Chapter 6, Problem 129RQ

To determine

The velocity of the nitrogen at the pipe’s inlet and outlet.

Explanation of Solution

Given:

The initial pressure $\left(P_1\right)$ is $100\text{ psia}\text{.}$

The initial temperature $\left(T_1\right)$ is $120°\text{F}$.

The final pressure $\left(P_2\right)$ is $50\text{ psia}\text{.}$

The final temperature $\left(T_2\right)$ is $70°\text{F}$.

Calculation:

Write the formula for mass flow rate of nitrogen gas at initial and final.

  ${\dot{m}}_1=\frac{A_1{V}_1{P}_1}{R{T}_1}$        (I)

Write the formula for mass flow rate of nitrogen gas at initial and final.

  ${\dot{m}}_2=\frac{A_2{V}_2{P}_2}{R{T}_2}$        (II)

Here, the cross sectional area of the pipe is $A$, the velocity is $V$, the gas constant is $R$ and the temperature is $T$; the subscripts 1 and 2 indicates the inlet and outlet condition of nitrogen gas.

Here, the inlet and exit mass flow rates are equal and  the diameter of the pipe is constant $\left(A_1=A_2\right)$.

Equate the Equations (I) and (II) to obtain the exit velocity $V_2$ .

  $\begin{array}{c}\frac{A_1{V}_1{P}_1}{R{T}_1}=\frac{A_2{V}_2{P}_2}{R{T}_2}\\ V_2=\frac{A_1{V}_1{P}_1}{R{T}_1}\times \frac{R{T}_2}{A_2{P}_2}\\ =\frac{A_1{P}_1{T}_2}{A_2{P}_2{T}_1}{V}_1\\ =\frac{P_1{T}_2}{P_2{T}_1}{V}_1\end{array}$        (III)

Refer Table A-2E (a), “Ideal-gas specific heats of various common gases”.

The specific heat of nitrogen gas is $0.248\text{Btu/lbm}\cdot \text{R}$.

Substitute $P_1=100\text{psia}$, $T_2=70\text{°F}$, $P_2=50\text{psia}$, and $T_1=120°\text{F}$ in Equation (III).

  $\begin{array}{c}{V}_2=\frac{\left(100\text{psia}\right)\left(70\text{°F}\right)}{\left(50\text{psia}\right)\left(120°\text{F}\right)}{V}_1\\ =\frac{\left(100\text{psia}\right)\left(70+460\right)\text{R}}{\left(50\text{psia}\right)\left(120+460\right)\text{R}}{V}_1\\ =\frac{53000}{29000}\\ =1.8276V_1\end{array}$        (IV)

Consider the nitrogen gas flows at steady state through one inlet and one exit system (pipe and duct flow). Hence, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the energy rate balance equation for one inlet and one outlet system.

  $\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$        (V)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The nitrogen gas flows at steady state through the system. Hence, the rate of change in net energy of the system becomes zero.

  $\Delta {\dot{E}}_{\text{system}}=0$

Neglect the heat transfer (adiabatic process), work transfer and potential energy changes i.e. ${\dot{W}}_{\text{1}}={\dot{W}}_{\text{2}}=0,{\dot{Q}}_1={\dot{Q}}_2=0,\Delta \text{PE}=0$.

The Equations (I) reduced as follows.

  $\begin{array}{l}\left[0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[0+0+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)=\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ h_1+\frac{V_1{}^2}{2}=h_2+\frac{V_2{}^2}{2}\\ \frac{V_1{}^2}{2}-\frac{V_2{}^2}{2}=h_2-h_1\end{array}$        (VI)

The change in enthalpy is expresses as follow.

  $h_2-h_1=c_p\left(T_2-T_1\right)$

Here, the specific heat is $c_p$, the exit temperature is $T_2$ and the inlet temperature is $T_1$.

Substitute $c_p\left(T_2-T_1\right)$ for $h_2-h_1$ in Equation (II) and rearrange it to obtain $V_1$.

  $\begin{array}{l}\frac{V_1{}^2}{2}=c_p\left(T_2-T_1\right)+\frac{V_2{}^2}{2}\\ V_1{}^2=2\left[c_p\left(T_2-T_1\right)+\frac{V_2{}^2}{2}\right]\\ V_1{}^2=2c_p\left(T_2-T_1\right)+V_2{}^2\end{array}$        (VII)

Substitute $c_p=0.248\text{Btu/lbm}\cdot \text{R}$, $T_2=70\text{°F}$, $T_1=120°\text{F}$ and $V_2=1.8276V_1$ in Equation (VII).

  $\begin{array}{l}{V}_1{}^2=2\left(0.248\text{Btu/lbm}\cdot \text{R}\right)\left(70\text{°F}-120°\text{F}\right)+{\left(1.8276V_1\right)}^2\\ V_1{}^2=0.496\text{Btu/lbm}\cdot \text{R}\left[\left(70+460\right)\text{R}-\left(120+460\right)\text{R}\right]+3.34V_1{}^2\\ V_1{}^2=0.496\text{Btu/lbm}\cdot \text{R}\left(-50\text{R}\right)+3.34V_1{}^2\\ V_1{}^2=-24.8\text{Btu/lbm}+3.34V_1{}^2\end{array}$

  $\begin{array}{l}{V}_1{}^2-3.34V_1{}^2=-24.8\text{Btu/lbm}\\ -2.34V_1{}^2=-24.8\text{Btu/lbm}\\ V_1=\sqrt{\frac{24.8\text{Btu/lbm}}{2.34}\times \frac{25037{\text{ft}}^2{\text{/s}}^2}{1\text{}\text{Btu/lbm}}}\\ V_1=515.1208\text{ft/s}\end{array}$

  $\simeq 515\text{ft/s}$

Thus, the velocity of the nitrogen at the pipe’s inlet is $515\text{ft/s}$.

Substitute $V_1=515\text{ft/s}$ in Equation (IV).

  $\begin{array}{c}{V}_2=1.8276\left(515\text{ft/s}\right)\\ =941.4347\text{ft/s}\\ \simeq 941\text{ft/s}\end{array}$

Thus, the velocity of the nitrogen at the pipe outlet is $941\text{ft/s}$.