Rocket Equation

tahanson43206 · 2024-05-28 18:31:21

For SpaceNut ...

This forum contained no topics with both the words "rocket" and 'equation"

We have many posts that make use of or refer to the Rocket Equation.

This topic is offered for NewMars members who would like to show various scenarios that can be addressed with the Rocket Equation.

This topic leads off with a scenario posted by RGClark in one of his many topics about SSTO...

In post #3 of this topic, I will post a calculation done by ChatGPT4o.

We used the figures from RGClark's scenario, with the slight variation that instead of 200,000 kg for the entire vehicle, ChatGPT4o used 200,000 kg for the propellant, and came up with 25 tons and change for the airframe and payload delivered to orbit.

I am surprised by this figure, and look forward to analysis by our members.

(th)

tahanson43206 · 2024-05-28 18:31:55

This post is reserved for an index to posts that may be contributed by NewMars members over time.

(th)

tahanson43206 · 2024-05-28 18:34:41

I asked ChatGPT4o to compute a flight to orbit for a vehicle weighting 200 tons at launch, using the Rocket Equation.

ChatGPT4o used the 200,000 kg figure for propellant and it did not specify the propellant. Instead, it used a figure for exhaust velocity that i assume must be well established as a standard that can be achieved with chemical propellant.

[h1]Rocket Equation Analysis for Spacecraft Payload[/h1]
Introduction
The Tsiolkovsky rocket equation, or simply the rocket equation, is fundamental in understanding the relationship between a rocket's mass, the mass of its fuel, and the change in velocity (delta-v) it can achieve. This analysis will demonstrate the application of the rocket equation to determine if a spacecraft with an initial mass of 200 metric tons can place a 13-ton payload into orbit.
Rocket Equation
The rocket equation is given by:
Δv = v_e * ln(m_0 / m_f)
Where:
Δv is the change in velocity (delta-v).
v_e is the effective exhaust velocity.
m_0 is the initial total mass (including fuel).
m_f is the final mass (after the fuel is burned).
Initial Setup
Initial mass of the spacecraft (m_0): 200 metric tons (200,000 kg).
Desired payload to orbit: 13 metric tons (13,000 kg).
Calculation Example
Assumptions:
Effective exhaust velocity (v_e): 4,500 m/s (typical for chemical rockets).
Required Δv for Low Earth Orbit (LEO): 9,300 m/s.
Using the rocket equation:
9,300 = 4,500 * ln(200,000 / m_f)
Solving for m_f:
9,300 / 4,500 = ln(200,000 / m_f)
2.067 = ln(200,000 / m_f)
Exponentiate both sides to remove the natural logarithm:
e^2.067 = 200,000 / m_f
7.9 = 200,000 / m_f
m_f = 200,000 / 7.9 ≈ 25,316 kg
Therefore, the remaining mass after burning the fuel is approximately 25,316 kg.
Discussion
If the inert mass (dry mass without fuel) is set to zero, the payload mass would be close to the final mass.
In reality, the inert mass cannot be zero. However, with minimal inert mass, the payload mass will be slightly less than the final mass.
Achieving 13 tons of payload to orbit with 200 metric tons at liftoff is challenging but possible with higher specific impulse or more efficient propellant use.
Conclusion
By understanding and adjusting key parameters such as exhaust velocity and initial mass, we can better evaluate the feasibility of our payload goals. This analysis shows that placing a 13-ton payload into orbit with a 200-ton spacecraft is theoretically possible but requires careful consideration of the rocket's design and propellant efficiency.

I would greatly appreciate forum members running their own calculations and reporting the results.

(th)

SpaceNut · 2024-05-28 19:05:20

tsiolkovsky rocket equation
has been posted about in topics.

https://sentinelmission.org/rocketry-pr … t-equation

The equation is expressed as: Δv = Ve * ln (Mi/Mf)
Where:
Δv = velocity change of the rocket
Ve = exhaust velocity of the propellant
Mi = initial mass of the rocket (including propellant)
Mf = final mass of the rocket (after propellant is expelled)

To orbit 125 miles (200 kilometers) above Earth, a spacecraft must travel at a screaming 17,400 mph (28,000 km/h).

