A room is at atmospheric pressure, 101 kPa, and a temperature of 293 K. If the air from the room is drawn into a 100-mm-diameter pipe isentropically, such that it has an absolute pressure of p1 = 80 kPa as it enters the pipe, determine the mass flow and the stagnation temperature and stagnation

Question

A room is at atmospheric pressure, 101 kPa, and a temperature of 293 K. If the air from the room is drawn into a 100-mm-diameter pipe isentropically, such that it has an absolute pressure of [latex] p_1 [/latex] = 80 kPa as it enters the pipe, determine the mass flow and the stagnation temperature and stagnation pressure at the location L = 0.9 m. The average friction factor is f = 0.03. Also, what is the total friction force acting on this 0.9-m length of pipe?

Accepted Answer

Fluid Description. We assume that adiabatic steady compressible (Fanno) flow occurs along the pipe.

Mass Flow. The mass flow can be determined at the entrance to the pipe using [latex] \dot{m}= ho_1 V_1 A_1 [/latex], but we must first determine [latex] V_1 ext { and } ho_1 [/latex]. Since the flow into the pipe is isentropic, and the pressure is [latex] p_1=80 kPa [/latex], while the stagnation pressure is [latex] p_0=101 kPa [/latex], we can determine the Mach number of the air and its temperature at the entrance using Eq. 13–27 and Eq. 13–26.

[latex] T_0=T\left(1+\frac{k-1}{2} M ^2 ight) [/latex] (13-26)

[latex] p_0=p\left(1+\frac{k-1}{2} M ^2 ight)^{k /(k-1)} [/latex] (13-27)

[latex] \begin{gathered} \frac{p_1}{p_0}=\frac{80 kPa }{101 kPa }=0.792 \ M _1=0.5868 \quad ext { and } \quad \frac{T_1}{T_0}=0.93557 \end{gathered} [/latex]

Therefore, [latex] T_1= 0.935571(293 K_2) [/latex] = 274.12 K, and so

[latex] \begin{aligned} V_1= M _1 \sqrt{k R T_1} & =0.5868 \sqrt{1.4(286.9 J / kg \cdot K )(274.12 K )} \ & =194.71 m / s \end{aligned} [/latex]

Using the ideal gas law to obtain [latex] ho_1 [/latex], we have

[latex] \begin{gathered} p_1= ho_1 R T_1 ; \quad 80\left(10^3 ight) Pa = ho_1(286.9 J / kg \cdot K )(274.12 K ) \ ho_1=1.0172 kg / m ^3 \end{gathered} [/latex]

The mass flow along the pipe is then

[latex] \begin{aligned} \dot{m}= ho_1 V_1 A_1= & \left(1.0172 kg / m ^3 ight)(194.71 m / s )\left[\pi(0.05 m )^2 ight] \ & \dot{m}=1.5556 kg / s =1.56 kg / s \end{aligned} [/latex]

Stagnation Temperature and Pressure. Because the flow is adiabatic through the pipe, the stagnation temperature remains constant at

[latex] \left(T_0 ight)_2=\left(T_0 ight)_1=293 K [/latex]

Friction will change the stagnation pressure throughout the pipe because the flow is nonisentropic. We can determine [latex] \left(p_0 ight)_2 ext { at } L=0.9 [/latex] m by using Eq. 13–51 (or Table B–2). First, however, we must find the length [latex] L_{\max } [/latex] needed to choke the flow. Using [latex] M _1 [/latex] = 0.5868, Eq. 13–45 gives [latex] f L_{\max } / D=0.03 L_{\max } / 0.1=0.5455 ext {, and so } L_{\max }=1.8183 m [/latex].
At this location, Eqs. 13–48, 13–51, and 13–50 give

[latex] \frac{f L_{\max }}{D}=\frac{1- M ^2}{k M ^2}+\frac{k+1}{2 k} \ln \left[\frac{\left[\frac{1}{2}(k+1) ight] M ^2}{1+\frac{1}{2}(k-1) M ^2} ight] [/latex] (13-45)

[latex] \frac{V}{V^*}=\frac{ M \sqrt{k R T}}{(1) \sqrt{k R T^*}}= M \left[\frac{\frac{1}{2}(k+1)}{1+\frac{1}{2}(k-1) M ^2} ight]^{1 / 2} [/latex] (13-48)

