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.

Answer and Explanation:

(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.


Learn more about this topic:

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What is Static Pressure? - Definition & Formula

from Introduction to Engineering

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