Question 13.14: A room is at atmospheric pressure, 101 kPa, and a temperatur......
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 p_1 = 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?
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 \dot{m} = \rho_1 V_1 A_1, but we must first determine V_1 and \rho_1. Since the flow into the pipe is isentropic, and the pressure is p_1 = 80 kPa, while the stagnation pressure is p_0 = 101 kPa, we can determine the Mach number of the air and its temperature at the entrance using Eq. 13–27 and Eq. 13–26.
\frac{p_1}{p_0} = \frac{80 kPa}{101 kPa} = 0.792M_1 = 0.5868 and \frac{T_1}{T_0} = 0.93557
Therefore, T_1 = 0.93557 (293 K) = 274.12 K, and so V_1 = M_1\sqrt{kRT_1} = 0.5868 \sqrt{1.4 (286.9 J/kg . K) (274.12 K)}
= 194.71 m/s
Using the ideal gas law to obtain \rho_1, we have
p_1 = \rho_1 RT_1; 80(10^3) Pa = \rho_1 (286.9 J/kg . K) (274.12 K)
\rho_1 = 1.0172 kg/m^3
The mass flow along the pipe is then
\dot{m} = 1.5556 kg/s = 1.56 kg/s
Stagnation Temperature and Pressure. Because the flow is adiabatic through the pipe, the stagnation temperature remains constant at (T_0)_2 = (T_0)_1 = 293 K
Friction will change the stagnation pressure throughout the pipe because the flow is nonisentropic. We can determine (p_0)_2 at L = 0.9 m by using Eq. 13–51 (or Table B–2). First, however, we must find the length L_{max} needed to choke the flow. Using M_1 = 0.5868, Eq. 13–45 gives fL_{max}/D = 0.03L_{max}/0.1 = 0.5455, and so L_a{max} = 1.8183 m.
At this location, Eqs. 13–48, 13–51, and 13–50 give
\frac{(p_0)_1}{p_0^*} = 1.2043; p_0^* = \frac{101 kPa}{1.2043} = 83.87 kPa
\frac{V_1}{V^*} = 1.8057; p^* = \frac{80 kPa}{1.8057} = 44.30 kPa
Since L_{max} locates the critical point, then at section 2, Fig. 13–30a, fL’/D = 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
\frac{(p_0)_2}{p_0^*} = 1.1188; (p_0)_2 = = 1.1188(83.87 kPa) = 93.8 kPaFriction 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 p_2 and the velocity V_2 at fL’/D = 0.27548,
\frac{p_2}{p^*} = 1.5689; p_2 = 1.5689(44.30 kPa) = 69.51 kPa\frac{p_2}{p^*} = 0.7021; V_2 = 0.7021(313.16 m/s) = 219.86 m/s
Therefore,
\overset{+}{\rightarrow } ∑F= \frac{∂}{∂t} \int_{cv} V_{\rho} d \sout{V} + \int_{cs} V_{\rho} V_{f/cs} · dA -F_f + p_1 A – p_2 A = 0 + V_2 \dot{m} + V_1 (-\dot{m}) -F_f + [80(10^3) N/m^2 ] [\pi (0.05 m)^2 ] – [69.51(10^3) N/m^2 ] [\pi (0.05 m)^2]= 0 + 1.5556 kg/s (219.86 m/s – 194.71 m/s)
F_f = 43.3 N
