EXTROVERTSpace Propulsion 02 1 Thrust, Rocket Equation, Specific Impulse, Mass Ratio

EXTROVERT Space Propulsion 02

1

Thrust, Rocket Equation, Specific Impulse, Mass Ratio


2

Thrust comes from:

a) Increase in momentum of the propellant fluid (momentum thrust)

b) Pressure at the exit plane being higher than the outside pressure (pressure thrust).

Where does the thrust act?

In the rocket engine, the force is felt on the nozzle and the combustor walls, and is transmitted through the engine mountings to the rest of the vehicle.

Effective Exhaust Velocity

( )ee e e a

Ac U p p

m= + −

&

Thrust( )f e e a eT m U P P A= + −&


3

Consider a rocket with effective exhaust velocity ce. As propellant is blasted out the exhaust nozzle, the mass of the vehicle decreases. This is substantial in the case of the rocket as compared to air-breathing engines, because all the propellant comes from inside the vehicle. From Newton's Second Law,

where M1 is the initial mass, which includes the propellant, and M2 is the mass after the propellant has been used up to achieve the velocity increment DV.

edM

dV cM

=−2

1

M

eM

dMV c

MΔ =− ∫

1

2e e

Mv c Log

M

⎛ ⎞Δ = ⎜ ⎟

⎝ ⎠

Delta V and Mass Ratio


4

Specific Impulse of the system is

where g is the standard value of acceleration due to gravity at sea-level (9.8m/s2). Note that the unit of Specific Impulse is seconds. Using this definition,

Mass Ratio of a rocket is

€

M1

M2

= eΔV

gIsp

Specific Impulse and Mass Ratio

Note: Some organizationsexpress Specific Impulse without dividing by g

€

ΔV = gIsp logM1

M2

⎛

⎝ ⎜

⎞

⎠ ⎟€

Isp =ce

g


5

For missions from Earth's surface to escape from earth's gravitational field, Mass Ratio is large.

For specific impulse of 390 s, g = 9.8 m/s2, and DV = 11,186 m/s (36700 fps), the mass ratio is 18.67.

This means that the rocket at launch time must be at least 18.67 times as big as the spacecraft which is left after all the fuel is burned. To get a high specific impulse like 390 s, we have to use a costly system like liquid hydrogen - liquid oxygen.

For earth orbit, the velocity increment DV needed is 25,000 fps, while 36,700fps will enable escape from Earth's gravitational field.

Example:


6

To find the velocity increment required for various missions, we must calculate trajectories and orbits. This is done using Newton's Law of Gravitation:

Here the lhs is the "radial force" of attraction due to gravitation, between two bodies; the big one of mass M, and the little one of mass m.

The universal gravitational constant G is 6.670 * 10-11 Nm2/kg2.

Newton's Law of Gravitation


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Rocket Equation Including Drag and Gravity:

edM D

du c dt gCos dtM M

θ=− − −

Ref: Hill & Peterson, Chapter 10.

.

The rate of acceleration of the vehicle is

Neglecting the air drag and gravity terms, we get the Ideal Rocket Equation

edM

du cM

=−


8

Drag Term in the Rocket Equation20.5 f DD U A Cρ=

With density in kg/m3 and h in meters, a = 1.2 and b = 2.9 x 10-5

Roughly, density at 30,000 meters is about 1% of its sea-level value.

1.15( ) ( )h aExp bhρ = −

where atmospheric density above Earth varies roughly as

Drag Coefficients (typical)

Note: drag coefficient peak is reached at around Mach 1.2.

Inclination, deg. to flt. direction, CD low-speed CD peak @1.2 CD @ Mach 2

0 0.06 0.15 0.13

4 0.185 0.16

8 0.23 0.2

mailto:CD@Mach


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Gravity Term

2

ee

e

Rg g

R h

⎛ ⎞= ⎜ ⎟+⎝ ⎠

At 100 miles above the surface the change from the surface is still only about 5%.


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Specific impulse of 390 s, g0 = 9.8 m/s2,

and DV = 11186 m/s (36700 fps),

Mass ratio is 18.67.

This means that the rocket at launch time must be at least 18.67 times as big as the spacecraft which is left after all the fuel is burned. To get a high specific impulse like 390 s, we have to use a costly system like liquid hydrogen - liquid oxygen.