[latex] \frac{p}{p^*}=\frac{1}{ M }\left[\frac{\frac{1}{2}(k+1)}{1+\frac{1}{2}(k-1) M ^2} ight]^{1 / 2} [/latex] (13-50)

[latex] \frac{p_0}{p_0^*}=\frac{1}{ M }\left[\left(\frac{2}{k+1} ight)\left(1+\frac{k-1}{2} M ^2 ight) ight]^{(k+1) /[2(k-1)]} [/latex] (13-51)

[latex] \begin{array}{ll} \frac{V_1}{V^*}=0.6218 ; & V^*=\frac{194.71 m / s }{0.6218}=313.16 m / s \ \frac{\left(p_0 ight)_1}{p_0^*}=1.2043 ; & p_0^*=\frac{101 kPa }{1.2043}=83.87 kPa \ \frac{p_1}{p^*}=1.8057 ; & p^*=\frac{80 kPa }{1.8057}=44.30 kPa \end{array} [/latex]

Since [latex] L_{\max } [/latex] locates the critical point, then at section 2, Fig. 13–30a, [latex] fL^{\prime}/D[/latex] = 0.03(1.8183 m - 0.9 m)/0.1 m = 0.27548. From Eq. 13–45, M = 0.6690, and then from Eq. 13–51 the stagnation pressure at this location is

[latex] \frac{\left(p_0 ight)_2}{p_0^*}=1.1188 ; \quad\left(p_0 ight)_2=1.1188(83.87 kPa )=93.8 kPa [/latex]

Friction Force. The resultant friction force is obtained using the momentum equation applied to the free-body diagram of the control volume, shown in Fig. 13–30b. First we must determine the static pressure [latex] p_2 ext { and the velocity } V_2 ext { at } f L^{\prime} / D=0.27548 [/latex],

[latex] \begin{array}{ll} \frac{p_2}{p^*}=1.5689 ; & p_2=1.5689(44.30 kPa )=69.51 kPa \ \ \frac{V_2}{V^*}=0.7021 ; & V_2=0.7021(313.16 m / s )=219.86 m / s \end{array} [/latex]

Therefore,

[latex] \begin{gathered} \stackrel{+}{ ightarrow} \Sigma F =\frac{\partial}{\partial t} \int_{ c ext{v} } V ho d \sout{V}+\int_{ cs } V ho V _{f / cs } \cdot d A \ -F_f+p_1 A-p_2 A=0+V_2 \dot{m}+V_1(-\dot{m}) \ -F_f+\left[80\left(10^3 ight) N / m ^2 ight]\left[\pi(0.05 m )^2 ight]-\left[69.51\left(10^3 ight) N / m ^2 ight]\left[\pi(0.05 m )^2 ight] \ =0+1.5556 kg / s (219.86 m / s -194.71 m / s ) \ F_f=43.3 N \end{gathered} [/latex]

Question 13.14: A room is at atmospheric pressure, 101 kPa, and a temperatur...

Related Answered Questions

The nozzle in Fig. 13–39a is connected to a large reservoir where the absolute air pressure is 350 kPa. Determine the range of outside backpressures that cause a shock wave to form within the nozzle and just outside of it. ...

Verified Answer:

Verified Answer:

Air flows over a surface at a speed of 900 m/s, where the absolute pressure is 100 kPa and the temperature is 30°C. Expansion waves occur at the sharp transition shown in Fig. 13–47a. Determine the velocity, temperature, and pressure of the flow just to the right of the wave. ...

Verified Answer:

A jet plane is traveling at M = 1.5, where the absolute air pressure is 50 kPa and the temperature is 8°C. At this speed, a shock forms at the intake of the engine, as shown in Fig. 13–38a. Determine the pressure and velocity of the air just to the right of the shock. ...

Verified Answer:

The pipe in Fig. 13–37 transports air at a temperature of 20°C having an absolute pressure of 30 kPa and a speed of 550 m/s, measured just behind a standing shock wave. Determine the temperature, pressure, and speed of the air just in front of the wave. ...

Verified Answer:

Verified Answer:

Verified Answer:

Air flows through the 100-mm-diameter pipe in Fig. 13–23 having an absolute pressure of p1 = 90 kPa. Determine the diameter d at the end of the nozzle so that isentropic flow occurs out of the nozzle at M2 = 0.7. The air within the pipe is taken from a large reservoir at standard atmospheric ...

Verified Answer:

Verified Answer:

Verified Answer: