4
$\begingroup$

I've attempted this question but once again got this wrong. Here is what I did

I knew that the thrust equation was mass flow * (Vj - V0).

I also knew that the mass flow was Pressure * Area * Velocity.

However, the engine was idle, so I cancelled out the velocity. Therefore the mass flow was

101325 * 0.05 = 5066.25 kg/s.

This is where I get confused; If I input this into the thrust equation, I get 5066.25 * (Vj - 0).

What is the question's answer and could I please have a step by step?

Here's the question and following parameters screenshot of falcon and a table of values

Variable Value
Pressure in combustion chamber 9.7 MPa
Temperature in combustion chamber 3685 K
Pressure at nozzle exit 1 atm
Throat area 0.05 $m^2$
$\gamma$ 1.14
R 378
$\endgroup$
9
  • $\begingroup$ I dont have the reputation to comment, so this isn't a full answer, but it doesn't look like you took into account R or γ in your calculation. $\endgroup$
    – f_n_lyre
    Commented Jul 14, 2021 at 11:58
  • $\begingroup$ Oh..how would I add this to my answer? And what would the answer be? $\endgroup$
    – Umer Riaz
    Commented Jul 14, 2021 at 12:00
  • 2
    $\begingroup$ Velocity doesn't refer to the velocity of the craft, but to the velocity of the gas leaving the nozzle. Try writing out the full equation and see what parameters are needed. You may have to perform additional calculations to derive some of the parameters. $\endgroup$ Commented Jul 14, 2021 at 12:18
  • $\begingroup$ @UmerRiaz Someone else will come along with a real answer, but see grc.nasa.gov/WWW/K-12/airplane/mflchk.html. I think you also need to account for the velocity of the gas leaving the nozzle. You have a pressure difference between tank and the atmosphere. $\endgroup$
    – f_n_lyre
    Commented Jul 14, 2021 at 12:22
  • 2
    $\begingroup$ The formula for mass flow is probably wrong. Pressure x Area x velocity has units of N/m^2 . m^2 . m/s = kg.m/s^2 . m/s =/= kg/s. What are the definitions or descriptions of Vj and V0 ? $\endgroup$
    – AJN
    Commented Jul 14, 2021 at 13:07

1 Answer 1

9
$\begingroup$

You have a lot of powerful information in that table. You also need to check your equations as they all seem a little screwy for rocket science.

You should start with the thrust equation for a rocket engine:

$$F = \dot{m}V_e + (P_e-P_o)A_e$$

Here, $V_e$ is the nozzle exit velocity, $\dot{m}$ is the propellant mass flow rate, $P_e$ and $P_o$ are the nozzle exit static pressure and ambient static pressure respectively, and $A_e$ is the nozzle exit area.

You can see that there are two major components that affect a rocket engine's thrust: the momentum transfer from ejecting the propellant mass, and a pressure force which will either add or subtract from this equation based on the pressures. It is worth noting that despite the "added thrust" in the case where the exit pressure is greater than the ambient pressure (a case called "underexpanded flow") the maximum thrust will occur when these two pressures are equal. This is because the exit velocity also varies with ambient and exit pressures and because there are cosine losses from non-axial propellant velocities in an underexpanded nozzle (Picture falcon 9 right before stage separation: the plume balloons out many times wider than the vehicle. This shows the sideways components of exit velocity which don't contribute to thrust!)

Enough of that. Now that we know the equation for thrust, we can start mathin this boi. You are told that the exit pressure is 1 atm. You are not told the ambient pressure or the nozzle exit area, so it is safe to assume that this is asking for sea-level thrust. This means the ambient pressure is also 1 atm. This makes $P_e-P_o$ equal to zero, so the thrust equation simplifies to the following:

$$F = \dot{m}V_e$$

Now you just need to figure out how to compute the mass flow and the exit velocity. Mass flow is easy because you know something very powerful about rocket engines. They typically have choked flow at the throat! The equation for choked mass flow is the following:

$$\dot{m} = \frac{AP_t}{\sqrt{T_t}}\sqrt{\frac{\gamma}{R}}(\frac{\gamma+1}{2})^{-\frac{\gamma+1}{2(\gamma-1)}}$$

Here, A is the throat area, and the subscript "t" implies total flow conditions. You are given everything you need in the table there to compute this value!

Now we just need to compute the exit velocity. There is probably an elegant way to do this, but I am not elegant, so my approach is the following:

The isentropic flow equations are very powerful. To find the exit velocity, my approach will be to find the exit mach number and the exit static temperature which will allow the velocity to be computed with the following equation:

$$V_e = M_ea_e = M_e\sqrt{\gamma R T_e}$$

Here, a is the speed of sound.

To compute the exit Mach number, you can rearrange the static to total pressure ratio equation into the following form:

$$M_e = \sqrt{\frac{2}{\gamma-1}\left(\left(\frac{P_e}{P_t}\right)^{-\frac{\gamma-1}{\gamma}}-1\right)}$$

Once you have computed the exit Mach number, It is easy to compute the exit temperature with the static to total temperature equation:

$$\frac{T}{T_t} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{-1}$$

Now you can plug the exit Mach and static temperature into the equation for exit velocity and wam bam you thrustin boi.

The equations I used are the most commonly taught compressible isentropic flow equations and can all be found on the NASA GRC Website.

Following my solution, I compute a mass flow of 262 kg/s, an exit Mach number of 5, an exit static temperature of 1,340 K, and an exit velocity of 3,800 m/s. Plugging mass flow and exit velocity into the thrust equation gives a thrust of 994 kN! Which is pretty close to the vacuum thrust of the actual merlin engine (981 kN). The surplus of thrust is likely due to the unrealistic exit pressure of 1 atm. The merlin engine's exit pressure is much lower for optimized performance at higher altitudes.

EDIT: I forgot to subtract the 1 in the Mach from pressure ratio equation. I have failed you all.

With the (now) fixed equation, the exit Mach is 3.28, the exit static temperature is 2,100 K, and the exit velocity is 3,120 m/s. This gives a thrust of 816 kN!. At the risk of further embarrassment I will not speculate why this value is markedly lower than the Merlin's sea-level thrust of 854 kN.

$\endgroup$
1
  • 1
    $\begingroup$ Nicely done! Once I figured out what "mathin this boi" was :) $\endgroup$ Commented Jul 14, 2021 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.