A water rocket of length 20 cm and cross sectional area 100 cm 2 has half its vo

Question

A water rocket of length 20 cm and cross sectional area 100 cm 2 has half its volume initially filled with water. The air inside the rocket is pressurized to 10 atm and the area of the exhaust nozzle is 1 cm2

(A) What is the initial exhaust of the rocket?

(B)What is the initial volume flow rate for the rocket?

(C) What is the initial rate of decrease of mass of the rocket?

(D) What is the initial thrust on the rocket?

(E) What is the initial acceleration of the rocket neglecting the mass of the container?

Explanation / Answer

A)to find initial exhaust veloicty of the rocket, we have to use bernoulii principle.

as here height remains constant throughout,

bernoulli principle becomes :

p+0.5*pho*v^2=constant

where p=pressure
pho=density of water
v=speed

initial pressure=10 atm

final pressure=1 atm

pho=1000 kg/m^3

initial speed=0

final speed=v1(to be calculated)

hence 10+0.5*1000*0^2=1+0.5*1000*v1^2

==>9=500*v1^2

==>v1=0.1342 m/s

hence initial exhaust veloicty is 0.1342 m/s

part B:

initial volume flow rate of rocket=nozzle area*exhaust veloicty

=10^(-4) m^2 * 0.1342 m/s

=1.342*10^(-5) m^3/s

part C:

initial rate of decrease of mass of the rocket=volume flow rate*density of water

=1.342*10^(-5)*1000=0.01342 kg/sec

part D:

initial thrust on the rocket=rate of mass flow*exhaust veloicty

=0.01342*0.1342=1.8*10^(-3) kg.m/s

part E:

initial force on the rocket=difference in pressure*nozzle area

=(10-1) atm*10^(-4) m^2

=
(9*101325 N/m^2)*10^(-4) m^2

=91.1925 N

hence initial acceleration=force/total mass

=91.1925/(volume*density)

=91.1925/(0.2*100*10^(-4)*1000)

=45.596 m/s^2

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