Your 33,480 km/hr is at a higher orbit

the x-33 was 129 mt with 2 engines of 1800 kN of trust each with LH2/LOX fuels and was to have a payload of 5.4 mT.

tahanson43206 · 2024-05-28 19:12:32

For SpaceNut ... in this topic I'm looking for members to show they can run the rocket equation.

You have shown that you can find a copy of it and paste it into a post, and that is an important skill.

However, it is not sufficient for ** this ** topic.

Please take the numbers quoted in post number 3, and see if you get the same result.

(th)

tahanson43206 · 2024-05-29 11:05:57

This is an experiment to see if GW Johnson's work (done in Microsoft Word) can be converted to bbcode for display here in the forum.

The document we will attempt to deliver is an answer to RGClark's SSTO scenario. I was surprised to see that the scenario would work at all, but it clearly does using GW's spreadsheet. The numbers for inert mass were set to zero, so that anything delivered to orbit shows up as payload. Obviously in real life, the inert mass would have value, and the payload would be set to zero. In that case, the ship itself ** is ** the payload.

Weight statement, metric tons
GWJ 5-29-2024
Payload: 13
Inert: 0 (ridiculous!)
Burnout: 13
Propellant: 200
Ignition: 213
Mass ratio = ign/bo = 213/13 = 16.3846, ln(MR) = 2.7963
Propulsion models (Vex = 9.80667*Isp/1000)
Isp, s
Vex, km/s
350
3.432
400
3.923
450
4.413
Rocket equation prediction at indicated mass ratio
dV = Vex ln(MR):
Isp, s
dV, km/s
350
9.597
400
10.970
450
12.340
Bear in mind these are ridiculous predictions:
It is NOT possible to build anything that has zero inert mass!
Look at the problem the other way around:
Mission (LEO eastward at low inclination, rendezvous, deorbit)
Required dV capability ~ 9.3 km/s
Propulsion
Isp, s
Vex, km/s
(dV/Vex)
350
3.432
2.70979
400
3.923
2.37063
450
4.413
2.10741
Rocket equation (reversed as MR = exp(dV/Vex))
Isp, s
MR
350
15.022
400
10.706
450
8.227
Weight statement-related:

Propellant mass fraction Wp/Wig = 1 – 1/MR
“allowance” = sum of payload and inert fractions = 1 – propellant mass fraction
Isp, s
Wp/Wig
allowance (= Wpay/Wig + Winert/Wig)
350
0.9334
0.0666 (everything that is “not propellant”)
400
0.9066
0.0934
450
0.8784
0.1216
Design considerations:

the “allowance” is your “design space”
Whatever you think your inert fraction is, subtract that from the allowance for payload fraction
Size the weight statement masses from the fractions and the payload you want to deliver
Size the thrust requirement using T/Wig = 1.5+
Size inert masses for engines, tanks, and equipment, then iterate until it matches inert fractions

The conversion was performed by ChatGPT4o, given the original text from a docx file, that had been rendered as odt format.

(th)

GW Johnson · 2024-05-29 13:25:46

The rocket equation as seen in all the books is quite simple: dV = Vex LN(MR) where MR = Wig/Wbo. It presumes a weight statement where Wpay + Winert = Wburnout, and W burnout + Wpropellant = Wignition. Vex is not the actual nozzle expansion velocity, but an effective exhaust velocity as if the exit plane expanded pressure were zero (so that only a momentum term appears in thrust). (In point of fact, Vex is the specific impulse value, if one uses dimensionally-consistent units of measure in thrust/flow rate.)

Everyone is taught to predict performance using the rocket equation "forward", with mass ratio MR and Vex as inputs. But for sizing designs, you need to use it backwards, in the reversed form MR = exp(dV/Vex). The mass ratio relates to the propellant mass fraction in a very simple way: Wpropellant/Wignition = 1 - 1/MR. So your mission sets the needed dV, and your propulsion sets the necessary value of Isp (and from it Vex = gc*Isp).

