HW1 olution-.nb plugging thi into 1.16 h P = Tgc m ov +1+1 h P J T m ov + N = h P V J T m o h P = V V h P = + V N = Tgc m o Tgc m o +V +V and, finally, thi goe into equation 1. to get TSFC = Tgc m o +V h T h PR c) For V = and 5 ft/, plot the preceding equation for TSFC [in (lbm/h)/lbf] v pecific thrut T/m o [in lbf/(lbm/)] for value of pecific thrut from to 1. Ue h T =.4and h PR = 18, 4 Btu/lbm. It i very eay to me up the unit of thi problem. Note that the input variable, T/m o, i in lbf/(lbm/ec), but the output variable i in (lbm/hour)/lbf. You mut multiply by the number of econd in an hour to output the correct unit. Alo, you mut ue the converion 1btu= ft lbf. You can et up an equation of the form TSFC = I36 ec hour M JX input variable Here' Matlab code you can ue to make the plot: X=:.1:1; TSFC=36*(X*3.174+*)/(*.4*184*778.16); TSFC5=36*(X*3.174+*5)/(*.4*184*778.16); plot(x,tsfc,x,tsfc5) lbf lbmÿft ft NJ3.174 N+J or 5 lbfÿ H.4LJ18,4 Btu ftÿlbf N lbm Btu lbmê N, wherex = T m i the o TSFC (lbm/hr)/lbf Specific Thrut lbf/(lbm/)

3 MAE 113 HW1 olution-.nb 3 The blue line i the V = line and the green line i the V = 5 ftê line. d) explain the trend You're on your own here. Say omething about how TSFC i alway higher when the inlet velocity i higher, and that TSFC increae linearly with pecific thrut.

4 4 MAE 113 HW1 olution-.nb 1.7 The JT9D high-bypa-ratio turbofan engine with V =, P = pia, T = R, and m C = 47 lbmê, m B = 148 lbmê, V Ce = 119 ftê, V Be = 885 ftê, m f = lbmê hr. Etimate the following auming P =P e : a) Thrut From Problem 1.5, we learn that thrut for a bypa engine i equal to the um of the thrut from the core and the bypa tream. F =F C +F B F C = 1 AIm C+m fmv Ce -m CV E F B = m B HV Be -V L Putting all thee together, we get an equation for the thrut for a bypa engine F = 1 AIm C+m fmv Ce -m CV +m BV Be -m BV E We have value for all the variable on the right hand ide, o we can calculate thrut, but again be very careful with unit. F = lbmÿft lbfÿ AI47 lbm lbm hour hour ft lbm M HL+148 lbm 885 ft lbm HLE 36 F = lbf b) Thermal efficiency, h T, with h PR º 18, 4 Btu lbm Equation 1.13 tell u that for a ingle inlet and ingle exhaut h T = W out Q in W out, ingle exhaut = 1 AIm o+m fmv e -m ov E Q in =m fh PR here, though, we have two inlet and two exhaut, therefore but V =, o W out, with bypa =W Cout+W Bout W out = 1 AIm C+m fmv Ce -m CV E+ 1 AHm BLV Be -m BV E W out = 1 AIm C+m fmv Ce +m BV Be E inerting thi into the firt equation give u the reult that h T = AIm C+m fmv Ce +m BV Be E m fh PR

5 MAE 113 HW1 olution-.nb 5 and now we can put in value h T = lbm BJ lbm hr hr 36 J3.174 lbmÿft lbfÿ NJ15 75 lbm h ft NJ119 N +148 lbm J885 ft h 36 h T =.338 h T = 33.8% NJ18,4 Btu lbm N F ftÿlbf Btu N c) Propulive efficiency,h P, and unintalled thrut pecific fuel conumption, S A defined in equation 1.14, And ince V =, From equation 1.1, h P = TV W out h P = S = lbm hr lbf S = m f F S =.361 lbmêhr lbf = 1.8ÿ 1-4 lbmê lbf

6 6 MAE 113 HW1 olution-.nb 1.14 An aircraft with wind area 8 ft in level flight at maximum C L êc D. C D =., K =, K 1 =., find a) The maximum C L êc D and correponding C L and C D Equation 1.48 i ued here Alo, equation 1.47 ay that J C L C D N * = J C L C D N * = 1 C D K 1 +K 1 H.L H.L + J C L C D N * = 7.96 C L * = C L * = C D K 1.. C L * =.316 now we can find C D *.316 C D * = 7.96 C D * =.4 b) The flight altitude and drag for aircraft weight of 45, lbf and Mach.8. Ue eqn 1.9 and 1.3b. Equation 1.9 and 1.3b are Solving for d and combining the two equation, Now we can plug in value L =nw =C L qs w q = g PM = g dp refm d = d = d = q gp ref M nw J CLSw N gp ref M nw gp ref M C L S w d = 1.4J14.7 lbf in H1LH45 lbfl 144 in ft d =.1876 NH.8L H.316LI8 ft M

