MAE 113, SS1 2009, HW #3 - University of California, San Diego

[Pages:15]MAE 113, SS1 2009, HW #3

2.35 An ideal ramjet is operated at 50,000 ft and Mach 3. The diffuser and nozzle are assembled to be isentropic, and the combustion is to be modeled as an ideal heat interaction at constant Mach number with constant total pressure. The cross-sectional area and Mach number for certain engine stations are given. The total temperature leaving the combustor Tr4 is 4000?R.

a) Determine the mass flow rate of air through the engine (lbm/s)

We do the usual

m? =

r1

A1 V1

=

P1 A1 R T1

M1

g R gc T1 = P1 A1 M1

g gc R T1

T1 = q Tstd = 0.7519 ? 518.7 ?R = 390.0 ?R

P1

=

d Pstd

= 0.1151 ? 2116.2

lb ft2

=

243.57

lb ft2

= 1.6914 psia

m?

=

243.57

lbf ft2

I4.235

ft2M H3L

1.4 ? 32.174 lbm ft

lbf s2

53.35 ft lbf H390.0 ?RL

lbm ?R

m? = 143.98 lbm ? s

b) Complete the table with flow areas, static pressures, static temperatures, and velocities.

Start by determining total temperature and pressure

Tt1

=

T1J1

+

g-1 2

M12N

=

390.0

?RI1

+

1.4-1 2

32M

=

1092.0

?R

g

1.4

Pt1

= P1J1 +

g-1 2

M12N g-1

= 243.57

lb ft2

I1 +

1.4-1 2

3 M2 1.4-1

= 8946.99

lb ft2

= 62.13 psia

The

total

pressure

will

remain

8,

946.99

lb ft2

for all stations, and the total temperature will be 1092.0?R for

stations 1, 2, and 3, and 4000?R for stations 4, 5, and 6.

We can find velocity by

V1 = M1

g R gc T1 = 3

1.4 I53.35

ft lbf M J32.174

lbm ?R

lbm ft lbf s2

N

390.0

?R

= 2904.27 ft ? s

So we've filled in station 1 of the table

Station 1

Area ft2 4.235

Mach

3

P psia 1.691

T ?R 390.

fps in V 2904

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2 MAE 113 HW3 solution.nb

We can use the isentropic tables to find A*. For M=3, A ? A* = 4.235. Since A=4.235, A* = 1. Since M=1 at station 2, A2 = 1 ft2. Also from the tables, we get

Tt2 T2

= 1.2 =

Tt1 T2

T2

=

Tt1 1.2

=

1092.0 ?R 1.2

= 910 ?R

Pt2 = 1.893 = Pt1

P2

P2

P2

=

62.13 psia 1.893

= 32.82 psia

And, for velocity

V2 = M2

g R gc T2 = 1

1.4 I53.35

ft lbf M J32.174

lbm ?R

lbm ft lbf s2

N

910

?R

= 1478.78 ft ? s

And the table is now

Station 1

2

Area ft2 4.235 1

Mach

3

1

P psia 1.691 32.82

T ?R 390. 910.

fps in V 2904 1479

Again using the isentropic tables for M=0.15:

A = 3.9103

A*

Tt3 = 1.0045 = Tt1

T3

T3

T3

=

Tt1 1.0045

=

1092.0 ?R 1.0045

= 1087.1 ?R

Pt3 = 1.0115 = Pt1

P3

P3

P3

=

62.13 psia 1.0115

= 61.16 psia

V3 = M3

g R gc T3 = .15

1.4 I53.35

ft lbf M J32.174

lbm ?R

lbm ft lbf s2

N

1087.1

?R

= 242.44 ft ? s

Station 1

2

3

Area ft2 4.235 1 3.9103

Mach

3

1 0.15

P psia 1.691 32.82 61.16

T ?R 390. 910. 1087.1

fps in V 2904 1479 242.44

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MAE 113 HW3 solution.nb 3

In going from 3 to 4, the total pressure stays the same but the total temperature increases. The pressure at 3 will be the same as the pressure at 4 since the Mach number and total pressure are the same.

