# How to calculate the pressure of an explosion?

• Thijske
In summary: I'm going to be honest: I made all of the mentioned numbers up. This is because I am yet to finish the calculations concerning these numbers.But yes, I am assuming 100% combustion, because I am using a premixed air-fuel mixture with a ratio of 14.7:1.
Thijske
Say there's 2 grams of gasoline (vapor) (E85) and sufficient air, how can you calculate the force/pressure of the explosion when this mixture is ignited?
The pressure of the mixture before ignition is 400 pascal. The calorific value of the fuel is 45 MJ/kg. The volume in which the explosion occurs is 9E-5 cubic meters.

Usually that kind of things are measured, not calculated. To calculate this accurately, you'd have to know how the heat capacities of the gaseous reaction products behave at temperatures over ##1000^\circ\text{C}##, as well as the equilibrium constants of reactions like ##2H_2 \text{ (g)}+ O_2\text{ (g)}\rightarrow 2H_2 O\text{ (g)}## at those temperatures.

Thijske said:
Say there's 2 grams of gasoline (vapor) (E85) and sufficient air, how can you calculate the force/pressure of the explosion when this mixture is ignited?
The pressure of the mixture before ignition is 400 pascal. The calorific value of the fuel is 45 MJ/kg. The volume in which the explosion occurs is 9E-5 cubic meters.

The idea is to calculate the temperature after the reaction from the stoichiometry of the reaction (assuming no excess air), the heat of the reaction, and the heat capacities of the products (water vapor and CO2). Once you know the final temperature, you can apply the ideal gas law to get the pressure.

Let's see your attempt to do this.

Thijske said:
Say there's 2 grams of gasoline (vapor) (E85) and sufficient air, how can you calculate the force/pressure of the explosion when this mixture is ignited?
The pressure of the mixture before ignition is 400 pascal. The calorific value of the fuel is 45 MJ/kg. The volume in which the explosion occurs is 9E-5 cubic meters.

First, E85 is up to 85% ethanol. You'll need to know the actual ratio. Second, your pressure seems very low for the 2 grams of reactants in that small space. Is the mixture very cold? Finally, are you assuming 100% combustion?

Before you go too far in making these calculations, you should make sure your initial conditions are right.

You have half a milligram of air at that pressure and volume. You have 4000x as much gasoline. Not only is the combustion incomplete, you are above the UEL for ethanol. It won't explode.

You have 4000x as much gasoline. Not only is the combustion incomplete, you are above the UEL for ethanol.
Agreed; is this for an internal combustion engine?

The answer in terms of a realistic confined explosion is that the pressure at a given position is a function of time. The concept of an explosion assumes a flame extending from a source of ignition sending pressure waves towards the confinement. It is possible that these will combine to form a detonation front. If not, the flame front travels into mixtures at different pressures, resulting in different temperatures. Only when an equilibrium is achieved throughout the vessel is a time and position independent temperature reached.

Bystander said:
Agreed; is this for an internal combustion engine?

It would explain why the Check Engine Light is on and the mileage is horrible.

berkeman
90 cc = 5.5 cubic inches. A moped? It would explain failure to start and run - the engine is flooded.

Then again, 400 Pa absolute is 29.80 In Hg vacuum. Two grams of gasoline would have to be in liquid form to have that absolute pressure in that volume. On the other hand, if the OP meant 400 Pa gauge pressure, then it might be possible to have a mixture that could fire. But at 400 Pa absolute, the piston would have to be at the bottom of the stroke, so the exhaust port/valve would be open. Result: A FOOM or BANG out the exhaust, but not much pressure.

I read OPs post to mean that the mass of the gasoline and air combined was about 2 grams. Also, 45 MJ/kg is for gasoline/diesel not E85.

Thanks to everyone for taking time to write a response!

Chestermiller said:
The idea is to calculate the temperature after the reaction
This was my idea as well. The problem is; I assume a part of the energy will be lost to the cylinder walls and other stuff. Therefore, not all generated energy will go into heating the air-fuel mixture. I don't know how to calculate the final temperature. Perhaps you do?

Dr_Nate said:
First, E85 is up to 85% ethanol. You'll need to know the actual ratio. Second, your pressure seems very low for the 2 grams of reactants in that small space. Is the mixture very cold? Finally, are you assuming 100% combustion?
I'm going to be honest: I made all of the mentioned numbers up. This is because I am yet to finish the calculations concerning these numbers.
But yes, I am assuming 100% combustion, because I am using a premixed air-fuel mixture with a ratio of 14.7:1.

Before you go too far in making these calculations, you should make sure your initial conditions are right.

