A liquid-fueled rocket is in steady operation on a test stand at sea level as shown in Figure 7.18A. What is the force acting on the support pylon at the rocket attachment point? What is the thrust produced by the rocket? Assume the velocity, pressure, and temperature of the exhaust

Question

A liquid-fueled rocket is in steady operation on a test stand at sea level as shown in Figure 7.18A. What is the force acting on the support pylon at the rocket attachment point? What is the thrust produced by the rocket? Assume the velocity, pressure, and temperature of the exhaust stream are uniform and known at the nozzle exit plane and that the combustion gas obeys the perfect gas law. Neglect the effects of the ﬂows of fuel and oxidizer.

Accepted Answer

Figure 7.18A shows the physical arrangement, and Figure 7.18B is an idealized representation of the pressure and shear stress acting inside the combustion chamber and nozzle of the rocket. Both questions relate to the forces acting on the rocket, so we will construct a force balance on the rocket as shown in Figure 7.18C. The relevant forces include the force applied by the high pressure combustion gas to the interior surfaces of the engine, the force of atmospheric air on the outside, the force of gravity, and the external force F_E applied to the rocket by the support pylon. Thus we have F_gas + F_air+ F_gravity + F_E = 0, and the external force acting on the rocket is F_E = −F_gas − F_air − F_gravity. However, in this example we are asked to ﬁnd the force F_pylon acting on the support pylon. Since this force is given by F_pylon = −F_E, the force balance can be written as

F_pylon = F_gas + F_air+ F_gravity (A)

To ﬁnd this force, we must ﬁnd the forces applied to the rocket by the combustion gas, the air, and gravity. The force of gravity on the rocket is F_gravity = −W_rocketk. The force of the air is found by using the trick involving the negative of the integral of atmospheric pressure over the open surface of the engine. In this case, the only open surface is the exit plane of the nozzle, so assuming that the nozzle wall is thin we have F_air = − [latex]\int_{exit}^{}{} [/latex] −p_An dS = p_AA_Ej. The force acting on the pylon is now

F_pylon = F_gas + p_AA_Ej − W_rocketk (B)

The force of the gas on the rocket is deﬁned by the integral of the stress vector over the interior surfaces of the engine, [latex]F_{gas}=\int_{engine}^{}{\sum{dS}}=\int_{engine}^{}{(−pn+\pmb{ au} )dS}[/latex]. Since we are not given any information about the pressure or shear stress distributions inside the engine, this integral cannot be evaluated directly. Instead we will use a decal surface and a momentum balance to ﬁnd this force.

To evaluate the force of the combustion gas on the interior surface of the engine, we will deﬁne a CV with a decal surface adjacent to the interior wall of the engine as shown in Figure 7.18D. Applying a steady ﬂow momentum balance to this CV, we have

[latex]\int_{CS}^{}{ (ρu)(u • n)dS}=\int_{CV}^{}{ ho fdV}+\int_{exit}^{}{\sum{dS}}+\int_{decal}^{}{\sum{dS}}[/latex]

The terms on the exit port are u = V_Ej, n = j, and u • n = V_E. Thus,

[latex]\int_{exit}^{}{ (ρu)(u • n)dS}=(ρV_Ej)(V_E)A_E = ρV^2 _E A_Ej [/latex]

The body force integral is simply the weight of the gas in the CV or −W_CVk. The stress on the exit port is due to the pressure, so we ﬁnd that [latex]\int_{exit}^{}{\sum{dS}}=\int_{exit}[/latex]− p_En dS = −p_EA_Ej. The integral over the decal surface is the reaction force. By the principle of action–reaction, we know that R = −F_gas; thus, [latex]\int_{decal}\sum{dS}[/latex] = R = −F_gas. The completed momentum balance is then

[latex]ρ_EV^2 _E A_Ej=−W_{CV} k− p_E A_Ej−F_{gas}[/latex] (C)

We now have two equations, (B) and (C), to determine F_pylon. Solving for this force, we ﬁnd

[latex]F_{pylon}=−\left[ρ_EV^2 _E A_E +(p_E − p_A)A_E ight]j−(W_{CV} +W_{rocket})k [/latex] (D)

This is our answer. Since we know the pressure and temperature at the exit, we can use the perfect gas law and the speciﬁc gas constant of the combustion products to ﬁnd the gas density. The exit velocity is typically very large (supersonic) for a rocket. We see that our answer is reasonable. It predicts that the total weight of the rocket acts downward on the pylon and that the engine also applies a horizontal force on the pylon to the left, as expected.

The thrust produced by a rocket engine is the force available to accelerate the vehicle containing the engine. In this case we conclude from (D) that the thrust is

[latex]F_{thrust} =−\left[ρ_EV^2 _E A_E +(p_E − p_A)A_E ight]j [/latex] (E)

Combining (D) and (E) gives

F_pylon= F_thrustj − (W_CV + W_rocket)k

which sensibly states that the force on the pylon is due to the thrust and the total weight.

Our analysis shows that the pressure at which the rocket exhaust stream exits the nozzle inﬂuences the amount of thrust produced. There is a complex relationship between the combustion chamber pressure, exit velocity, exit pressure, and ambient pressure of a rocket nozzle, so the exit velocity and exit pressure are not independent. The analysis of isentropic compressible ﬂow through a converging–diverging nozzle, a topic not covered in this text, shows that the maximum thrust occurs for an exit pressure equal to the ambient pressure or p_E = p_A, thus the maximum thrust is given by [latex]F_{thrust}=-ρ_EV^2 _E A_Ej [/latex].

We can also solve this problem using a mixed CV with a decal surface to isolate the force applied to the pylon. This CV is shown in Figure 7.18E. Writing a steady ﬂow momentum balance for this CV, we have [latex]\int_{exit}^{}{} [/latex] (ρu)(u • n)dS =[latex]\int_{CV}^{}{} [/latex] ρf dV + [latex]\int_{decal}^{}{\sum} [/latex]dS +[latex]\int_{air}^{}{\sum} [/latex]dS +[latex]\int_{exit}^{}{\sum} [/latex]dS. Evaluating the various terms and noting that by action–reaction [latex]\int_{decal}^{}{\sum} [/latex]dS = −F_pylon, we ﬁnd

[latex]ρ_EV^2 _E A_Ej=−(W_{CV} +W_{rocket} )k−F_{pylon}+ p_AA_Ej+(−p_E A_Ej)[/latex]

which is identical to the earlier result (D), and much quicker.

Question 7.10: A liquid-fueled rocket is in steady operation on a test stan...

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