Chapter 12, Problem 109P

Air enters a 15-cm-diameter adiabatic duct with inlet conditions of $V_1\text{ = }150m/s,T_1\text{ = }500K,and{P}_1\text{ = 200 }kPa.$ For an average friction factor of 0.014, determine the duct length from the inlet where the inlet velocity doubles. Also determine the pressure drop along that section of the duct.

To determine

The duct length when velocity is doubled and the pressure drop along the length of duct.

Answer to Problem 109P

The duct length for velocity to be doubled is $39.583\text{m}$.

The pressure drop along the length of duct is $106.728\text{kPa}$.

Explanation of Solution

Given information:

The diameter of pipe is $15\text{cm}$, inlet velocity is $150\text{m}/\text{s}$, inlet temperature is $500\text{K}$, inlet pressure is $200\text{kPa}$, average friction factor is $0.014$.

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 pressure drop

   $\Delta P=P_1-P_2$     ...... (III)

Here, inlet pressure is $P_1$.

Expression for length required for velocity double

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

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

Calculation:

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 $150\text{m}/\text{s}$ 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).

   $\begin{array}{c}{\text{Ma}}_{\text{1}}=\frac{150\text{m}/\text{s}}{\sqrt{1.4\left(287\text{J}/\text{kg}\cdot \text{K}\right)500\text{K}}}\\ =\frac{150\text{m}/\text{s}}{\sqrt{200900\text{J}/\text{kg}\times \frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/\text{kg}\cdot {\text{s}}^{\text{2}}}{1\text{J}/\text{kg}}}}\\ =\frac{150\text{m}/\text{s}}{448.2186\text{m}/\text{s}}\\ =0.33465\end{array}$

Refer to Table-A-15 “Rayleigh flow function for an ideal gas with $k=1.4$ ” at Mach number $0.33465$ to obtain ratio of pressure temperature and velocity.

Relation of velocity at initial state and sonic state

   $\frac{V_1}{V^{*}}=0.362553$     ...... (V)

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

Substitute $150\text{m}/\text{s}$ for $V_1$ in Equation (VI).

   $\begin{array}{l}\frac{150\text{m}/\text{s}}{V^{*}}=0.362553\\ V^{*}=\frac{150\text{m}/\text{s}}{0.362553}\\ V^{*}=413.732\text{m}/\text{s}\end{array}$

Relation of pressure at initial state and sonic state

   $\frac{P_1}{P^{*}}=3.237351$     ...... (VI)

Here, inlet pressure at sonic state is $P^{*}$.

Substitute $200\text{kPa}$ for $P_1$ in Equation (VI).

   $\begin{array}{l}\frac{200\text{kPa}}{P^{*}}=3.237351\\ P^{*}=\frac{200\text{kPa}}{3.237351}\\ P^{*}=61.778\text{kPa}\end{array}$

Substitute $0.33465$ for ${\text{Ma}}_{\text{1}}$, $1.4$ for $k$ in Equation (II).

   $\begin{array}{c}\frac{f{L}_1^{*}}{D_h}=\frac{1-{\left(0.33465\right)}^2}{1.4{\left(0.33465\right)}^2}+\frac{1.4+1}{2\times 1.4}\ln \frac{\left(1.4+1\right){\left(0.33465\right)}^2}{2+\left(1.4-1\right){\left(0.33465\right)}^2}\\ =\frac{1-{\left(0.33465\right)}^2}{1.4{\left(0.33465\right)}^2}+\frac{6}{7}\ln \frac{\left(2.4\right){\left(0.33465\right)}^2}{2+\left(0.4\right){\left(0.33465\right)}^2}\\ =3.9245\end{array}$

Substitute $300\text{m}/\text{s}$ for $V_2$ and $413.732\text{m}/\text{s}$ for $V^{*}$ in the ratio $\frac{V_2}{V^{*}}$.

   $\begin{array}{c}\frac{V_2}{V^{*}}=\frac{300\text{m}/\text{s}}{413.732\text{m}/\text{s}}\\ =0.7251\times \frac{\text{1}\text{m}/\text{s}}{\text{1}\text{m}/\text{s}}\\ =0.7251\end{array}$

Refer to Table-A-16 “Fanno flow function for an ideal gas with $k=1.4$ ” at $\frac{V_2}{V^{*}}$ as $0.7251$ to obtain Mach number at exit as $0.693$, $\frac{f{L}_2^{*}}{D_h}$ as $0.23$ and pressure ratio as $1.5098$.

Relation of pressure at exit and sonic state

   $\frac{P_2}{P^{*}}=1.5098$     ...... (VII)

Substitute $61.778\text{kPa}$ for $P^{*}$ in Equation (VII).

   $\begin{array}{l}\frac{P_2}{61.778\text{kPa}}=1.5098\\ P_2=61.778\text{kPa}\times 1.5098\\ P_2=93.272\text{kPa}\end{array}$

Substitute $93.272\text{kPa}$ for $P_2$ and $200\text{kPa}$ for $P_1$ in Equation (III).

   $\begin{array}{c}\Delta P=200\text{kPa}-93.272\text{kPa}\\ =106.728\text{kPa}\end{array}$

Substitute $0.23$ for $\frac{f{L}_2^{*}}{D_h}$, $3.9245$ for $\frac{f{L}_1^{*}}{D_h}$, $0.014$ for $f$ and $15\text{cm}$ for $D_h$ in Equation (IV).

   $\begin{array}{c}L=\frac{\left(3.9245-0.23\right)}{\left(\frac{0.014}{15\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)}\\ =\frac{\left(3.9245-0.23\right)}{\left(\frac{0.014}{0.15\text{m}}\right)}\\ =39.583\text{m}\end{array}$

Conclusion:

The duct length for velocity to be doubled is $39.583\text{m}$.

The pressure drop along the length of duct is $106.728\text{kPa}$.