Use that to determine the propellant mass fraction required of the stage. Whatever is left after you subtract it from one is the sum of your inert and payload fractions, a sort of "allowance" for the non-propellant items. And THAT is the space in which you do your vehicle designs. What inert fractions those designs turn out to have, subtract from the allowance to determine payload fraction.

If your as-designed inert fraction exceeds the allowance, then payload fraction is inescapably negative, and your design is fundamentally infeasible, no matter what other requirements it might meet. The rocket equation feasibility requirement comes first and foremost.

GW

tahanson43206 · 2024-05-29 19:25:43

For GW Johnson... I gave your post to ChatGPT4o, and I am not surprised that it confirmed your observations. Here is the result:

The first try did not work... I'll try again.

Conclusion
The post by GW provides an accurate and practical explanation of the rocket equation and its application in rocket design. The focus on using the equation both forward for performance prediction and backward for design sizing is essential for practical rocketry. The method of determining propellant mass fraction and using it to allocate the remaining mass between inert structure and payload is a standard and effective approach in the industry.
Document 1: Review and Confirmation of GW's Work
Review of GW's Rocket Design Explanation
Conclusion: GW's explanation is accurate and practical. The focus on both forward and backward use of the rocket equation, along with determining propellant mass fraction and designing within constraints, aligns with industry standards in rocketry.
Document 2: Observations on Practical Rocket Design
Observations on Practical Rocket Design by GW
[]The remaining mass after accounting for propellant is allocated to inert structure and payload.
[]Ensuring the inert fraction is within allowable limits is critical for feasible design.
Conclusion: GW's approach to rocket design using the rocket equation is sound and reflects standard practices in the industry. Determining propellant mass fraction and designing within these constraints ensures practical and feasible rocket designs.

(th)

tahanson43206 · 2024-05-29 19:27:01

Here is another try:

Review of GW's Rocket Design Explanation
Key Points:
The rocket equation: \( \Delta V = V_{ex} \ln(\text{MR}) \)
Mass Ratio (MR) = \( \frac{W_{ig}}{W_{bo}} \)
Weight relationships: \( W_{pay} + W_{inert} = W_{burnout} \) and \( W_{burnout} + W_{propellant} = W_{ignition} \)
Effective exhaust velocity \(V_{ex}\) = \( I_{sp} \times g_0 \)
Forward and backward use of the rocket equation
Propellant mass fraction: \( \frac{W_{propellant}}{W_{ignition}} = 1 - \frac{1}{\text{MR}} \)
Designing within mass fraction constraints
Conclusion: GW's explanation is accurate and practical. The focus on both forward and backward use of the rocket equation, along with determining propellant mass fraction and designing within constraints, aligns with industry standards in rocketry.

and here is ** another ** try:

Review of GW's Rocket Design Explanation
Key Points:
The rocket equation: ΔV = Vex ln(MR)
Mass Ratio (MR) = Wig/Wbo
Weight relationships: Wpay + Winert = Wburnout and Wburnout + Wpropellant = Wignition
Effective exhaust velocity Vex = Isp * g0
Forward and backward use of the rocket equation
Propellant mass fraction: Wpropellant/Wignition = 1 - 1/MR
Designing within mass fraction constraints
Conclusion: GW's explanation is accurate and practical. The focus on both forward and backward use of the rocket equation, along with determining propellant mass fraction and designing within constraints, aligns with industry standards in rocketry.

(th)

tahanson43206 · 2024-05-29 19:27:55

And trying again for Document 2:

Observations on Practical Rocket Design by GW
Key Principles:
The rocket equation is fundamental in predicting performance and designing rockets.
Forward use of the rocket equation involves calculating \(\Delta V\) from known parameters.
Backward use involves determining the mass ratio from mission \(\Delta V\) and effective exhaust velocity.
Propellant mass fraction: \( \frac{W_{propellant}}{W_{ignition}} = 1 - \frac{1}{\text{MR}} \)
The remaining mass after accounting for propellant is allocated to inert structure and payload.
Ensuring the inert fraction is within allowable limits is critical for feasible design.
Conclusion: GW's approach to rocket design using the rocket equation is sound and reflects standard practices in the industry. Determining propellant mass fraction and designing within these constraints ensures practical and feasible rocket designs.

