Page: Basic Equations of compressible flow
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1) Continuity equation

2) Bernoulli's equation

3) Momentum equation

4) Equation of state

Continuity equation:-

1) This equation is based on the law of conservation of mass.

2) The law of conservation of mass states that matter can neither be created nor it can be destroyed.

3) That mean mass is constant.

For one dimensional steady flow, the mass per second = $\rho AV$

Where, $\rho$ = mass density

A= area of cross-section

V=Velocity

As mass or mass per second is constant.

$\therefore \rho A V=\text{Constant}$

$d(\rho AV)=0$

$\rho d(AV)+AVd\rho=0$

$\rho[AdV+VdA]+AVd\rho=0$

$\rho AdV+\rho VdA+AVd\rho=0$

Dividing by $\rho AV$, we get

$\left[\dfrac{dV}V+\dfrac {dA}{A}+\dfrac{d\rho}{\rho}=0\right]$

This is the continuity equation.

Bernoulli's equation:-

$\dfrac {dP}\rho+VdV+gdz=0$

Integrating the above equation, we get

$\int \dfrac {dP}\rho+\int VdV+\int gdz=\text{constant}$

or, $\int \dfrac {dP}\rho+\dfrac {V^2}V+gz=\text{constant}$

In case, of incompressible flow, the density $\rho$ is constant and hence the integration of $\dfrac {dP}\rho$ is equal to $\dfrac P\rho$. But in the case of compressible flow, the density is not constant. Hence 'P' cannot be taken outside the integration sign.

A) Bernoulli's equation for the isothermal process:-

$\dfrac P\rho=\text{constant}=c_1...................(1)$

$\therefore \rho=\dfrac P{c_1}$

Hence, $\int \dfrac {dP}\rho=\int \dfrac{dP}{\dfrac P{c_1}}=\int\dfrac{c_1dP}{P}=c_1\int\dfrac{dP}{P}$

($\because c_1$ is constant)

$=c_1log_eP$

$=\dfrac P\rho log_eP$

$\left(\because c_1=\dfrac P\rho\right)$

Substituting the value in the equation of Bernoulli's we get,

$\dfrac P\rho log_eP+\dfrac {V^2}{V}+gz=\text{constant}$

Dividing by 'g',

$\left[\dfrac{P}{\rho g} log_eP+\dfrac {V^2}{2g}+z=\text{constant}\right]$

Bernoulli's equation for compressible flow undergoing the isothermal process.

B) Bernoulli's equation for an adiabatic process:-

$\dfrac P{\rho^K}=\text{constant}c_2$

$\therefore \rho^K=\dfrac P{c_2}$

or, $\rho=\left(\dfrac P{c_2}\right)^{1/K}$

Hence, $\int\dfrac {dP}\rho=\int{\dfrac {dP}{\left(\dfrac P{c_2}\right)^{1/K}}}$

$=\int\dfrac{c_2^{1/K}}{P^{1/4}}$

$d\rho=c_2^{1/K}\int \dfrac 1{P^{1/K}}dP$

$=c_2^{1/k}\int P^{-1/K}dP=c_2^{1/K}.\dfrac {P^{\left(-1\dfrac 1K+1\right)}}{\left(-\dfrac 1K+1\right)}$

$=\left(\dfrac K{K-1}\right)\left(\dfrac P{\rho^K}\right)^{1/K}P^{\left(\dfrac {K-1}K\right)}$

$=\left(\dfrac K{K-1}\right).\dfrac {P^{1/K}}{\rho^{K\times 1/K}}P^{\left(\dfrac {K-1}K\right)}$

$=\left(\dfrac K{K-1}\right).\dfrac {P^{\left(\dfrac 1K+\dfrac {K-1}K\right)}}{\rho}$

$=\left(\dfrac K{K-1}\right).\dfrac {P}{\rho}$

Substituting the value of $\int \dfrac {dP}{\rho}=\left(\dfrac K{K-1}\right).\dfrac {P}{\rho}$, we get

$\left(\dfrac K{K-1}\right).\dfrac {P}{\rho}+\dfrac {V^2}{2}+gz=\text{constant}$

Dividing by 'g',

$\left(\dfrac K{K-1}\right).\dfrac {P}{\rho g}+\dfrac {V^2}{2g}+z=\text{constant}$


Numericals

Q1) A gas is flowing through a horizontal pipe at a temperature of $4^\circ C$. The diameter of the pipe is 8cm and its section 1-1 in this pipe, the pressure is $30.3N/cm^2$(gauge). The diameter of pipe changes from 8cm to 4cm at the section 2-2, where the pressure is $20.3N/cm^2$ (gauge). Find the velocities of gas at these sections assuming an isothermal process. Take $R=287.14Nm/kgK$, and atmospheric pressure $=10N/m^2$.

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Solution: Given:

For section 1-1

Temperature, $t_1=4^\circ C$

Absolute temp, $T_1=4+273=277K$

$D_1=8cm=0.08m$

Therefore, Area, $A_1=\dfrac \pi 4\times D_1^2=\dfrac\pi 4\times (0.08)^2=0.005026m^2$

Pressure, $P_1=30.3N/cm^2$ (gauge)

$P_1=30.3+10=40.3N/cm^2$ (absolute)

$P_1=40.3\times10^4N/m^2$

For section 2-2,

$D_2=4cm=0.04m$

Area, $A_2=\dfrac \pi 4\times D_2^2=\dfrac\pi 4\times (0.04)^2=0.001256m^2$

Pressure, $P_2=20.3N/cm^2$ (gauge)

$P_2=20.3+10=30.3N/cm^2$ (absolute)