7 MAE 113 HW1 olution-.nb 7 Ue Appendix A to ee that thi correpond to an altitude of about 39, 8 ft. Next, equation 1.31 give u drag D =C D qs w D =C D nw D =.4H1L C L H45 lbfl.316 D = lbf c) Flight altitude and drag for an aircraft of weight 35, lbf and Mach.8 Analyi i nearly identical to part b, with only a change in weight. d = d = 1.4J14.7 lbf in nw gp ref M C L S w H1LH35 lbfl 144 in ft NH.8L H.316LI8 ft M d =.1459 Again, ue Appendix A to ee that thi correpond to an altitude of about 45, ft. Alo, D =C D nw D =.4H1L C L H35 lbfl.316 D = lbf d) Range for an intalled engine TSFC rate of.8 (lbm/hr)/lbf, if the 1,-1bf difference in aircraft weight between part b and c i due only to fuel conumption. Firt tart with equation 1.43 for range factor RF = C L C D V TSFC We need to calculate velocity in ft/ rather than Mach. Ue appendix A and the note that peed of ound a=a td q. V = 1116 ft g V =am V = 774 ft

8 8 MAE 113 HW1 olution-.nb Now we can go back to RF RF = Next, we ue equation 1.45a to find the range, 774 ft.8 lbmêhr lbf hr lbmÿft lbfÿ lbmÿft lbfÿ nm RF = ft 68 ft RF = nm W f W i = expi- RF M = RF lnj W i W f N 45 lbf = nm lni M 35 lbf = 1138 nm = mi

9 MAE 113 HW1 olution-.nb 9.3 Conider the flow hown in Figure P.. It ha radiu, velocity V 1, and preure P 1. The fluid leave with a velocity V =V max B1-I r M F with uniform preure P. Ue the conervation of ma and momentum equation to how that the force neceary to hold the pipe in place can be written a Start with equation. F = p JP 1 -P + rv 1 3 N F = 1 I dm dt +M out-m inm by auminonervation of ma (ince there i no ink or ource), dm dt =. F = 1 IM out-m inm The um of the force i F = -F+HP 1 -P LA Where F i the force required to keep the pipe in place and A i the area of the pipe. We can combine thee two equation to get F = 1 IM out-m inm = -F+HP 1 -P LA F = 1 IM in-m outm+hp 1 -P LA Thee equation are wrong. However, the firt part of thi equation doe not make ene. Force hould not be proportional to M in-m out, becaue that would imply that if M in >M out, the force would be potive. However, M in >M out implie that air i moving from back to front, which hould decreae the force required to keep the pipe teady. Thi mean that we have defined our um of force to be in a different direction in each of the equation. Intead, the equation hould read: F = 1 IM out-m inm+hp 1 -P LA

10 1 MAE 113 HW1 olution-.nb Momentum flux can be obtained by integrating M out =M = Ÿ m V m M r = Ÿ rv pr r M = Ÿ rjv max B1-I r M FN pr r M = Ÿ rv max J1-I r M N pr r M = p rv max Ÿ J1-I r M +I r M 4 Nr r M = p rv max Ÿ Jr- r3 + r5 4N r M = p rv max B r - r4 + r F M = p rv max B - r 4 6 M = p rv max B M = prv max F - + F 6 Similarly, M in =M 1 = Ÿ m V 1 m 1 M r 1 = Ÿ rv 1 pr r And plugging into to the equation for force above Conervation of ma can be ued to find V max M 1 = rv 1 pb r F M 1 = rv 1 p F = 1 IM out-m inm+hp 1 -P LA F = pr :P 1 -P - r J V max -V g 1 N> c m 1 =m p rv 1 = rÿ V max :1-I r M > pr r V 1 = V max V 1 = V max V 1 = V max 3 Ÿ :r- r3 > r B r - r4 4 F B r - r 4 4 F V 1 = V max V max = V 1

11 MAE 113 HW1 olution-.nb 11 And we arrive at F = pr :P 1 -P + r J HV 1L -V g 1 N> c F = pr :P 1 -P + rv 1 I 4-1M> 3 F = pr :P 1 -P + rv 1 I 4-1M> 3 F = p :P 1 -P + rv 1 3 > 3

12 1 MAE 113 HW1 olution-.nb.6 Uing Figure P.5, with 15 lbm/ of air at 6 F and 14.7 pia entering the engine at a velocity of 45 ft/ and that 15 lbm/ of bypa air leave the engine at 6 to the horizontal, at a velocity of 89 ft/ and preure of 14.7 pia. The remaining 5 lbm/ leave the engine core at a velocity of 1 ft/ and preure of 14.7 pia. Determine the force on the trut, F x. Aume an ambient preure of 14.7 pia. By ymmetry, there i no net force in the y-direction. The momentum equation in the x-direction i Therefore, the force i F x =F x = 1 IM x core out+m x fan out-m x inm M x in = I15 lbm M x core out = I5 lbm M x fan out = I-15 lbm MI45 ft ftÿlbm M = 675 MI1 ft ftÿlbm M = 3 MI89 ft ftÿlbm M coh6 L = F x = lbmÿft lbfÿ B3 ftÿlbm +J ftÿlbm N-675 ftÿlbm F Thi mean a force of 8,945 lbf lowing the plane. F x = -8, 945 lbf