(th)

SpaceNut · 2024-06-01 14:51:09

Not sure if I have all of the numbers yet but wanted to see what I could do with stages that are already con structed.

The equation is expressed as: Δv = Ve * ln (Mi/Mf)

Natural log IN

Where:
Δv = velocity change of the rocket
Ve = exhaust velocity of the propellant
Mi = initial mass of the rocket (including propellant)
Mf = final mass of the rocket (after propellant is expelled)

Falcon 9 first stage modified for ssto.
Mass (without propellant)[39] 22,200 kg (48,900 lb)
Mass (with propellant) 433,100 kg (954,800 lb)
Liquid oxygen tank capacity 287,400 kg (633,600 lb)
Kerosene tank capacity 123,500 kg (272,300 lb)
Payload fairing 1,700 kg (3,700 lb)

Thrust (stage total)[4] 7,607 kN (1,710,000 lbf) (sea level)
Specific impulse Sea level: 282 seconds

Propulsion models (Vex = 9.80667*Isp/1000) 2.76548094
MR = 433,100 kg / 22,000 kg = 19.6863
Rocket equation (reversed as MR = exp(dV/Vex))
Δv = 2.76548094 * 2.97992296 = 8.24092

kbd512 · 2024-06-01 19:27:11

SpaceNut,

I ran numbers. Merlin-1D's Isp is too low. You either get no payload or very little, with 3,000kg being best case scenario. If the avg Isp is 338s (staged combustion), then you get 7,305kg to 17,799kg, with 12,060kg being the estimate. This is for an expendable SSTO, not reusable. I presume engine weight remains the same, Isp increases by using staged combustion, and CFRP tanks weigh 50% of what the Aluminum tanks weigh.

Falcon 9 expendable is 22,800kg.
Falcon 9 reusable is 17,500kg (requires landing the booster downrange on the drone ship).

SpaceNut · 2024-06-02 17:20:00

https://www.omnicalculator.com/physics/specific-impulse

Seems that the engine running on the RP/Lox are all about the same even the Russian models that we are building a replacement for.

Seems even methane / lox is not all that much better.

https://en.wikipedia.org/wiki/Compariso … et_engines

SpaceNut · 2024-06-23 19:23:28

Rocket Thrust Equations

tahanson43206 · 2025-04-02 17:16:52

This post is reserved for two slides contributed by GW Johnson on the subject of this topic:

1) Rocket performance to fixed expansiion

2) thrust coefficient form

(th)

tahanson43206 · 2025-04-02 17:46:05

This post is reserved for two slides contributed by GW Johnson on the subject of this topic:

1) Rocket performance to fixed expansiion

2) thrust coefficient form

(th)

tahanson43206 · 2025-04-02 17:50:56

In this post, I have asked ChatGPT4o to take the NASA document provided by SpaceNut in Post #14 into account.

You said:
Thank you for accepting the NASA document for reference, since that allows us to share the same document with our members who are trying to follow along. And thank you for considering the advice from our aerospace engineer! Rather than jump to the calculations, I'd like to see if we might be able to share the work you just did with our readers, by using the BBCode you have mastered so well. Please keep in mind we can't show latex. I understand it is a powerful tool that would work for many customers. Please remember that header is indicated by the symbol "h", and there needs to be an asterisk inside brackets when you are making a list. I can manually edit the BBCode if you miss anything.

