Combustion of an explosive liquid bubble

[DarkSkinDude]

Homework Statement

An explosive liquid at temperature 300 K contains a spherical bubble of radius 5 mm, full of its vapour. When a mechanical shock to the liquid causes adiabatic compression of the bubble, what radius of the bubble is required for combustion of the vapour, given that the vapour ignites spontaneously at 1100 degrees C? The ratio of $C_V/(nR)$ is 3.0 for the vapour

Relevant Equations

$$\frac{dP}{P}+\gamma\frac{dV}{V}=0, \ \gamma=\frac{C_P}{C_V}$$

Problem Statement: An explosive liquid at temperature 300 K contains a spherical bubble of radius 5 mm, full of its vapour. When a mechanical shock to the liquid causes adiabatic compression of the bubble, what radius of the bubble is required for combustion of the vapour, given that the vapour ignites spontaneously at 1100 degrees C? The ratio of $C_V/(nR)$ is 3.0 for the vapour
Relevant Equations: $$\frac{dP}{P}+\gamma\frac{dV}{V}=0, \ \gamma=\frac{C_P}{C_V}$$

Hey guys, I tried to solve this problem, got stuck, and noticed there weren't any threads with a solution. I ended up solving it shortly and so decided to share.

In this problem we consider a reversible adiabatic change happening to an ideal gas (the water vapour bubble). To solve this, we start with the following:

$$\frac{dP}{P}+\gamma\frac{dV}{V}=0, \ \gamma=\frac{C_P}{C_V}$$

The question involves the relationship between the volume and temperature of the bubble.
Since its an ideal gas, we can substitute $P=\frac{nRT}{V}$ into the above equation, resulting in:

$$-\frac{dV}{V}(\gamma-1)=\frac{dT}{T}$$

We are given a value for the ratio: $C_V/(nR)=3$. Using the fact that $C_P=nR+C_V$, this ratio is found in the above constant:

$$\gamma-1=\frac{C_P}{C_V}-1=\frac{C_V+nR}{C_V}-1=\frac{nR}{C_V}=1/3$$

The bubble is spherical, so we substitute $V=\frac{4}{3}\pi r^3$ and then integrate both sides from their initial values to their final values:

$$-3\frac{nR}{C_V}\int_{r_i}^{r_f}\frac{dr}{r}=\int^{T_f}_{T_i}\frac{dT}{T}$$

The constant on the LHS becomes -1 due to the value given for $C_V/(nR)$.
Evaluating the integrals and isolating for the final radius $r_f$, we get:

$$\frac{T_i}{T_f}r_i=r_f$$

Plugging in $r_i=5 \ mm$, $T_i=300 \ K$, $T_f=(1100+273) \ K=1373 \ K$, we get our final answer of:

$$r_f=1.092 \ mm$$

Thus, the vapour bubble must be compressed to 1.092 mm for the vapour to ignite spontaneously.

 

[kuruman]

Problem Statement: An explosive liquid at temperature 300 K contains a spherical bubble of radius 5 mm, full of its vapour. When a mechanical shock to the liquid causes adiabatic compression of the bubble, what radius of the bubble is required for combustion of the vapour, given that the vapour ignites spontaneously at 1100 degrees C? The ratio of $C_V/(nR)$ is 3.0 for the vapour
Relevant Equations: $$\frac{dP}{P}+\gamma\frac{dV}{V}=0, \ \gamma=\frac{C_P}{C_V}$$

Problem Statement: An explosive liquid at temperature 300 K contains a spherical bubble of radius 5 mm, full of its vapour. When a mechanical shock to the liquid causes adiabatic compression of the bubble, what radius of the bubble is required for combustion of the vapour, given that the vapour ignites spontaneously at 1100 degrees C? The ratio of $C_V/(nR)$ is 3.0 for the vapour
Relevant Equations: $$\frac{dP}{P}+\gamma\frac{dV}{V}=0, \ \gamma=\frac{C_P}{C_V}$$

Hey guys, I tried to solve this problem, got stuck, and noticed there weren't any threads with a solution. I ended up solving it shortly and so decided to share.

In this problem we consider a reversible adiabatic change happening to an ideal gas (the water vapour bubble). To solve this, we start with the following:

$$\frac{dP}{P}+\gamma\frac{dV}{V}=0, \ \gamma=\frac{C_P}{C_V}$$

The question involves the relationship between the volume and temperature of the bubble.
Since its an ideal gas, we can substitute $P=\frac{nRT}{V}$ into the above equation, resulting in:

$$-\frac{dV}{V}(\gamma-1)=\frac{dT}{T}$$

We are given a value for the ratio: $C_V/(nR)=3$. Using the fact that $C_P=nR+C_V$, this ratio is found in the above constant:

$$\gamma-1=\frac{C_P}{C_V}-1=\frac{C_V+nR}{C_V}-1=\frac{nR}{C_V}=1/3$$

The bubble is spherical, so we substitute $V=\frac{4}{3}\pi r^3$ and then integrate both sides from their initial values to their final values:

$$-3\frac{nR}{C_V}\int_{r_i}^{r_f}\frac{dr}{r}=\int^{T_f}_{T_i}\frac{dT}{T}$$

The constant on the LHS becomes -1 due to the value given for $C_V/(nR)$.
Evaluating the integrals and isolating for the final radius $r_f$, we get:

$$\frac{T_i}{T_f}r_i=r_f$$

Plugging in $r_i=5 \ mm$, $T_i=300 \ K$, $T_f=(1100+273) \ K=1373 \ K$, we get our final answer of:

$$r_f=1.092 \ mm$$

Thus, the vapour bubble must be compressed to 1.092 mm for the vapour to ignite spontaneously.

Hi @DarkSkinDude and welcome to PF.

Thank you for your post. I agree with your solution which I got without integrating by observing that $$T~V^{\gamma-1}=const.$$ for an adiabatic process. Since you know the initial volume and temperature, you know the value of the constant.

 

[DarkSkinDude]

Hi @kuruman, thanks for your reply.

Hahaha looks like I overthought this one a little too much. Of course if $TV^{\gamma-1}=const.$
then since $\gamma-1=1/3$, it immediately follows that:

$$\frac{T_1}{T_2}=\frac{r_2}{r_1}$$

I like this solution better. Nice and Clean.