You have half a milligram of air at that pressure and volume. You have 4000x as much gasoline. Not only is the combustion incomplete, you are above the UEL for ethanol. It won't explode.
Same as I said to Dr_Nate. Sorry for that...

Dr_Nate said:
45 MJ/kg is for gasoline/diesel not E85.
I searched for the calorific value of E85, but I couldn't find it. Do you know that value?

Thanks again to everyone, would mean the world if you'd help me out here!

Thijske said:
I searched for the calorific value of E85, but I couldn't find it. Do you know that value?
It's a weighted sum of the components.

It's bad form to just make up numbers like that. People waste their time trying to help you. Just to give you some direction, I'll let you know that you need to go learn about specific heat capacities, the actual chemical reaction and the stoichiometries, and the ideal gas law (and possibly the real gas law). Once you learn those, come back with background on what exactly you are trying to do and an attempt at a solution. The we can see if we can help guide you.

davenn
Dr_Nate said:
It's a weighted sum of the components.

It's bad form to just make up numbers like that. People waste their time trying to help you. Just to give you some direction, I'll let you know that you need to go learn about specific heat capacities, the actual chemical reaction and the stoichiometries, and the ideal gas law (and possibly the real gas law). Once you learn those, come back with background on what exactly you are trying to do and an attempt at a solution. The we can see if we can help guide you.

I genuinely apologize and I understand it's a bit irritating that my numbers are a bit off, but I want you to know that the numbers were educated guesses (400Pa should be 400kPa).
Also, I am no absolute rookie in science (done 6 years of pre-university), so I think I do know about the things you mentioned.

What I attempted was the following:
- wrote down chemical equation.
- made a chemical energy diagram.
- converted released energy to temperature by using specific heat.

Here's why I think above method is wrong/incomplete: not all energy will go into heating the combustion gasses. A part of the energy will go into heating the walls and other stuff. I think that loss can be calculated by solving Tinside in the following equation:
Q/t=Wall area*(Tinside-Toutside)/specific heat of walls
Any thoughts?

Thijske said:
Here's why I think above method is wrong/incomplete: not all energy will go into heating the combustion gasses. A part of the energy will go into heating the walls and other stuff.
How quickly? An explosion is fast. Heat transfer can be slow.

How quickly can heat diffuse through the hot combustion gasses to the cool walls? How quickly can heat be conducted from the heated cylinder walls to the rest of the cool engine block?

256bits
Thijske said:
solving Tinside in the following equation
As a guestimate of the amount of heat transfer to the walls.
How much mass.
You also have radiation exchange between the gas and the wall during the combustion, as well as some conduction between the gas and the wall.
The whole wall is not at a common temperature - as the gas expands it cools, the lower parts of the wall will be cooler than the top. The top, if it has heated to a higher temperature than the expanded gas may re-supply the gas with more heat.
Partitioning the heat exchange is the difficulty.

This is a VERY complex problem. A working engineer might do some initial back-of-the-envelope calculations, but then quickly move on to software designed specifically to simulate problems like this. Then they would build a prototype and then test it. Analysis and looping back around to previous steps would happen, and so on.

Now, all that being said, if I were you, I would first estimate the maximum pressure in the steady state with no heat transfer and no work outside the system. This would allow me to wrap my head around the quantities I'm dealing with.

jbriggs444 and 256bits

## 1. What is the formula for calculating the pressure of an explosion?

The formula for calculating the pressure of an explosion is P = F/A, where P is the pressure, F is the force, and A is the area over which the force is applied.

## 2. How do you determine the force of an explosion?

The force of an explosion can be determined by measuring the mass of the explosive material and the velocity at which it explodes. The force can also be estimated by the amount of damage caused by the explosion.

## 3. Can the pressure of an explosion be accurately calculated?

Yes, the pressure of an explosion can be accurately calculated using the formula P = F/A and by obtaining accurate measurements of the force and area.

## 4. Are there any factors that can affect the pressure of an explosion?

Yes, there are several factors that can affect the pressure of an explosion, such as the type and amount of explosive material, the containment of the explosion, and the distance from the explosion.

## 5. How is the pressure of an explosion measured?

The pressure of an explosion can be measured using specialized equipment, such as pressure gauges, which can be placed at various distances from the explosion to obtain accurate readings.

• Mechanics
Replies
4
Views
2K
• Mechanics
Replies
1
Views
2K
• Mechanics
Replies
2
Views
2K
• Mechanical Engineering
Replies
2
Views
552
• Mechanics
Replies
10
Views
4K
• Mechanics
Replies
15
Views
3K
• Mechanical Engineering
Replies
5
Views
2K
• General Engineering
Replies
15
Views
1K
• Mechanical Engineering
Replies
2
Views
3K
• Mechanics
Replies
11
Views
3K