Question

Please answer part c this question has been posted previously was given the wrong answer

To understand how the linear momentum equation is derived from Reynolds transport theorem and to use the equation to calculate forces. The Reynolds transport theorem(DNDt)syst-aatJcvηρdVtfcsqpVdA relates the change in an extensive quantity N for a system of Lagrangian particles (the left side) to the changes in an intensive quantity η:nm, where m is the mass of the system, in a Eulerian control volume that initially contains the system. The conservation of nomentum equation can be developed using N-mV, which is the linear momentum of the system of particles. Part A - The effecl of he propellarnl flow The mnmentum eIation for the horizontal compnnent o farce is Σ r v ρο V+ s /=ρν d Usin the entire rocket engine E13 the vontrol volume, there are two open control surfcs for the prope lent inleta Express your answer In KN to three signiicant tigures For a set of particles, the momentum can change. The change in momentum is described by Newton's second law IF-ddt(mV). The corresponding intensive property is V. Therefore, the conservation of linear momentum equation is View Available Hint(s) 5.44 kN Part B The effects of pressure The momentum equation includes a sum of forces on the control volume on the left side. The net pressure on the surface of the contral volume must be included here What is the force on the control volume due to these pressures? Let forces acting to the right be positive The first term in this equation represents the change in momentum in the control volume with time. The second term in this equation represents the flux or rate of flow of momentum out of the control volume. This is a vector equation. One scalar equation can be written for each component of the velocity It is helpful to note from this equation that units of force also can be interpreted as units of rate of change in momentum. In the context of an engine, the force, or rate of change of momentum of the control volume, is called the thrust. It is a combination of the pressure forces and momentum flows at the inlets and exits. Express your answer in N to three significant figures. View Available Hint(s) 137 kN Consider the rocket engine (Figure 1) mounted horizontally on a test stand. Fuel and oxidizer flow into the rocket engine on the left side at m fuel 50 kg/s and m ox 125 kg/s. The exhaust flows out the right side at atmospheric pressure and Te = 1300 K . The nozzle diameter is de-60cm . The diameters of the inlets are dfuel 5 cm and dox - 6.7 cm,the gage pressure in the propellant lines is 25 MPa, and the densities are pfuel 820 kg/m3 and 1140 kg/m3. The flow of the propellants through the engine will generate a force on the engine to the left. This force is called the thrust of the rocket engine. The test stand must generate an equal but opposite force to hold the engine stationary. Assume that the propellant tanks are not mounted on the stand and that the flow at each open control surface is uniform. Part C The effect of the exhaust Up to this point, the only known term that has not been calculated is the force due to the momentum flux out of the control volume at the exhaust. Because this is a rocket engine, this flux should have the largest contribution. This increase in the momentum fiux is created by the energy extracted from the combustion of the fuel inside the control volume. What is the momentum flow out of the control volume due to the exhaust flow? The gas constant for the exhaust is Rx 356.8 ku.K Express your answer in kN to three significant figures. kN Fstand dfval 2 exhaust oxidizer → dox Fstand de throat

Answer 1

Exhaust density is given as;

101.30.218 kg/m RT 356.8 x 1300

Exhaust mass flow rate;

nn 12550 175 kg/s

We know;

$\dot{m} =\rho AV$

175-0.218(π/4)(0.62)

2839.2 m/s

Momentum flow out;

$F = \dot{m}V = 175(2839.2)$

+ F- 496860 N

F= 497 kN