Rocket Equation Derivation Notes (1)

8/10/2019 Rocket Equation Derivation Notes (1)

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Rocket Equation Derivation Notes and Practice Problems

In class we looked at deriving the rocket equation, which provides a relationship for the change in velocity of a rocket to parameters such as the exhaust velocity and mass

fraction. The sketch below shows a rocket accelerating in a perfectly vertical orientation.

V(t is the velocity of the rocket relative to

an inertial observer

Ve is the exit velocity of the propellantrelative to the rocket and taken to be a

constant

The rocket has a mass !(t

"eglecting all forces acting on the rocket and #e$#a, the momentum equation in the y-

direction can be written as shown below%

( ) &' =⋅+ ∫ ∫ dS nuudV udt

d y

S

y

V

ρ ρ

)r, written explicitly employing a moving control surface (which is fixed to the rocket%

( ) &' =⋅−+ ∫ ∫ dS nuuudV udt

d cs y

S

y

V

ρ ρ

*

The volume integral (the first term in either case is evaluated as%

( ) ( )[ ]dt

dV M V

dt

dM t V t M

dt

d dV u

dt

d y

V

+==∫ ρ +

To make sure that you understand this derivation completely%

. rite down the result of the momentum flux term using the form of the y-

momentum equation expressed in quation .*. rite down the result of the momentum flux term using the form of the y-

momentum equation expressed in quation .

+. /se the results of #art and #art * to develop the Ideal 0ocket quation.1. 2ompare the results of quation and quation * to the 0ocket 2ar example

(rocket launched hori3ontally developed in lecture.

V

x

y

Ve

!



The key to these problems is to pay strict detail to the sign convention and deal with the

outward-normal on the control surface properly.

4elow is the solution for the momentum flux term, which is the second term of the

quations and *. 5lso, for simplicity, the integral of the area is 5e and the density is ρe,

both of which are constant.

Part 1: Solution using Equation 1

/sing quation %

( ) &' =⋅∫ dS nuu y

S

ρ 6.

Taking the density and area out of the integral we have%

( ) &' =⋅

nuu A yee ρ 6.*

e have to be very careful in evaluating each of these terms, but remember that we are

interested in velocities and fluxes across a surface. The surface that we are interested in is

the exit plane of the rocket. This is the only place on the control volume where a velocity,mass or momentum flux is crossing a surface.

"ow look at each term%

• The first one is uy, which is the velocity of the exit propellant in the inertial frame

(ground-based observer. The velocity of that exit propellant can be determined as

follows in an inertial frame.

o hat is the velocity of the quantity that is crossing the control surface7 Itis Ve, but it is negative based on our sign convention (it is in the negativey direction.

o "ext, what is the velocity of the control volume (or the control surface7

Its velocity is V and it is positive since it is in the positive y-direction.

o The sum of these is the inertial velocity of the propellant that is crossing

the control surface% -Ve8V, or V-Ve. This is the velocity that an inertial

observer would measure for the exit propellant.

o "ote% There is no dot product on this term, it is done.

• The next term is ( )nu '⋅ , which is the velocity of the propellant crossing the

control surface in the relative frame. 5lso note that this is the dot product of two

vectors, which means that we need to pay careful attention to the direction of thenormal vectors to the control surface (which always point outward. e canevaluate this term as follows%

o hat is the velocity of the propellant7 It is 9Ve, since it is in the negative

y-direction. The normal vector is drawn pointing outward and it is also in

the negative y-direction.

o The dot product of 9Ve in the direction of n is Ve.

*



o "otice that when Ve is in the direction of n (regardless of the sign

convention the result will be a positive quantity (compare to the rocket

car in class. This is by convention. e define positive momentum flux as

out of the control volume and negative momentum flux as into the controlvolume. This is by convention and is the reason that we draw the normal

unit vectors outward.

e can write this out completely now%

( ) ( )( ) ( ) &(' =−⋅−−=⋅ eeee yee V V V Anuu A ρ ρ 6.+

6implifying, we have%

( ) &' * =−=−=⋅ eeeeeeeee yee V mV mV AVV Anuu A ρ ρ ρ 6.1

The momentum flux term in the y-direction (the only direction with a flux term is%

( ) eee yee V mV mnuu A −=⋅ ' ρ 6.:

Part 2: Solution using Equation 2:

The quantity we are interested in evaluating can be written in the y-direction as%

( ) dS nuuu cs y

S

'⋅−∫ ρ 6.;

e have to be very careful in evaluating each of these terms, but remember that we areinterested in velocities and fluxes across a surface. The surface that we are interested in is

the exit plane of the rocket. This is the only place on the control volume where a velocity,

mass or momentum flux is crossing a surface.

"ow look at each term (same as above, this term never changes between the two forms%

• The first one is uy, which is the velocity of the exit propellant in the inertial frame

(ground-based observer. The velocity of that exit propellant can be determined as

follows in an inertial frame.

o hat is the velocity of the quantity that is crossing the control surface7 It

is Ve, but it is negative based on our sign convention (it is in the negative

y direction.

o "ext, what is the velocity of the control volume (or the control surface7

Its velocity is V and it is positive since it is in the positive y-direction.

o The sum of these is the inertial velocity of the propellant that is crossing

the control surface% -Ve8V, or V-Ve. This is the velocity that an inertialobserver would measure for the exit propellant.

+



o "ote% There is no dot product on this term, it is done. 5lso notice that it is

identical to the operations performed in #art .