ChatGPT said:
Absolutely! Here’s a BBCode version of our analysis so far, formatted for your forum’s style. I've replaced LaTeX-style math with text equations <snip>.
Rocket Thrust from Limited Data: Exploring NASA’s Guide
We began with a limited set of known variables:
Mass flow rate: 2 kg/s
Propellant: Hydrogen
Exhaust temperature: 3000 Kelvin
Our goal is to estimate thrust, but instead of jumping into calculations, we took a step back to analyze the NASA Thrust Equations Summary and consider how to fill in the missing variables.
The General Thrust Equation
From NASA, the thrust of a rocket in vacuum is given by:
F = (mass flow rate) × (exhaust velocity) + (exit pressure - ambient pressure) × (exit area)
In vacuum, the ambient pressure is zero, so the simplified equation becomes:
F ≈ mass flow rate × exhaust velocity
This means that if we can estimate the exhaust velocity, we can calculate thrust with reasonable accuracy.
Step 1: Estimate Exhaust Velocity
Using thermodynamic principles, we can estimate the exhaust velocity of a gas expanding from high temperature and pressure. For an ideal gas, the exhaust velocity can be estimated from:
Vₑ = sqrt( (2 × k × R × T) / (M × (k - 1)) )
Where:
Vₑ is the exhaust velocity in m/s
k is the ratio of specific heats (gamma), approximately 1.4 for hydrogen
R is the universal gas constant = 8314 J/kmol·K
T is the exhaust temperature in Kelvin = 3000 K
M is the molar mass of hydrogen = 2.016 kg/kmol
With this formula, we can estimate how fast the hydrogen is exiting the nozzle, assuming ideal gas behavior.
Step 2: Speed of Sound in Hydrogen
The speed of sound in a gas is given by:
a = sqrt( k × R × T / M )
This can help us estimate conditions at the nozzle throat (where the exhaust gas reaches sonic velocity). Knowing the speed of sound lets us work backward from the mass flow rate to estimate the throat area.
Step 3: Throat Area and Mass Flow
NASA’s equations also show that mass flow rate depends on the throat area:
mass flow = density × throat area × speed of sound
If we guess or estimate the density of the gas at the throat, and calculate the speed of sound, we can solve for the throat area needed to produce the known mass flow of 2 kg/s.
An Iterative Path Forward
We may need to make some assumptions and refine them over time. Here's one possible pathway:
Use the simplified exhaust velocity equation to estimate Vₑ
Multiply by mass flow to estimate thrust
Use the speed of sound formula to get conditions at the nozzle throat
Estimate throat area using mass flow = density × area × sonic speed
Adjust any assumed values (like density) based on updated estimates
What’s Next?
We can now choose whether to:
Run the actual numbers for speed of sound, exhaust velocity, and thrust
Create a spreadsheet or calculator tool for students to experiment with
Introduce refinement factors (like dissociation or non-ideal behavior)
If our readers are following along, we invite them to try using the formulas above with their own calculator or spreadsheet — and we’ll be happy to compare results and improve the estimates together.
Let me know if you'd like to continue with the calculations next, or if you'd like a second BBCode post to walk through the numeric example!

(th)

tahanson43206 · 2025-04-02 18:03:36

In this post, I asked ChatGPT4o to proceed to perform the calculations, and to show the work in BBCode so we could see it in the forum.

