Chapter 12, Problem 106P

Air enters a 15-m-long, 4-cm-diameter adiabatic duct at $V_1\text{ = }70m/s,T_1\text{ = }500K,and{P}_1\text{ = }300kPa.$ The average friction factor for the duct is estimated to be 0.023. Determine the Mach number at the duct exit, the exit velocity, and the mass flow rate of air.

To determine

The exit Mach number, the exit velocity and the mass flow rate of air.

Answer to Problem 106P

The exit Mach number is $0.1873$.

The exit velocity is $84.16\text{m}/\text{s}$.

The mass flow rate of air is $0.183\text{kg}/\text{s}$.

Explanation of Solution

Given information:

The length of pipe is $15\text{m}$, diameter of pipe is $4\text{cm}$, inlet velocity is $70\text{m}/\text{s}$, inlet temperature is $500\text{K}$, inlet pressure is $300\text{kPa}$, average friction factor is $0.023$.

Calculation:

Expression for inlet Mach number

   ${\text{Ma}}_{\text{1}}=\frac{V_1}{\sqrt{kR{T}_1}}$     ...... (I)

Here, inlet velocity is $V_1$, gas constant is $R$, specific heat ratio is $k$, inlet temperature is $T_1$.

Expression for length required for sonic flow for inlet condition

   $\frac{f{L}_1^{*}}{D_h}=\frac{1-{\text{Ma}}_1^2}{k{\text{Ma}}_1^2}+\frac{k+1}{2k}\ln \frac{\left(k+1\right){\text{Ma}}_1^2}{2+\left(k-1\right){\text{Ma}}_1^2}$     ...... (II)

Here, friction factor is $f$, diameter of pipe is $D_h$, inlet sonic length is $L_1^{*}$.

Expression for inlet density

   ${\rho }_1=\frac{P_1}{R{T}_1}$     ...... (III)

Here, inlet pressure is $P_1$.

Expression for length required for sonic flow for outlet condition

   $\frac{f{L}_2^{*}}{D_h}=\frac{f{L}_1^{*}}{D_h}-\frac{fL}{D_h}$     ...... (IV)

Here, length of pipe is $L$, outlet sonic length is $L_2^{*}$.

Expression for mass flow rate of air

   $\dot{m}={\rho }_1\times \left(\frac{\pi }{4}{D}_h{}^2\right)\times V_1$     ...... (V)

Refer to Table-A-1 “Molar mass, gas constant, and ideal gas specific heat of some substances” to obtain gas constant of air as $287\text{J}/\text{kg}\cdot \text{K}$ and specific heat ratio as $1.4$

Substitute $70\text{m}/\text{s}$ for $V_1$, $500\text{K}$ for $T_1$, $1.4$ for $k$ and $287\text{J}/\text{kg}\cdot \text{K}$ for $R$ in Equation (I).

   $\begin{array}{c}{\text{Ma}}_{\text{1}}=\frac{70\text{m}/\text{s}}{\sqrt{1.4\left(287\text{J}/\text{kg}\cdot \text{K}\right)500\text{K}}}\\ =\frac{70\text{m}/\text{s}}{\sqrt{200900\text{J}/\text{kg}\times \frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/\text{kg}\cdot {\text{s}}^{\text{2}}}{1\text{J}/\text{kg}}}}\\ =\frac{70\text{m}/\text{s}}{448.2186\text{m}/\text{s}}\\ =0.1561\end{array}$

Refer to Table-A-15 “Rayleigh flow function for an ideal gas with $k=1.4$ ” at Mach number $0.129$ to obtain ratio of pressure temperature and velocity.

Relation of velocity at initial state and sonic state

   $\frac{V_1}{V^{*}}=0.170$     ...... (VI)

Here, inlet velocity at sonic state is $V^{*}$.

Substitute $70\text{m}/\text{s}$ for $V_1$ in Equation (VI).

