MAE113 HW1 Solution
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Transcript of MAE113 HW1 Solution
MAE 113, Summer Session 1, 2009
HW #1
1.2, 1.7, 1.14, 2.3, 2.6
1.2 Develop the following analytical expressions for a turbojet engine:
a) When m°f << m
°o, Pe = Pa, and finlet = fnoz = 0, then the installed thrust is given by:
T =m°o
gcHVe -V0L
From equation 1.5,
F =Im° o +m° f M Ve -m° 0 V0
gc+ HPe - P0L Ae
when we apply m°f << m
°o and Pe = Pa, we get
F =m°o Ve -m
°0 V0
gc
From equation 1.9,
T = FH1 -finlet - fnozLbut, finlet = fnoz = 0, so T=F. Thus,
T =m°o
gcHVe - V0L
b) By the same conditions, show
TSFC =Tgcêm
°o + 2 V0
2 hT hPR
From equation 1.20,
TSFC = V0
hP hT hPR
and from equation 1.16,
hP =2
VeêV0+1
from part (a),
T =m°o
gcHVe -V0L
T gc
m°o
= Ve -V0
T gc
m°o V0
=Ve
V0- 1
Ve
V0=
T gc
m°o V0
+ 1
Printed by Mathematica for Students
plugging this into 1.16
hP =2
T gc
m°o V0
+1+1
hP J T gcm°o V0
+ 2N = 2
hP
V0J T gcm°o
+ 2 V0N = 2
hP
V0=
2T gc
m°o
+2 V0
V0
hP=
T gc
m°o
+2 V0
2
and, finally, this goes into equation 1.20 to get
TSFC =
T gc
m°o
+2 V0
2 hT hPR
c) For V0 = 0 and 500 ft/s, plot the preceding equation for TSFC [in (lbm/h)/lbf] vs specific thrust T/m°o [in
lbf/(lbm/s)] for values of specific thrust from 0 to 120. Use hT = 0.4and hPR = 18, 400 Btu/lbm.
It is very easy to mess up the units of this problem. Note that the input variable, T/m°o, is in lbf/(lbm/sec), but
the output variable is in (lbm/hour)/lbf. You must multiply by the number of seconds in an hour to output the
correct units. Also, you must use the conversion 1btu=778.16 ft·lbf.
You can set up an equation of the form TSFC = I3600sec
hourMJX lbf
lbmês N J32.174lbmÿft
lbf ÿs2N+2 J0 or 500
ft
sN
2 H0.4L J18,400Btu
lbm
778.16 ftÿlbf
BtuN
, where X =T
m°o
is the
input variable
Here's Matlab code you can use to make the plot:
X=0:0.1:120;
TSFC0=3600*(X*32.174+2*0)/(2*0.4*18400*778.16);
TSFC500=3600*(X*32.174+2*500)/(2*0.4*18400*778.16);
plot(X,TSFC0,X,TSFC500)
0 20 40 60 80 100 1200
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Specific Thrust lbf/(lbm/s)
TSFC
(lb
m/h
r)/lbf
2 MAE 113 HW1 solution-2.nb
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The blue line is the V0 = 0 line and the green line is the V0 = 500 ft ê s line.
d) explain the trends
You're on your own here. Say something about how TSFC is always higher when the inlet velocity is higher,
and that TSFC increases linearly with specific thrust.
MAE 113 HW1 solution-2.nb 3
Printed by Mathematica for Students
1.7 The JT9D high-bypass-ratio turbofan engine with V0 = 0, P0 = 14.696 psia, T0 = 518.7 °R, and
m°C = 247 lbm ê s, m
°B = 1248 lbm ê s, VCe = 1190 ft ê s, VBe = 885 ft ê s, m
°f = 15 750 lbm ê hr. Estimate the
following assuming P0 = Pe :
a) Thrust
From Problem 1.5, we learn that thrust for a bypass engine is equal to the sum of the thrust from the core and
the bypass stream.
F = FC + FB
FC =1
gcAIm° C +m° f M VCe -m
°C V0E
FB =m°B
gcHVBe -V0L
Putting all these together, we get an equation for the thrust for a bypass engine
F =1
gcAIm° C +m° f M VCe -m
°C V0 +m
°B VBe -m
°B V0E
We have values for all the variables on the right hand side, so we can calculate thrust, but again be very careful
with units.