• The next term is ( ) nuucs

'⋅− . This entire quantity is the velocity of the propellant

crossing the control surface in the relative frame. ithin this quantity the first uis inertial and the second is the velocity of the control surface. <ets look at it piece

by piece%o The first term, u , is the same as above. It is the velocity of the propellant

across the control surface as observed from an inertial frame, V-Ve

o The second term is the velocity of the control surface. The control surface

is moving in the positive y-direction, so csu is equation to V.

o "ow we need to take this sum =(V-Ve-V>, and dot it with the normal

vector. The sum is equal to 9Ve, and the normal vector is in the negative y-

direction. 6o we have ( ) ( ) ee V V =−⋅− (

e can write this out completely now%

( ) ( ) ( ) ( )( )[ ] ( )(' −⋅−−−=⋅− V V V V V Anuuu A eeeecs yee ρ ρ 6.?

6implifying, we have%

( ) ( )[ ] ( ) eeeeeeeeeeecs yee V V AVV AV V V Anuuu A ρ ρ ρ ρ −=−⋅−−=⋅− (' 6.@

The momentum flux term in the y-direction (the only direction with a flux term is%

( ) eeeeeeeeeecs yee

V mV mV V AVV Anuuu A −=−=⋅− ρ ρ ρ ' 6.A

2omparing quations 6.: and 6.A, we see that the result of using either form of themomentum equation will result in the same answer.

Part 3: Application o t!is result to t!e development o t!e idea rocket equation

These momentum flux terms can be combined with the part that I gave you in quation +

to derive the rocket equation. 2ombining these results%

&=−++ eee V mV m

dt

dV M V

dt

dM

6.&

e can write the term d!Bdt as%

emdt

dm

dt

dM −=

−= 6.

6ubstituting this into quation 6.&%

1



&=−++− eeee V mV mdt

dV M V m

6.*

e can simplify and show that%

eeV mdt

dV M =

6.+

mploying quation 6. and solving for the incremental velocity change dV%

M

dM V dt

M

V mdV e

ee −== 6.1

e can now integrate quation 6.1, from initial to final conditions, and making the

assumption that Ve is constant.

∫ ∫ −= f

i

f

i

M

M

e

V

V M

dM V dV

6.:

Integration gives%

=

−=−

f

i

e

i

f

ei f M

M V

M

M V V V lnln

6.;

This is quation &. shown in 6ection &.+ of the text book.

Part ": #o$ does t!is result compare to t!e rocket car s!o$n in lecture%

Cirst note, that this is exactly the same problem. The rocket car is traveling hori3ontally

and we are neglecting drag and assuming #e$#a. The rocket shown on page is travelingvertically with #e$#a, and we are neglecting drag as well as gravity.

The difference between the rocket in vertical flight and the rocket car is the sign

convention. "ote that for the rocket car, the car is traveling in the negative x-directionand the propellant is being exhausted in the positive x-direction. "otice also that, as

always, all the normal unit vectors to the control surface point outward.

<etDs look at the derivation we did in class again, and compare it term by term to the

derivation done here and see where the sign convention and normal vectors play

important roles.

The volume integral of the momentum equation, shown in quation + is evaluated as%

:



( ) ( )[ ]dt

dV M V

dt

dM t V t M

dt

d dV u

dt

d y

V

+==∫ ρ +

Cor the 0ocket 2ar, it is%

( ) ( )[ ][ ]dt dV M V

dt dM t V t M

dt d dV u

dt d

y

V

−−=−=∫ ρ +.02

This is because the rocket car is moving in the negative x-direction and hence its velocity

is in the negative x-direction. The structure of the terms is the same, the direction haschanged.

"ow look at the momentum flux terms%

Cor the vertically launched rocket, we showed using quation , that%

( ) eee yee V mV mnuu A −=⋅ ' ρ 6.:

Cor the rocket car, this equation would be modified to read%

( ) ( ) ( ) ( )[ ] eeeeeee xee

V mV mV V V Anuu A +−=⋅−=⋅ (' ρ ρ 6.: 02

It is almost the same as 6.:, except the signs are flipped, because the sign convention was

flipped. e can do the same thing using quation *, and the result that was shown for the

vertically launched rocket, from quation 6.? was%

( ) ( )[ ] ( ) eeeeeeeeeeecs yee V V AVV AV V V Anuuu A ρ ρ ρ ρ −=−⋅−−=⋅− ('

6.@

Cor the rocket car, we have%

( ) ( ) ( ) ( )[ ] ( ) eeeeeeeeeecs xee

V V AVV AV V V V V Anuuu A ρ ρ ρ ρ +−=⋅−−−−=⋅− (' 6.@ 60

"otice that Eust the signs switched, but notice that the sign of the momentum flux term

relative to the rocket or the rocket car didnDt change. This again is because of convention.henever there is an outward momentum flux it will yield a positive value.

If we proceed to combine these terms, we can derive the rocket equation for the car.

2ombining terms we have%

&=+−−− eee V mV mdt

dV M V

dt

dM

6.?

5gain employ quation 6., to yield%

;



&=+−− eeee V mV m

dt

dV M V m

6.@

hich can be simplified to%

eeV mdt dV M = 6.A

This is exactly the same result which was obtained in the vertical rocket case, shown in

6.+. That means that the same rocket equation will result when quation 6.A isintegrated.

4ut waitF If the coordinate system has switched signs, shouldnDt the resulting direction of

V change in the final result of the rocket equation7

The answer is no. This equation is predicting velocity in the direction you drew the

velocity vectors in. Cor example, in the rocket car, the velocity vector of the car wasdrawn in the 9x-direction. The result for a positive Ve will be a positive V from the

equation, but you have to look at your velocity vector to determine which direction the

car is going, which is always in the opposite direction of Ve. The same is true of thevertically launched rocket.

?

Rocket Equation Derivation Notes (1)

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