ChatGPT said:
Fantastic! I'm glad the BBCode version worked well in your forum. Now, let’s follow through with the calculations, using the same format to present the results clearly for your readers.
Example Calculation: Estimating Thrust from Limited Data
We’ll use the known variables from earlier:
Mass flow rate (ṁ): 2 kg/s
Propellant: Hydrogen
Exhaust temperature (T): 3000 K
Molar mass of hydrogen (M): 2.016 kg/kmol
Gas constant (R): 8314 J/kmol·K
Specific heat ratio (k): 1.4 (for H₂ under ideal conditions)
We’ll calculate:
Speed of sound at 3000 K
Exhaust velocity
Thrust
Estimate of throat area
Step 1: Speed of Sound in Hydrogen
Formula:
a = sqrt( k × R × T / M )
Substitute the values:
a = sqrt(1.4 × 8314 × 3000 / 2.016)
a = sqrt(1.4 × 24,942,000 / 2.016)
a = sqrt(17,334,168.25) ≈ 4162 m/s
Step 2: Exhaust Velocity
Formula:
Vₑ = sqrt( (2 × k × R × T) / (M × (k - 1)) )
Substitute the values:
Vₑ = sqrt( (2 × 1.4 × 8314 × 3000) / (2.016 × 0.4) )
Vₑ = sqrt(69,833,600 / 0.8064)
Vₑ = sqrt(86,605,878.2) ≈ 9306 m/s
Step 3: Thrust
Formula:
F = ṁ × Vₑ
F = 2 kg/s × 9306 m/s = 18,612 N (newtons)
That’s roughly 1.9 metric tons of thrust.
Step 4: Throat Area Estimate
Assuming sonic flow at the throat, we use:
ṁ = ρ × A × a
Rearranged:
A = ṁ / (ρ × a)
Let’s estimate density (ρ) using the ideal gas law:
ρ = P / (R_specific × T)
R_specific = R / M = 8314 / 2.016 ≈ 4125 J/kg·K
We’ll assume chamber pressure of 10 atmospheres (typical for test cases), or:
P = 10 × 101325 = 1,013,250 Pa
Then:
ρ = 1,013,250 / (4125 × 3000) ≈ 0.0819 kg/m³
Now plug in:
A = 2 / (0.0819 × 4162) ≈ 2 / 341 ≈ 0.00587 m²
That’s about 58.7 cm² for the nozzle throat area.
Summary of Results
Speed of sound (3000 K hydrogen): 4162 m/s
Estimated exhaust velocity: 9306 m/s
Thrust: 18,612 N
Throat area: ~58.7 cm²
These results are based on ideal gas assumptions and a guessed chamber pressure. In reality, effects like dissociation, non-ideal expansion, and nozzle efficiency would modify these values.
Would you like me to prepare a follow-up post that discusses possible refinements — such as real gas effects or using different chamber pressures and expansion ratios?

I asked about the 18K Newtons being the same as estimated earlier, but ChatGPT4o had forgotten that earlier work.

I am hoping our NewMars members can go through the calculations with a fine toothed comb and find any errors there may be.

(th)

kbd512 · 2025-04-02 21:43:35

tahanson43206,

Study on high-temperature hydrogen dissociation for nuclear thermal propulsion reactor

Boltzmann statistics - dissociation of molecular hydrogen into atomic hydrogen

root mean squared formula

At 3,000K, there will be some dissociation of Hydrogen molecules, which will affect their root mean squared speed (rms).

Ask the AI how this effect will affect exhaust velocity, thus the average kinetic energy of the atomic Hydrogen and thrust produced.

tahanson43206 · Yesterday 18:54:43

For kbd512 re #19

Thank you for the links to help our readers to learn about dissociation as it applies to our rocket design.

Your observation about dissociation at 3000 Kelvin matches up nicely with GW Johnson's recommendation of a change from 1.4 to 1.2 in the equations.

Because GW had multiple suggestions for improvement of the calculation, I decided to tackle only this one variable today.

Below is a transcript of a discussion with ChatGPT4o about the Specific Heat Ratio.

Please note that this change resulted in a higher value for thrust. GW had other suggestions that are ultimately going to reduce the thrust.