Velocity increment DV for Low Earth Orbit: ~ 25,000 fps,

Escape from Earth's gravitational field ~ 36,700fps

Example


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Single Stage Sounding RocketAltitude at burnout, assuming it goes straight up:

0

tb

bh Udt=∫

Neglecting drag

0lne e

MU c g t

M= −

If is constant,em&

( ) ( )b

b t

tMMMtM −−= 00

(Sounding rocket: not quite single-stage)NASA Goddard Space Flight centerhttp://www.gsfc.nasa.gov


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Single-Stage Sounding Rocket Going Straight Up (cont’d)

1ln 1 1e e

b

tU c g t

R t

⎡ ⎤⎛ ⎞=− − − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

2ln 1

1̀ 2b e b e b e bR

h c t c t g tR

=− + −−

Define Mass Ratio

€

R ≡M0

Mb

Substituting,


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Equating kinetic energy at burnout with change in potential energy of the final mass bM

( )bebb

b hhgMU

M −= max

2

2

e

bb g

Uhh

2

2

max +=

( )22

maxln

ln 12 1

ee b

e

c R Rh c t R

g R⎛ ⎞= − −⎜ ⎟−⎝ ⎠

Expression for Maximum Altitude ReachedNote that at burnout, the sounding rocket is still moving fast upward.


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( )22

maxln

ln 12 1

ee b

e

c R Rh c t R

g R⎛ ⎞= − −⎜ ⎟−⎝ ⎠

Example

Values given: Ce = 3000 m/sR = 10tb = 30s

Hmax = ??


15

Chemical Rockets

Burn time of existing rockets is ~ 30 to 200 seconds.

spL MMMM ++=0

Payload ratio

sLb MMM +=

€

R ≡M0

Mb

≈M0

ML + Ms

sp

L

L

L

MM

M

MM

M

+=

−=

0λ

Structure coefficient L

Lb

sp

s

MM

MM

MM

M

−−≈

+=

0ε

Thus, λελ

++=1

R

Definitions

LM

sM

pM

Payload Mass

Structure (incl. engine) Mass

Propellant Mass


16

Multistaging -1Total initial mass of i-th stage prior to firing, include its effective payload.

Total mass of i-th stage after burnout, include its effective payload.

Payload of last stage.

Structural mass of i-th stage; include engine controls, instruments.

iM0

biM

LM

siM


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Payload of stage is mass of all subsequent stages.

thi

( )100 +−

=ii

isi MM

Mε

thi

( )

( )100

10

+

+

−

−=

ii

iibi MM

MMε

thi

Structural coefficient of stage:

If stage contains no propellant at burnout:

Multistaging - 2


18

Mass ratio of stagethi

ib

ii M

MR

0=

i.e.,

ii

iiR λε

λ++= 1

iiei RCU ln=Δ

∑==

n

iiien RCU

1ln

.

λεSimilar stages: same and

⎟⎠⎞

⎜⎝⎛

++=λελ1

lnen nCU

Multistaging - 3


19

1

1

02

01 1

λλ+=

MM

2

2

03

02 1

λλ+=

MM

n

n

L

n

M

M

λλ+=10 ∏ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=⇒=

n

i i

i

LM

M

1

01 1

λ

λ

iλIf are equal,

n

LM

M⎟⎠⎞

⎜⎝⎛ +

=λλ101

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

11

ln1

01

1

01

n

i

n

i

e

n

M

M

M

M

nC

U

ε

Multistaging - 4


20

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛+−−=

Δ−

eC

U

p

kengineL eM

M

M

M

M

M111 tan

00

Structural coefficient

0

tan

0

tan

tan

tan

11

1

M

M

M

Me

M

Me

M

M

MMM

MM

engine

p

keCU

engineeCU

p

k

enginekp

enginek

+⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

+++

=

Δ−

Δ−

ε

Multistaging - 5


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Engine J2 H1

Thrust, kN 1023 1023

Fuel Hydrogen Hydrocarbon

Engine Mass, Millions of grams 1.622 0.921

Engine Mass Fraction 0.024 0.014

Tank Mass/ Propellant Mass 0.046 0.016

Eq. Exhaust Vel. m/s 4175 2891

Specific Impulse, seconds 426 295

Apollo engines (Source: Hill Peterson, page 479)


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Stage 1 2 3

Engine F-1 J-2 J-2

Fuel RP1 Hydrocarbon LH2 LH2

Number of engines 5 5 1

Total thrust, Newton 33 Million 4.45 Million 0.89 Million

Total Initial Mass, Kg

2.780 Million 0.677 Million 0.215 Million

Propellant mass, kg 1.997 Million 0.429 Million 0.109 Million

Structure & engine, kg

0.106 Million 0.0326 Million 0.0257 Million

Structure mass fraction

0.05 0.071 0.191

Payload fraction 0.321 0.466 0.603

Saturn V Apollo 11 Flight Configuration


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The idea of “Thrust Coefficient”

where F is the thrust, At is the nozzle throat area and P0 is the combustion chamber stagnation pressure. Basically a higher thrust coefficient means a better usage of the available stagnation pressure in converting to thrust.