F =1
32.174lbmÿft
lbf ÿs2
AI247lbm
s+ 15 750
lbm
hour
hour
3600 sM 1190
ft
s- 247
lbm
sH0L+ 1248
lbm
s885
ft
s- 1248
lbm
sH0LE
F = 43 626 lbf
b) Thermal efficiency, hT , with hPR º 18, 400Btu
lbm
Equation 1.13 tells us that for a single inlet and single exhaust
hT =W°
out
Q°
in
W°
out, single exhaust =1
2 gcAIm° o +m° f M Ve2 -m° o V0
2EQ°
in = m°f hPR
here, though, we have two inlets and two exhausts, therefore
W°
out, with bypass = W°
Cout +W°
Bout
W°
out =1
2 gcAIm° C +m° f M VCe
2 -m°C V0
2E+ 1
2 gcAHm° BL VBe
2 -m°B V0
2E
but V0 = 0, so
W°
out =1
2 gcAIm° C +m° f M VCe
2 +m°B VBe
2 E
inserting this into the first equation gives us the result that
hT =AIm° C+m° f M VCe
2 +m°B VBe
2 E2 gc m
°f hPR
4 MAE 113 HW1 solution-2.nb
Printed by Mathematica for Students
and now we can put in values
hT =BJ247
lbm
s+15 750
lbm
hr
hr
3600 sN J1190
ft
sN2+1248
lbm
sJ885
ft
sN2F
2 J32.174lbmÿft
lbf ÿs2N J15 750
lbm
h
h
3600 sN J18,400
Btu
lbm
778.16 ftÿlbf
BtuN
hT = 0.3308
hT = 33.08 %
c) Propulsive efficiency, hP, and uninstalled thrust specific fuel consumption, S
As defined in equation 1.14,
hP =T V0
W°
out
And since V0 = 0,
hP = 0
From equation 1.10,
S =m°f
F
S =15 750
lbm
hr
43 626 lbf
S = 0.361lbmêhr
lbf= 1.0028 ÿ 10-4 lbmês
lbf
MAE 113 HW1 solution-2.nb 5
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1.14 An aircraft with wind area 800 ft2 in level flight at maximum CL êCD. CD0 = 0.02, K2 = 0, K1 = 0.2, find
a) The maximum CL êCD and corresponding CL and CD
Equation 1.48 is used here
J CLCD
N* = 1
2 CD0 K1 +K2
J CLCD
N* = 1
2 H0.02L H0.2L +0
J CLCD
N* = 7.906
Also, equation 1.47 says that
CL* =
CD0
K1
CL* =
0.02
0.2
CL* = 0.3162
now we can find CD*
0.3162
CD* = 7.906
CD* = 0.04
b) The flight altitude and drag for aircraft weight of 45,000 lbf and Mach 0.8. Use eqns 1.29 and 1.30b.
Equations 1.29 and 1.30b are
L = n W = CL q Swq =
g
2P M0
2 =g
2d Pref M0
2
Solving for d and combining the two equations,
d =2 q
g Pref M02
d =2 J n W
CL SwN
g Pref M02
d =2 nW
g Pref M02 CL Sw
Now we can plug in values
d =2 H1L H45 000 lbfL
1.4 J14.7lbf
in2
144 in2
ft2N H0.8L2 H0.3162L I800 ft2M
d = 0.1876
6 MAE 113 HW1 solution-2.nb
Printed by Mathematica for Students
Use Appendix A to see that this corresponds to an altitude of about 39, 800 ft .
Next, equation 1.31 gives us drag
D = CD q Sw
D = CDnW
CL
D = 0.04 H1L H45 000 lbfL0.3162
D = 5692.6 lbf
c) Flight altitude and drag for an aircraft of weight 35,000 lbf and Mach 0.8
Analysis is nearly identical to part b, with only a change in weight.
d =2 nW
g Pref M02 CL Sw
d =2 H1L H35 000 lbfL
1.4 J14.7lbf
in2
144 in2
ft2N H0.8L2 H0.3162L I800 ft2M
d = 0.1459
Again, use Appendix A to see that this corresponds to an altitude of about 45, 000 ft .
Also,
D = CDnW
CL
D = 0.04 H1L H35 000 lbfL0.3162
D = 4427.6 lbf
d) Range for an installed engine TSFC rate of 0.8 (lbm/hr)/lbf, if the 10,000-1bf difference in aircraft weight
between parts b and c is due only to fuel consumption.
First start with equation 1.43 for range factor
RF =CL
CD
V
TSFC
gc
g0
We need to calculate velocity in ft/s rather than Mach. Use appendix A and the note that speed of sound
a=astd q .
V = a M
V = 1116ft
s0.7519 0.8
V = 774ft
s
MAE 113 HW1 solution-2.nb 7
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Now we can go back to RF
RF =0.3162
0.04
774ft
s
0.8lbmêhr
lbf
hr
3600 s
32.174lbmÿft
lbf ÿs2
32.174lbmÿft
lbf ÿs2
RF = 27 533 115 ftnm
6080 ft
RF = 4528.5 nm
Next, we use equation 1.45a to find the range, s
W f
Wi= expI- s
RFM
s = RF ln J WiW f
N
s = 4528.5 nm lnI 45 000 lbf
35 000 lbfM
s = 1138 nm = 1309.6 mi
8 MAE 113 HW1 solution-2.nb
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2.3 Consider the flow shown in Figure P2.2. It has radius r0, velocity V1, and pressure P 1. The fluid leaves
with a velocity
V2 = VmaxB1 - I rr0M2F
with uniform pressure P2. Use the conservation of mass and momentum equations to show that the force
necessary to hold the pipe in place can be written as
F = p r02 JP1 - P2 +
r V12
3 gcN
Start with equation 2.20
⁄Fs =1
gcI dMs
dt+M
°out -M
°inM
by assuming conservation of mass (since there is no sink or source), dMs
dt= 0.