You said:
Hello Again! And thank you (again) for providing the analysis of our problem, and the carefully worked out solution that arrived at an estimate of 18,600 Newtons of thrust. Our resident aerospace engineer looked carefully at your work. He had several suggestions that I hope will help us to improve our estimate. The first was the value you selected for Specific heat ratio (k). Your value of 1.4 is a bit higher than the 1.2 he said he would have used. Please explain (for my sake) what that value represents, and what the difference would be between 1.2 and 1.4
***
ChatGPT said:
Welcome back! It's always a pleasure to continue this thoughtful collaboration — and I'm very glad your aerospace engineer is reviewing the work closely and providing feedback. His suggestion to revisit the specific heat ratio
k (also known as gamma, γ) is spot on. Let's explore what it represents and what the impact of using 1.2 instead of 1.4 would be.
What is the Specific Heat Ratio (k or γ)?
The specific heat ratio, denoted as k or γ, is the ratio of:
k = Cₚ / Cᵥ
Where:
Cₚ is the specific heat at constant pressure
Cᵥ is the specific heat at constant volume
It measures how a gas stores and transfers energy. Gases with more internal degrees of freedom (like vibrations and rotations) tend to have lower k, because more energy goes into internal motion rather than temperature increase.
Typical Values of k
The value of k depends on:
The type of gas (monatomic vs. diatomic vs. polyatomic)
The temperature — especially at high temperatures when dissociation or ionization occurs
For hydrogen (H₂):
At low temperatures (room temp), it's considered diatomic, and k ≈ 1.4
At high temperatures (~3000 K), H₂ may partially dissociate into atomic hydrogen (H), increasing degrees of freedom
Dissociation and excitation reduce k → k drops toward 1.2 or even lower
So your engineer is exactly right — at 3000 K, assuming k = 1.2 better reflects real behavior.
How Does k Affect Our Calculations?
The exhaust velocity formula is:
Vₑ = sqrt( (2 × k × R × T) / (M × (k - 1)) )
Let’s compare the effect of two values:
Case 1: k = 1.4
Vₑ ≈ 9306 m/s (as previously calculated)
Case 2: k = 1.2
Let’s recalculate:
Numerator: 2 × 1.2 × 8314 × 3000 = 59,860,800
Denominator: 2.016 × (1.2 − 1) = 2.016 × 0.2 = 0.4032
Vₑ = sqrt(59,860,800 / 0.4032) = sqrt(148,481,481) ≈ 12,184 m/s
This is significantly higher! Why? Because reducing k−1 in the denominator increases the velocity more than the small reduction in k in the numerator.
Recalculating Thrust with k = 1.2
F = ṁ × Vₑ = 2 kg/s × 12,184 m/s = 24,368 N
So with this more realistic value of k = 1.2, we estimate:
Thrust = 24,368 newtons, an increase of about 31% compared to our previous estimate.
Summary
k (gamma) is the ratio of specific heats, representing how energy is stored in a gas
High-temperature hydrogen has more internal motion (and possible dissociation), leading to lower k values
Using k = 1.2 instead of 1.4 results in higher exhaust velocity and thrust
Updated thrust estimate: ~24,368 N (up from 18,612 N)
Would you like a full BBCode version of this post for sharing with your group? I'd be happy to provide it next!
***
You said:
Please do prepare the updated report including the discussion. Let's work with one item at a time.

ChatGPT said:
Perfect! Let’s start with a BBCode-formatted report that focuses specifically on the meaning and effect of the specific heat ratio k, just as you requested. We’ll walk through its role in the rocket thrust equation and then show the impact of changing it from 1.4 to 1.2.
Here’s the BBCode version, ready to paste into your forum:
Refining the Rocket Thrust Estimate: Specific Heat Ratio (k)
Our earlier thrust estimate used a value of k = 1.4 for hydrogen, which produced:
Estimated thrust: ~18,600 newtons
However, after reviewing the calculation, our aerospace engineer suggested using a lower value: k = 1.2. This small change has a significant impact on the results.
What is the Specific Heat Ratio?
The specific heat ratio, usually written as k or γ, is defined as:
k = Cₚ / Cᵥ
Where:
Cₚ is the specific heat at constant pressure
Cᵥ is the specific heat at constant volume
It tells us how a gas stores thermal energy and how it behaves during expansion or compression. The value of k depends on:
The number of ways molecules can move and store energy (degrees of freedom)
The type of gas (monatomic, diatomic, or complex)
The temperature — especially if the gas begins to dissociate or ionize
Why Does k Decrease at High Temperatures?
Hydrogen gas (H₂) behaves as a diatomic gas at lower temperatures, with a typical value of k ≈ 1.4.
But at very high temperatures (like 3000 K), hydrogen molecules may:
Gain access to more vibrational modes
Begin to dissociate into atomic hydrogen (H)
Both effects increase internal degrees of freedom, causing k to drop toward 1.2 or even lower.
Our engineer advised that k = 1.2 is a more realistic choice at 3000 K — and using that value significantly improves the accuracy of our estimate.
How Does k Affect Thrust?
Let’s revisit the ideal exhaust velocity formula:
Vₑ = sqrt( (2 × k × R × T) / (M × (k − 1)) )
Where:
R is the universal gas constant = 8314 J/kmol·K
T is exhaust temperature = 3000 K
M is molar mass of hydrogen = 2.016 kg/kmol
ṁ is mass flow = 2 kg/s
Updated Calculations with k = 1.2
Exhaust velocity:
Numerator: 2 × 1.2 × 8314 × 3000 = 59,860,800
Denominator: 2.016 × (1.2 − 1) = 0.4032
Vₑ = sqrt(59,860,800 / 0.4032) ≈ 12,184 m/s
Thrust:
F = ṁ × Vₑ = 2 kg/s × 12,184 m/s = 24,368 newtons
Conclusion
By using k = 1.2, which better reflects the physics of hydrogen at 3000 K, we improved our thrust estimate from 18,600 N to ~24,368 N, a difference of more than 30%.
This highlights how a single thermodynamic parameter can significantly influence performance predictions — and how expert insight helps refine our model.
Let me know when you’re ready to move to the next item or enhancement — I’ll be glad to help with it!

