# Consider a simple model of combustion inside a cavity. A cylinder of mass M can freely move up...

## Question:

Consider a simple model of combustion inside a cavity. A cylinder of mass {eq}M {/eq} can freely move up and down inside a cavity with a cross sectional area {eq}A {/eq}. The connection between the cylinder wall and the cavity wall is airtight. Below the cylinder lies an air cavity, which contains an air mass {eq}m {/eq}. Initially, the entire system is in thermal and mechanical equilibrium at the ambient temperature {eq}T_a {/eq}. The air within the cavity is then suddenly heated to a temperature {eq}T_{hot} {/eq}. During this initial heating the cylinder does not move.

a) Show that the pressure in the air cavity immediately following the initial heating is {eq}P= (P_a + \frac {Mg}{A}) \frac {T_{hot}} {T_a} {/eq}

b) Assuming that there is no friction between the cylinder and the cavity wall and that the cavity and cylinder are insulated, show that the position x, of a slowly moving cylinder satisfies the following differential equation (where h is the initial position of the bottom of the cylinder): {eq}\frac {T_{hot}}{T_a} ( P_a + \frac {Mg}{A}) (\frac {h} {x}) ^{\gamma} A-{P_a}{A} -Mg = M \frac {d^2x}{dt^2} {/eq}

c) Show that the final equilibrium position of the cylinder is {eq}x = (\frac {T_{hot}}{T_a})^{1/{\gamma}} h {/eq}

d) Is this process reversible (yes or no)? Explain why>

## Reversible Process:

The process where the path can be retraced back to its initial condition, this process is known to be reversible process. For a thermodynamic process to be reversible the work done in the process should be zero.

(a) Initial pressure for the given figure can be known by free body diagram

So, initial pressure is

{eq}P_a+\frac{mg}{A} {/eq}

Also from Ideal gas equation

PV=nRT

Since, cylinder doesn't move in heating means volume is constant.

So,

{eq}\frac{P_1}{T_1}=\frac{P_2}{T_2}\\ P_2=P_1\frac{T_2}{T_1}\\ T_2=T_{hot}\\ T_1=T_a\\ P_2=P_1\frac{T_{hot}}{T_a}\\ P_2=(P_a+\frac{mg}{A})\frac{T_{hot}}{T_a} {/eq}

(b) The pressure developed after heating will be

{eq}P_2=\frac{T_{hot}}{T_a}(P+\frac{mg}{A}) {/eq}

Effective area is {eq}A_{eff.}={({\frac{h}{x}})^{\gamma}}A {/eq}

The net force for expansion will be as

{eq}P_{new}=P_2A_{eff.}-P_aA {/eq}

Balancing of force

Net force = mass x accelration

So,

{eq}(P+\frac{mg}{A})\frac{T_{hot}}{T_a}(\frac{h}{x})^{\gamma}A-P_aA-Mg=M\frac{d^2x}{dt^2} {/eq}

(c) For equilibrium condition

{eq}M\frac{d^2x}{dt^2}=0\\ (P+\frac{mg}{A})\frac{T_{hot}}{T_a}(\frac{h}{x})^{\gamma}A=P_aA-mg\\ (P+\frac{mg}{A})\frac{T_{hot}}{T_a}(\frac{h}{x})^{\gamma}=(P+\frac{mg}{A})\\ \frac{T_{hot}}{T_a}(\frac{h}{x})^{\gamma}=1\\ x=(\frac{T_{hot}}{T_a})^{1/\gamma}h {/eq}

(d) No the process is not reversible because the process cannot be taken back to its initial start stage.