⁄Fs =1
gcIM° out -M
°inM
The sum of the forces is
⁄Fs = -F + HP1 - P2L AWhere F is the force required to keep the pipe in place and A is the area of the pipe.
We can combine these two equations to get
⁄Fs =1
gcIM° out -M
°inM = -F + HP1 - P2L A
F =1
gcIM° in -M
°outM+ HP1 - P2L A
These equations are wrong.
However, the first part of this equation does not make sense. Force should not be proportional to M°
in -M°
out,
because that would imply that if M°
in > M°
out, the force would be postive. However, M°
in > M°
out implies that air
is moving from back to front, which should decrease the force required to keep the pipe steady. This means that
we have defined our sum of forces to be in a different direction in each of the equations.
Instead, the equation should read:
F =1
gcIM° out -M
°inM+ HP1 - P2L A
MAE 113 HW1 solution-2.nb 9
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Momentum flux can be obtained by integrating
M°
out = M°
2 = Ÿm° 2V2 „m
°
M°
2 = Ÿ0r0r V2
2 2 p r „ r
M°
2 = Ÿ0r0r JVmaxB1- I r
r0M2FN2 2 p r „r
M°
2 = Ÿ0r0r Vmax
2J1- I rr0M2N2 2 p r „r
M°
2 = 2 p r Vmax2 Ÿ0
r0J1 - 2 I rr0M2 + I r
r0M4N r „ r
M°
2 = 2 p r Vmax2 Ÿ0
r0Jr- 2r3
r02 +
r5
r04 N „r
M°
2 = 2 p r Vmax2B r2
2- 2
r4
4 r02 +
r6
6 r04 F
0
r0
M°
2 = 2 p r Vmax2B r02
2- 2
r04
4 r02 +
r06
6 r04 F
M°
2 = 2 p r Vmax2B r02
2-r0
2
2+r0
2
6F
M°
2 =p r Vmax
2 r02
3
Similarly,
M°
in = M°
1 = Ÿm° 1V1 „m
°
M°
1 = Ÿ0r0r V1
2 2 p r „ r
M°
1 = r V12 2 p B r2
2F
0
r0
M°
1 = r V12 p r0
2
And plugging into to the equation for force above
F =1
gcIM° out -M
°inM+ HP1 - P2L A
F = p r02 :P1 - P2 -
r
gcJ Vmax
2
3- V1
2 N>
Conservation of mass can be used to find Vmax
m°
1 = m°
2
p r02 r V1 = r Ÿ0
r0Vmax :1 - I r
r0M2> 2 p r „r
V1 =2 Vmax
r02 Ÿ0
r0:r- r3
r02 > „ r
V1 =2 Vmax
r02 B r2
2-
r4
4 r02 F
0
r0
V1 =2 Vmax
r02
B r022-
r04
4 r02F
V1 =Vmax
2
Vmax = 2 V1
10 MAE 113 HW1 solution-2.nb
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And we arrive at
F = p r02 :P1 - P2 +
r
gcJ H2 V1L2
3- V1
2 N>F = p r0
2 :P1 - P2 +r V1
2
gcI 4
3- 1M>
F = p r02 :P1 - P2 +
r V12
gcI 4
3- 1M>
F = p r02 :P1 - P2 +
r V12
3 gc>
MAE 113 HW1 solution-2.nb 11
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2.6 Using Figure P2.5, with 1500 lbm/s of air at 60°F and 14.7 psia entering the engine at a velocity of 450 ft/s
and that 1250 lbm/s of bypass air leaves the engine at 60° to the horizontal, at a velocity of 890 ft/s and pres-
sure of 14.7 psia. The remaining 250 lbm/s leaves the engine core at a velocity of 1200 ft/s and pressure of 14.7
psia. Determine the force on the strut, Fx. Assume an ambient pressure of 14.7 psia.
By symmetry, there is no net force in the y-direction. The momentum equation in the x-direction is
⁄Fx = Fx = 1
gcIM° x core out +M
°x fan out -M
°x inM
M°x in = I1500
lbm
sM I450
ft
sM = 675 000
ftÿlbm
s2
M°x core out = I250
lbm
sM I1200
ft
sM = 300 000
ftÿlbm
s2
M°x fan out = I-1250
lbm
sM I890
ft
sM cosH60 °L = -556 250
ftÿlbm
s2
Therefore, the force is
Fx =1
32.174lbmÿft
lbf ÿs2
B300 000ftÿlbm
s2+ J-556 250
ftÿlbm
s2N- 675 000
ftÿlbm
s2F
Fx = -28, 945 lbf
This means a force of 28,945 lbf slowing the plane.
12 MAE 113 HW1 solution-2.nb
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