A gas is flowing through a horizontal pipe at a temperature of 4∘C4∘C. The diameter of the pipe is 8cm and its section 1-1 in this pipe, the pressure is 30.3N/cm230.3N/cm2(gauge). The diameter of pipe changes from 8cm to 4cm at the section 2-2, where the pressure is 20.3N/cm220.3N/cm2 (gauge). Find the velocities of gas at these sections assuming an isothermal process. Take R=287.14Nm/kgKR=287.14Nm/kgK, and atmospheric pressure =10N/m2=10N/m2.

A gas is flowing through a horizontal pipe at a temperature of 4∘C$4^{\circ }C$. The diameter of the pipe is 8cm and its section 1-1 in this pipe, the pressure is 30.3N/cm2$30.3N/c{m}^2$(gauge). The diameter of pipe changes from 8cm to 4cm at the section 2-2, where the pressure is 20.3N/cm2$20.3N/c{m}^2$ (gauge). Find the velocities of gas at these sections assuming an isothermal process. Take R=287.14Nm/kgK$R=287.14Nm/kgK$, and atmospheric pressure =10N/m2$=10N/m^2$.

Solution: 

Given:

For section 1-1

Temperature, t1=4∘C$t_1=4^{\circ }C$

Absolute temp, T1=4+273=277K$T_1=4+273=277K$

D1=8cm=0.08m$D_1=8cm=0.08m$

Therefore, Area, A1=π4×D21=π4×(0.08)2=0.005026m2$A_1={\displaystyle \frac{\pi }{4}}\times D_1^2={\displaystyle \frac{\pi }{4}}\times (0.08{)}^2=0.005026m^2$

Pressure, P1=30.3N/cm2$P_1=30.3N/c{m}^2$ (gauge)

P1=30.3+10=40.3N/cm2$P_1=30.3+10=40.3N/c{m}^2$ (absolute)

P1=40.3×104N/m2$P_1=40.3\times {10}^{4}N/m^2$

For section 2-2,

D2=4cm=0.04m$D_2=4cm=0.04m$

Area, A2=π4×D22=π4×(0.04)2=0.001256m2$A_2={\displaystyle \frac{\pi }{4}}\times D_2^2={\displaystyle \frac{\pi }{4}}\times (0.04{)}^2=0.001256m^2$

Pressure, P2=20.3N/cm2$P_2=20.3N/c{m}^2$ (gauge)

P2=20.3+10=30.3N/cm2$P_2=20.3+10=30.3N/c{m}^2$ (absolute)

P2=30.3×104N/m2$P_2=30.3\times {10}^{4}N/m^2$

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

Ratio of specific heat, K=1.4$K=1.4$

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

ρ1A1V1=ρ2V2A2${\rho }_1{A}_1{V}_1={\rho }_2{V}_2{A}_2$

V2V1=ρ1A1ρ2A2${\displaystyle \frac{V_2}{V_1}}={\displaystyle \frac{{\rho }_1{A}_1}{{\rho }_2{A}_2}}$

V2V1=ρ1×0.005026ρ2×0.001256${\displaystyle \frac{V_2}{V_1}}={\displaystyle \frac{{\rho }_1\times 0.005026}{{\rho }_2\times 0.001256}}$

V2V1=ρ1ρ2×4.................(1)${\displaystyle \frac{V_2}{V_1}}={\displaystyle \frac{{\rho }_1}{{\rho }_2}}\times \mathrm{4.................}(1)$

For isothermal process,

P1ρ1=P2ρ2${\displaystyle \frac{P_1}{{\rho }_1}}={\displaystyle \frac{P_2}{{\rho }_2}}$

ρ1ρ2=P1P2${\displaystyle \frac{{\rho }_1}{{\rho }_2}}={\displaystyle \frac{P_1}{P_2}}$

ρ1ρ2=40.3×10430.3×104${\displaystyle \frac{{\rho }_1}{{\rho }_2}}={\displaystyle \frac{40.3\times {10}^4}{30.3\times {10}^4}}$