GW Johnson has suggestions for rework involving pressure inside the Hot Bulb, the size of the throat, and several other aspects of the system.

If we have a reader who got this far, please note that trying to work backward from the three input values provided by kbd512 is NOT how most engineers would approach this problem. ChatGPT4o had to make assumptions, and GW Johnson did not agree with several of them.

(th)

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#1 2024-05-28 18:31:21

Rocket Equation

#2 2024-05-28 18:31:55

Re: Rocket Equation

#3 2024-05-28 18:34:41

Re: Rocket Equation

#4 2024-05-28 19:05:20

Re: Rocket Equation

#5 2024-05-28 19:12:32

Re: Rocket Equation

#6 2024-05-29 11:05:57

Re: Rocket Equation

Weight statement, metric tons

Propulsion models (Vex = 9.80667*Isp/1000)

Rocket equation prediction at indicated mass ratio

Propulsion

Rocket equation (reversed as MR = exp(dV/Vex))

Weight statement-related:

Design considerations:

#7 2024-05-29 13:25:46

Re: Rocket Equation

#8 2024-05-29 19:25:43

Re: Rocket Equation

#9 2024-05-29 19:27:01

Re: Rocket Equation

#10 2024-05-29 19:27:55

Re: Rocket Equation

#11 2024-06-01 14:51:09

Re: Rocket Equation

#12 2024-06-01 19:27:11

Re: Rocket Equation

#13 2024-06-02 17:20:00

Re: Rocket Equation

#14 2024-06-23 19:23:28

Re: Rocket Equation

#15 2025-04-02 17:16:52

Re: Rocket Equation

#16 2025-04-02 17:46:05

Re: Rocket Equation

#17 2025-04-02 17:50:56

Re: Rocket Equation

Rocket Thrust from Limited Data: Exploring NASA’s Guide

The General Thrust Equation

Step 1: Estimate Exhaust Velocity

Step 2: Speed of Sound in Hydrogen

Step 3: Throat Area and Mass Flow

An Iterative Path Forward

#18 2025-04-02 18:03:36

Re: Rocket Equation

Example Calculation: Estimating Thrust from Limited Data

Step 1: Speed of Sound in Hydrogen

Step 2: Exhaust Velocity

Step 3: Thrust

Step 4: Throat Area Estimate

Summary of Results

#19 2025-04-02 21:43:35

Re: Rocket Equation

#20 Yesterday 18:54:43

Re: Rocket Equation

What is the Specific Heat Ratio (k or γ)?

Typical Values of k

How Does k Affect Our Calculations?

Recalculating Thrust with k = 1.2

Summary

Refining the Rocket Thrust Estimate: Specific Heat Ratio (k)

What is the Specific Heat Ratio?

Why Does k Decrease at High Temperatures?

How Does k Affect Thrust?

Updated Calculations with k = 1.2

Conclusion

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