Question

. The mass flow rate through the nozzle of a rocket engine is 200 kg/s. The areas of the nozzle inlet and exit planes are 0.7 and 2.4 m2, respectively. On the nozzle inlet plane, the pressure and velocity are 1600 kPa and 150 m/s, respectively, whereas on the nozzle exit plane, the pressure and velocity are 80 kPa and 2300 m/s, respectively. Find the thrust force acting on the nozzle.

Answer 1

To solve this problem we will define the data obtained in each of the sections. We know that the Net Force is equivalent to the Force in section 1, which can be found through mass flow and velocity, plus the force in section two, which can be found as the product between pressure and Area, so therefore we have

State 1:

$\dot{m} = 200kg/s$

$v_1 = 150m/s$

$P_1 = 1600kPa$

State 2

$v_2 = 2300m/s$

$A_2 = 2.4m^2$

$P_2 = 80kPa$

We have that net force is equal to

$F_{net} = F_1 + F_2$

$F_1 = \dot{m} (v_2-v_1)$

$F_2 = (P_1-p_2)A_2$

Replacing,

$F = \dot{m} (v_2-v_1) + (P_1-p_2)A_2$

$F = 200(2300-150)+(1600-80)*10^3(2.4)$

$F = 4078kN$

Therefore  the thrust force acting on the nozzle is 4078kN