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 cm^2. (a) What is the initial exhaust velocity 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 consider)?

Explanation / Answer

the rocket can be assumed to be vertical for the cvalculations, with pressure inside the rocket for air, P = 10 atm = 10*1.01*10^5 Pa
so, pressure difference at the nozzle = 10atm + rho*g*h - 1 atm [ rho is density of water, h is the height of the water in the bottle]
so, dP = 9 atm + 1000*9.81*0.1 = 9*1.01*10^5 + 981 = 909,981 Pa

now if exit flow velocity is v, from bernoullis theorem
dP = 0.5rho*v^2
909,981 = 0.5*1000*v^2
v = 42.661 m/s

a. so initial exhaust velocity, v = 42.661 m/s
b. initial volume flow rate = Av = 100*10^-4*41.661 = 0.4266 m^3/s
c. mass flow rate = rho*A*v = 426.66 kg/s = initial rate of mass decrease of rocket
d. initial thrust = rate of change of momentum = mass flow rate * vexhaust velociuty = 426.66*42.661 = 18199.614 N
e. initial acceleration = thrust/mass of rocekt = 18199.614/rho*V = 18199.614/1000*(100*10^-4)*0.1 = 18199.614 m/s/s

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