The thrust coefficient has values ranging from 0.8 to 1.9.

Note also that a plot of thrust coefficient vs. altitude for a given nozzle will give the variation of thrust with altitude for a given chamber pressure and nozzle throat area.

The thrust coefficient is also used to compare different nozzle designs for given constraints.

In the following we will use gas dynamics to derive expressions for the thrust coefficient in termsof gas properties.

€

CF ≡F

P0At

⎛

⎝ ⎜

⎞

⎠ ⎟


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Thrust Coefficient - 1

where At is nozzle throat area and p0 is chamber pressure (N/m2)

t tu RTγ=Thus,

For sonic conditions at the throat,

1/( 1)

02

1t

γ

ρ ργ

−⎛ ⎞

= ⎜ ⎟+⎝ ⎠and

( ) ( ) 1/ 21 /1/ 1

00 0

0

2 21 ( )

1 1t e

t e eTT p

F A p R p p Ap

γ γγ

γγ γ

−−⎧ ⎫⎡ ⎤⎛ ⎞⎛ ⎞⎪ ⎪⎢ ⎥= − + −⎨ ⎬⎜ ⎟⎜ ⎟− + ⎢ ⎥⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

€

CF ≡F

P0At

⎛

⎝ ⎜

⎞

⎠ ⎟


25

Using isentropic flow relations, 1

10

02

1tt

pTT

R

γγ

ρ γ

⎛ ⎞+⎜ ⎟−⎝ ⎠⎛ ⎞⎛ ⎞

=⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠21

2 20

1 2 1

1t

γ

γρ ρ+⎛ ⎞=⎜ ⎟+⎝ ⎠

1/ 2111

00

2 21 ( )

1 1e

t e a ep

F A p p p Ap

γγγγ

γγ γ

−+−

⎧ ⎫⎡ ⎤⎛ ⎞⎛ ⎞⎪ ⎪= − + −⎨ ⎬⎢ ⎥⎜ ⎟⎜ ⎟− +⎝ ⎠ ⎝ ⎠⎣ ⎦⎪ ⎪

⎩ ⎭and Thrust Coefficient

1/ 2112 1

0 0

2 2 ( )1

1 1e e a e

Ft

p p p AC

p p A

γγγγγ

γ γ

−+−

⎧ ⎫⎡ ⎤⎛ ⎞⎛ ⎞ −⎪ ⎪= − +⎨ ⎬⎢ ⎥⎜ ⎟⎜ ⎟− +⎝ ⎠ ⎝ ⎠⎣ ⎦⎪ ⎪

⎩ ⎭Depends entirely on nozzle characteristics. The thrust coefficient is used to evaluate nozzle performance.

Thrust Coefficient - 2


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Characteristic Exhaust Velocity c*Used to characterize the performance of propellants and combustion chambers independent of the nozzle characteristics.

( )1

2 10 0

0 0

2

1t tp A p A

mRT RT

γ

γγ

γγ γ

+

−⎧ ⎫⎛ ⎞⎪ ⎪

= = Γ⎨ ⎬⎜ ⎟+⎝ ⎠⎪ ⎪⎩ ⎭

&

where is the quantity in brackets. Note: Γ 0 0a RTγ=

So 0

0

tp Am

a= Γ&

Characteristic exhaust velocity 0* tp A

cm

=&

Assuming steady, quasi-1-dimensional, perfect gas. The condition for maximum thrust is ideal expansion: nozzle exit static pressure being equal to the outside pressure. In other words,

e ap p=


27

We’ll end Lecture 2 here, and go on to discuss orbits before getting back to compressibleFlow and chemistry considerations. The purposes are:

1. To enable mission calculations. 2. To give everyone a chance to look at the content so far and see how much they

need review of compressible flow material. Please browse the web links in the first lecture.

End of Section 2

Documents

EXTROVERTSpace Propulsion 02 1 Thrust, Rocket Equation, Specific Impulse, Mass Ratio