$P_2=30.3\times10^4N/m^2$

Gas constant, $R=287.14N-m/Kg-K$

Ratio of specific heat, $K=1.4$

Applying the continuity equation at sections 1 and 2, we get

$\rho_1A_1V_1=\rho_2V_2A_2$

$\dfrac {V_2}{V_1}=\dfrac {\rho_1A_1}{\rho_2A_2}$

$\dfrac {V_2}{V_1}=\dfrac {\rho_1\times 0.005026}{\rho_2\times 0.001256}$

$\dfrac {V_2}{V_1}=\dfrac {\rho_1}{\rho_2}\times 4.................(1)$

For isothermal process,

$\dfrac {P_1}{\rho_1}=\dfrac {P_2}{\rho_2}$

$\dfrac {\rho_1}{\rho_2}=\dfrac {P_1}{P_2}$

$\dfrac {\rho_1}{\rho_2}=\dfrac {40.3\times 10^4}{30.3\times 10^4}$

$\dfrac {\rho_1}{\rho_2}=1.33$

Substitute the value of $\left(\dfrac {\rho_1}{\rho_2}\right)$ in equation (1),

$\dfrac {V_2}{V_1}=1.33\times 4$

$V_2=5.32V_1....................(2)$

Applying Bernoulli's equation at section 1-1 and 2-2 for isothermal process,

$\dfrac {P_1}{\rho_1g}log_eP_1+\dfrac {V_1^2}{2g}+z_1=\dfrac {P_2}{\rho_2 g}log_eP_2+\dfrac {V_2^2}{2g}+z_2$

$z_1=z_2$ (for horizontal pipe)

$\therefore \dfrac {P_1}{\rho_1g}log_eP_1+\dfrac {V_1^2}{2g}=\dfrac {P_2}{\rho_2 g}log_eP_2+\dfrac {V_2^2}{2g}$

$\therefore \dfrac {P_1}{\rho_1g}log_eP_1-\dfrac {P_2}{\rho_2 g}log_eP_2=\dfrac {V_2^2}{2g}-\dfrac {V_1^2}{2g}$

But for isothermal process, $\dfrac {P_1}{\rho_1}=\dfrac {P_2}{\rho_2}$

$\dfrac {P_1}{\rho_1g}log_eP_1-\dfrac {P_1}{\rho_1 g}log_eP_2=\dfrac {V_2^2}{2g}-\dfrac {V_1^2}{2g}$

$\dfrac {P_1}{\rho_1g}\left[log_e\dfrac {P_1}{P_2}\right]=\dfrac {(5.32V_1)^2}{2g}-\dfrac {V_1^2}{2g}$

$\dfrac {P_1}{\rho_1g}\left[log_e 1.33\right]=27.30\dfrac {V_1^2}{2g}$

$\dfrac {P_1}{\rho_1}=\dfrac {27.30}{2\times 0.285}v_1^2=27.30\dfrac {v_1^2}{2g}................(3)$

Now, $\dfrac P\rho=RT$ from the equation of state.

Forsection (1-1):-

$\dfrac {P_1}{\rho_1}=RT_1=287.14\times 277$

$\dfrac {P_1}{\rho_1}=79537.4$

Substituting the value in equation 3, we get

$79537.4=47.894V_1^2$

$V_1^2=\dfrac {79537.4}{47.894}$

$V_1=\sqrt{\dfrac {79537.4}{47.894}}$

$V_1=40.75m/s$

Therefore, equation (2),

$V_2=5.32\times V=5.32\times 40.75$

$V_2=216.79m/s$


Q2) A gas is flowing through a horizontal pipe which is having an area of cross-section as $40cm^2$, where the pressure is $40N/cm^2$ and temperature is $15^\circ C$. At another section, the area of cross section is $20cm^2$ and pressure is $30N/cm^2$ (gauge). If the mass rate of flow of gas through the pipe is 0.5 kg/s. Find the velocities of gas at these sections, assuming an isothermal change. $R=292N-m/kg-K$ and atmospheric pressure $= 10N/cm^2$

Solution: Given:

For section 1:

Temperature, $t_1=15^\circ C$

Absolute temp, $T_1=15+273=288K$

Area, $A_1=40cm=40\times 10^{-4}m^2$

Pressure, $P_1=40N/cm^2$ (gauge)

$P_1=40+10=50N/cm^2$ (absolute)

$P_1=50\times10^4N/m^2$

For section 2:

Area, $A_2=20cm=20\times 10^{-4}m^2$

Pressure, $P_2=30N/cm^2$ (gauge)

$P_2=30+10=40N/cm^2$ (absolute)

$P_2=40\times10^4N/m^2$

Gas constant, $R=292N-m/kg-K$

$\therefore \dfrac {P_1}{\rho_1}=RT_1$

$\rho_1=\dfrac {P_1}{RT_1}$

$\rho_1=\dfrac {50\times 10^4}{292\times 288}$

$\rho_1=5.945kg/m^3$

Mass rate of flow $=\rho_1A_1V_1=0.5$

$0.5=5.945\times 40\times 10^{-4}\times V_1$

$V_1=\dfrac {0.5}{5.945\times 40\times 10^{-4}}$

$V_1=21.02m/s$

For isothermal process, temperature is constant and hence temperature at section 2 is also 288K.

$T_2=288K$

$\therefore \dfrac {P_2}{\rho_2}=RT_2$

$\rho_2=\dfrac {P_2}{RT_2}$

$\rho_2=\dfrac {40\times 10^4}{292\times 288}$

$\rho_2=4.756kg/m^3$

Therefore, mass rate of flow = $\rho_2A_2V_2=0.5$

$\therefore 0.5=\rho_2A_2V_2$

$V_2=\dfrac {0.5}{4.756\times 20\times 10^{-4}}$

$V_2=52.565 \ m/s$

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modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 270 written 7 months ago by gravatar for Syedahina Mohi Syedahina Mohi0
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