ρ1ρ2=1.33${\displaystyle \frac{{\rho }_1}{{\rho }_2}}=1.33$

Substitute the value of (ρ1ρ2)$\left({\displaystyle \frac{{\rho }_1}{{\rho }_2}}\right)$ in equation (1),

V2V1=1.33×4${\displaystyle \frac{V_2}{V_1}}=1.33\times 4$

V2=5.32V1....................(2)$V_2=5.32V_1....................(2)$

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

P1ρ1glogeP1+V212g+z1=P2ρ2glogeP2+V222g+z2${\displaystyle \frac{P_1}{{\rho }_{1}g}}lo{g}_e{P}_1+{\displaystyle \frac{V_1^2}{2g}}+z_1={\displaystyle \frac{P_2}{{\rho }_{2}g}}lo{g}_e{P}_2+{\displaystyle \frac{V_2^2}{2g}}+z_2$

z1=z2$z_1=z_2$ (for horizontal pipe)

∴P1ρ1glogeP1+V212g=P2ρ2glogeP2+V222g$\therefore {\displaystyle \frac{P_1}{{\rho }_{1}g}}lo{g}_e{P}_1+{\displaystyle \frac{V_1^2}{2g}}={\displaystyle \frac{P_2}{{\rho }_{2}g}}lo{g}_e{P}_2+{\displaystyle \frac{V_2^2}{2g}}$

∴P1ρ1glogeP1−P2ρ2glogeP2=V222g−V212g$\therefore {\displaystyle \frac{P_1}{{\rho }_{1}g}}lo{g}_e{P}_1-{\displaystyle \frac{P_2}{{\rho }_{2}g}}lo{g}_e{P}_2={\displaystyle \frac{V_2^2}{2g}}-{\displaystyle \frac{V_1^2}{2g}}$

But for isothermal process, P1ρ1=P2ρ2${\displaystyle \frac{P_1}{{\rho }_1}}={\displaystyle \frac{P_2}{{\rho }_2}}$

P1ρ1glogeP1−P1ρ1glogeP2=V222g−V212g${\displaystyle \frac{P_1}{{\rho }_{1}g}}lo{g}_e{P}_1-{\displaystyle \frac{P_1}{{\rho }_{1}g}}lo{g}_e{P}_2={\displaystyle \frac{V_2^2}{2g}}-{\displaystyle \frac{V_1^2}{2g}}$

P1ρ1g[logeP1P2]=(5.32V1)22g−V212g${\displaystyle \frac{P_1}{{\rho }_{1}g}}\left[lo{g}_e{\displaystyle \frac{P_1}{P_2}}\right]={\displaystyle \frac{(5.32V_1{)}^2}{2g}}-{\displaystyle \frac{V_1^2}{2g}}$

P1ρ1g[loge1.33]=27.30V212g${\displaystyle \frac{P_1}{{\rho }_{1}g}}\left[lo{g}_{e}1.33\right]=27.30{\displaystyle \frac{V_1^2}{2g}}$

P1ρ1=27.302×0.285v21=27.30v212g................(3)$$P1ρ1=27.302×0.285v12=27.30v122g................(3)

Now, Pρ=RT${\displaystyle \frac{P}{\rho }}=RT$ from the equation of state.

Forsection (1-1):-

P1ρ1=RT1=287.14×277${\displaystyle \frac{P_1}{{\rho }_1}}=R{T}_1=287.14\times 277$

P1ρ1=79537.4${\displaystyle \frac{P_1}{{\rho }_1}}=79537.4$

Substituting the value in equation 3, we get

79537.4=47.894V21$79537.4=47.894V_1^2$

V21=79537.447.894$V_1^2={\displaystyle \frac{79537.4}{47.894}}$

V1=79537.447.894−−−−−−−√$V_1=\sqrt{{\displaystyle \frac{79537.4}{47.894}}}$

V1=40.75m/s$V_1=40.75m/s$

Therefore, equation (2),

V2=5.32×V=5.32×40.75$V_2=5.32\times V=5.32\times 40.75$

V2=216.79m/s