A gas is flowing through a horizontal pipe which is having an area of cross-section as 40cm240cm2, where the pressure is 40N/cm240N/cm2 and temperature is 15∘C15∘C. At another section, the area of cross section is 20cm220cm2 and pressure is 30N/cm230N/cm2 (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−KR=292N−m/kg−K and atmospheric pressure =10N/cm2=10N/cm2

 A gas is flowing through a horizontal pipe which is having an area of cross-section as 

40cm2$40c{m}^2$, where the pressure is 40N/cm2$40N/c{m}^2$ and temperature is 15∘C${15}^{\circ }C$. At another section, the area of cross section is 20cm2$20c{m}^2$ and pressure is 30N/cm2$30N/c{m}^2$ (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$R=292N-m/kg-K$ and atmospheric pressure =10N/cm2$=10N/c{m}^2$

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Solution: 

Given:

For section 1:

Temperature, t1=15∘C$t_1={15}^{\circ }C$

Absolute temp, T1=15+273=288K$T_1=15+273=288K$

Area, A1=40cm=40×10−4m2$A_1=40cm=40\times {10}^{-4}{m}^2$

Pressure, P1=40N/cm2$P_1=40N/c{m}^2$ (gauge)

P1=40+10=50N/cm2$P_1=40+10=50N/c{m}^2$ (absolute)

P1=50×104N/m2$P_1=50\times {10}^{4}N/m^2$

For section 2:

Area, A2=20cm=20×10−4m2$A_2=20cm=20\times {10}^{-4}{m}^2$

Pressure, P2=30N/cm2$P_2=30N/c{m}^2$ (gauge)

P2=30+10=40N/cm2$P_2=30+10=40N/c{m}^2$ (absolute)

P2=40×104N/m2$P_2=40\times {10}^{4}N/m^2$

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

∴P1ρ1=RT1$\therefore {\displaystyle \frac{P_1}{{\rho }_1}}=R{T}_1$

ρ1=P1RT1${\rho }_1={\displaystyle \frac{P_1}{R{T}_1}}$

ρ1=50×104292×288${\rho }_1={\displaystyle \frac{50\times {10}^4}{292\times 288}}$

ρ1=5.945kg/m3${\rho }_1=5.945kg/m^3$

Mass rate of flow =ρ1A1V1=0.5$={\rho }_1{A}_1{V}_1=0.5$

0.5=5.945×40×10−4×V1$0.5=5.945\times 40\times {10}^{-4}\times V_1$

V1=0.55.945×40×10−4$V_1={\displaystyle \frac{0.5}{5.945\times 40\times {10}^{-4}}}$

V1=21.02m/s$V_1=21.02m/s$

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

T2=288K$T_2=288K$

∴P2ρ2=RT2$\therefore {\displaystyle \frac{P_2}{{\rho }_2}}=R{T}_2$

ρ2=P2RT2${\rho }_2={\displaystyle \frac{P_2}{R{T}_2}}$

ρ2=40×104292×288${\rho }_2={\displaystyle \frac{40\times {10}^4}{292\times 288}}$

ρ2=4.756kg/m3${\rho }_2=4.756kg/m^3$

Therefore, mass rate of flow = ρ2A2V2=0.5${\rho }_2{A}_2{V}_2=0.5$

∴0.5=ρ2A2V2$\therefore 0.5={\rho }_2{A}_2{V}_2$

V2=0.54.756×20×10−4$V_2={\displaystyle \frac{0.5}{4.756\times 20\times {10}^{-4}}}$

V2=52.565 m/s