Speed of sound

Given equation:

$$\dot{m}=\rho A\vec{v}$$

 

$$\rho A\vec{v}=(\rho+\Delta\rho)A(c-\Delta\vec{v})$$

$$\rho c=(\rho+\Delta\rho)(c-\Delta\vec{v})$$

$$\rho c = \rho c - \rho\Delta\vec{v}+\Delta\rho c-\Delta\rho\Delta\vec{v}$$

 

We can neglect the last term of equation

$$0=-\rho\Delta\vec{v}+\Delta\rho c$$

$$c\Delta\rho=\rho\Delta\vec{v}$$

 

$$c\frac{\Delta\rho}{\rho}=\Delta\vec{v}\cdots(1)$$

 

Given equation:

$$PA+\rho \vec{v} \vec{v} A=(p+\Delta p)A+(\rho+\Delta \rho)(\vec{v}-\Delta\vec{v})(\vec{v}-\Delta\vec{v})A$$

 

As we can rearrange the two term like below:

$$(\rho+\Delta\rho)(\vec{v}-\Delta\vec{v})=\rho\vec{v}$$

 

$$PA+\rho \vec{v} \vec{v} A = (p+\Delta p)A+\rho \vec{v}(\vec{v}-\Delta\vec{v})A$$

 

We can omit \(A\) and others...

$$P+\rho\vec{v}\vec{v}=(p+\Delta p)+\rho\vec{v}(\vec{v}-\Delta\vec{v})$$

$$\rho\vec{v}\vec{v}=\Delta p+\rho\vec{v}(\vec{v}-\Delta\vec{v})$$

 

$$\rho\vec{v}\vec{v}=\Delta p+\rho\vec{v}\vec{v}-\rho\vec{v}\Delta\vec{v}$$

$$0=\Delta p-\rho\vec{v}\Delta\vec{v}$$

 

$$\rho\vec{v}\Delta\vec{v}=\Delta p$$

$$\Delta\vec{v}=\frac{1}{\rho\vec{v}}\Delta p$$

 

We can replace \(\vec{v}\) to \(c\) which means 'speed of sound'

$$\Delta \vec{v}=\frac{1}{\rho c}\Delta p\cdots(2)$$

 

Now use equation \((1)\) and \((2)\)$$\Delta\vec{v}=c\frac{\Delta p}{\rho}\cdots(1)$$$$\Delta\vec{v}=\frac{1}{\rho c}\Delta p\cdots(2)$$

 

$$\frac{1}{\rho c}\Delta p=c\frac{\Delta \rho}{\rho}$$$$\frac{1}{c}\Delta p=c\Delta\rho$$$$\frac{\Delta p}{\Delta \rho}=c^{2}$$

 

$$a^{2}=\left(\frac{\Delta p}{\Delta \rho}\right)_{s}=\left(\frac{\partial p}{\partial \rho}\right)_{s}$$