   $\begin{array}{l}\frac{70\text{m}/\text{s}}{V^{*}}=0.170\\ V^{*}=\frac{70\text{m}/\text{s}}{0.170}\\ V^{*}=411.7647\text{m}/\text{s}\end{array}$

Substitute $0.1561$ for ${\text{Ma}}_{\text{1}}$, $1.4$ for $k$ in Equation (II).

   $\begin{array}{c}\frac{f{L}_1^{*}}{D_h}=\frac{1-{\left(0.1561\right)}^2}{1.4{\left(0.1561\right)}^2}+\frac{1.4+1}{2\times 1.4}\ln \frac{\left(1.4+1\right){\left(0.1561\right)}^2}{2+\left(1.4-1\right){\left(0.1561\right)}^2}\\ =\frac{1-{\left(0.1561\right)}^2}{1.4{\left(0.1561\right)}^2}+\frac{6}{7}\ln \frac{\left(2.4\right){\left(0.1561\right)}^2}{2+\left(0.4\right){\left(0.1561\right)}^2}\\ =25.56\end{array}$

Substitute $25.56$ for $\frac{f{L}_1^{*}}{D_h}$, $15\text{m}$ for $L$, $0.023$ for $f$ and $4\text{cm}$ for $D_h$ in Equation (IV).

   $\begin{array}{c}\frac{f{L}_2^{*}}{D_h}=25.56-\frac{0.023\times 15\text{m}}{4\text{cm}}\\ =25.56-\frac{0.023\times 15\text{m}}{4\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\\ =25.56-\frac{0.023\times 15\text{m}}{0.04\text{m}}\\ =16.933\end{array}$

Refer to Table-A-16 “Fanno flow function for an ideal gas with $k=1.4$ ” at $\frac{f{L}_2^{*}}{D_h}$ as $16.933$ to obtain Mach number at exit as $0.1873$ and velocity ratio as $0.2044$.

Relation of velocity at exit and sonic state

   $\frac{V_2}{V^{*}}=0.2044$     ...... (VII)

Substitute $411.7647\text{m}/\text{s}$ for $V^{*}$ in Equation (VII).

   $\begin{array}{l}\frac{V_2}{411.7647\text{m}/\text{s}}=0.2044\\ V_2=411.7647\text{m}/\text{s}\times 0.2044\\ V_2=84.16\text{m}/\text{s}\end{array}$

Substitute $300\text{kPa}$ for $P_1$, $287\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $500\text{K}$ for $T_1$ in Equation (III).

   $\begin{array}{c}{\rho }_1=\frac{300\text{kPa}}{287\text{J}/\text{kg}\cdot \text{K}\times 500\text{K}}\\ =\frac{300\text{kPa}\times \frac{1000\text{Pa}}{1\text{kPa}}}{143500\text{J}/\text{kg}\times \frac{1\text{N}\cdot \text{m}/\text{kg}}{1\text{J}/\text{kg}}}\\ =\frac{300000\text{Pa}\times \frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}}{143500\text{N}\cdot \text{m}/\text{kg}}\\ =2.09059\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $2.09059\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_1$, $4\text{cm}$ for $D_h$ and $70\text{m}/\text{s}$ for $V_1$ in Equation (V).

   $\begin{array}{c}\dot{m}=2.09059\text{kg}/{\text{m}}^{\text{3}}\times \left(\frac{\pi }{4}{\left(4\text{cm}\right)}^2\right)\times 70\text{m}/\text{s}\\ =2.09059\text{kg}/{\text{m}}^{\text{3}}\times \left(\frac{\pi }{4}{\left(4\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\right)\times 70\text{m}/\text{s}\\ =2.09059\text{kg}/{\text{m}}^{\text{3}}\times \left(\frac{\pi }{4}{\left(0.04\text{m}\right)}^2\right)\times 70\text{m}/\text{s}\\ =0.183\text{kg}/\text{s}\end{array}$

Conclusion:

The exit Mach number is $0.1873$.

The exit velocity is $84.16\text{m}/\text{s}$.

The mass flow rate of air is $0.183\text{kg}/\text{s}$.