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


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


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



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



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



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



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



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



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



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>



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.



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MAE113 HW1 Solution