P4 = 61.16 psia

Tt4 = 1.0045

T4

T4

=

Tt4 1.0045

=

4000 ?R 1.0045

=

3982.1 ?R

V4 = M4

g R gc T4 = .15

1.4 I53.35

ft lbf M J32.174

lbm ?R

lbm ft lbf s2

N

3982.1

?R

= 464.01 ft ? s

Now, we can return to mass flow rate

m? = P4 A4 M4

g gc R T4

A4

=

m? P4 M4

R T4 g gc

=

143.98 lbm?s 61.16 psia J 144 in2 N H0.15L

ft2

53.35 ft lbf H3982.1 ?RL

lbm ?R

1.4 J32.174 lbm ft N

lbf s2

= 7.485 ft2

Station 1

2

3

4

Area ft2 4.235 1 3.9103 7.485

Mach

3

1 0.15 0.15

P psia 1.691 32.82 61.16 61.16

T ?R 390. 910. 1087.1 3982.1

fps in V 2904 1479 242.44 464.01

Chugging on, we note that from station 4, we could find the new A*

A A*

=

3.9103

A*

=

A5

=

A4 3.9103

=

7.485 ft2 3.9103

= 1.914 ft2

Then the usual

Tt5 = 1.2 = Tt4

T5

T5

T5

=

Tt4 1.2

=

4000 ?R 1.2

=

3333.3 ?R

Pt5 = 1.893 = Pt1

P5

P5

P5

=

62.13 psia 1.893

= 32.82 psia

V5 = M5

g R gc T5 = 1

1.4 I53.35

ft lbf M J32.174

lbm ?R

lbm ft lbf s2

N

3333.3

?R

= 2830.24 ft ? s

Station 1

2

3

4

5

Area ft2 4.235 1 3.9103 7.485 1.914

Mach

3

1 0.15 0.15

1

P psia 1.691 32.82 61.16 61.16 32.82

T ?R 390. 910. 1087.1 3982.1 3333.3

fps in V 2904 1479 242.44 464.01 2830.24

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4 MAE 113 HW3 solution.nb

And, lastly

A A*

=

4.235

A6 = 4.235 A* = 4.235 ? 1.914 ft2 = 8.106 ft2

Tt6 T6

= 2.8 =

Tt4 T6

T6

=

Tt4 2.8

=

4000 ?R 2.8

=

1428.57 ?R

Pt6 = 36.73 = Pt1

P6

P6

P6

=

62.13 psia 36.73

= 1.691 psia

V6 = M6

g R gc T6 = 3

1.4 I53.35

ft lbf M J32.174

lbm ?R

lbm ft lbf s2

N

1428.57

?R

= 5558.48 ft ? s

Such that, finally,

Station 1

2

3

4

5

6

Area ft2 4.235 1 3.9103 7.485 1.914 8.106

Mach

3

1 0.15 0.15

1

3

P psia 1.691 32.82 61.16 61.16 32.82 1.691

T ?R 390. 910. 1087.1 3982.1 3333.3 1428.6

fps in V 2904 1479 242.44 464.01 2830.24 5558.48

c) Find the thrust (magnitude and direction) of the diffuser, combustor, and nozzle.

The forces are

Fdiffuser

=

m? gc

HV3 - V1L + HP3 - P1L A3

Fdiffuser

=

143.98 lbm?s 32.174 lbm ft

H242.44 ft ? s - 2904 ft ? sL + H61.16 psia - 1.691 psiaL

144 in2 ft2

3.9103 ft2

lbf s2

Fdiffuser = 21 575.4 lbf

Fcombustor

=

m? gc

HV4 - V3L + HP4 - P1L A4 - HP3 - P1L A3

Fcombustor

=

143.98 lbm?s 32.174 lbm ft

H464.01 ft ? s - 242.44 ft ? sL +

lbf s2

H61.16

psia - 1.691

psiaL

144 in2 ft2

7.485

ft2

- H61.16

psia - 1.691

psiaL

144 in2 ft2

3.9103

ft2

Fcombustor = 31 603.6 lbf

Fnozzle =

m? gc

HV6 - V4L - HP4 - P1L A4

Fnozzle

=

143.98 lbm?s 32.174 lbm ft

H5558.48 ft ? s - 464.01 ft ? sL - H61.16 psia - 1.691 psiaL

144 in2 ft2

7.485 ft2

lbf s2

Fnozzle = -41 300.1 lbf

The nozzle therefore has a thrust to the right (a drag)

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MAE 113 HW3 solution.nb 5

d) Find the thrust of the ramjet This is done by summing all the components

Framjet = Fdiffuser + Fcombustor + Fnozzle Framjet = 21 575.4 lbf + 31 603.6 lbf + - 41 300.1 lbf

Framjet = 11 878.9 lbf

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6 MAE 113 HW3 solution.nb

2.38 Air at 20 kPa, 260K, and Mach 3 passes through a normal shock. Determine

a) Total temperature and pressure upstream of shock.

Start with the relation from class

T01 T1

= 1+

g-1 2

M2

T01

=

T1J1

+

g-1 2

M 2N

=

260

KI1 +

1.4-1 2

32M

T01 = 728 K

Now use the pressure relation presented in class

g

= I M P1

T1 Ig-1M

P01

T01

g

1.4

P01

=

P I M T01 Ig-1M

1 T1

=

20 kPaI 728 K M 1.4-1

260 K

P01 = 734.65 kPa

b) Total temperature and pressure downstream of shock.

Since the flow through the shock is adiabatic, the total temperature is a constant

T02 = T01 = 728 K

Thre pressure downstream of the shock can be found using

g

1.4

p02 p01

= BJ g+1 2

M12

N?J1+

g-1 2

M12NF g-1

1

J 2g

g+1

M1

2

-

g-1 g+1

N

g-1

BJ 1.4+1 32N?J1+ 1.4-1 32NF 1.4-1

=2

2

1

J 2 H1.4L 32- 1.4-1 N 1.4-1

1.4+1

1.4+1

= 0.3283

p02 = 0.3283 p01 = 0.3283 ? 734.65 kPa

P02 = 241.22 kPa

c) Static temperature and pressure downstream of the shock.

Start by finding downstream Mach number.

Back to and again,

M2 =

M12+

2 g-1

2g g-1

M12-1

=

32+ 2

1.4-1

2 H1.4L 1.4-1

32-1

= 0.4752

T02 = 1 + g-1 M 2 = 1 + 1.4-1 H0.4752L2 = 1.0452

T2

2

2

T2

=

T02 1.0452

=

728 K 1.0452

T2 = 696.5 K

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g

= I M P2

T2 Ig-1M

P02

T02

g

1.4

P2

=

P I M T2 Ig-1M

02 T02

=

241.22 kPaI 696.5 K M 1.4-1

728 K

P2 = 206.67 kPa

MAE 113 HW3 solution.nb 7

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8 MAE 113 HW3 solution.nb

3.11 A rocket nozzle has an ideal thrust coefficient CFi of 1.5, a chamber pressure Pc of 100 atm, and a throat area At of 0.15m2. Determine the ideal thrust Fi.

Use equation 3.43

CFi

=

Fi Pc At

Fi

=

CFi

Pc

At

=

H1.5L I100

atm

101 325 atm

Pa

M

I0.15

m2M

Fi = 2, 279, 812.5

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