Question 7.8: The steady turbulent flow of a constant density fluid through ...
The steady turbulent flow of a constant density fluid through a sudden expansion in a round pipe is shown in Figure 7.16A. Find an expression for the pressure change across the expansion, and use it to determine the loss coefficient. Assume a uniform flow at the inlet and exit stations as shown, and neglect the effects of gravity.
In Example 7.7 we saw that pressure and shear stress appear in the surface force terms of a momentum balance. Since we are asked to derive an expression for the pressure change across the sudden expansion, we will perform a steady flow momentum balance and choose a CV as shown in Figure 7.16B. Note the placement of control surfaces at the inlet and exit of the expansion and two decal surfaces adjacent to the walls of the expansion. The latter are a wall decal surface adjacent to the round walls and a washer decal surface adjacent to the washer-shaped shoulder. Upon applying Eq. 7.19, neglecting the body force term, we have
\int_{CS}^{}{(ρu)(u • n)dS=F_B +F_S } (7.19a)
\int_{CS}^{}{(ρu)(u • n)dS}=\int_{CV}^{}{ρfdV}+\int_{CS}^{}{\sum{} dS } (7.19b)
\int_{CS}^{}{(ρu)(u • n)dS}=\int_{CS}^{}{\sum{} dS }
Momentum transport occurs only on the inlet and exit ports. We will use cylindrical coordinates. On the inlet we have u = V1k, n = −k, and u • n = −V1. On the exit, u = V2k, n = k, and u • n = V2. The momentum transport at the inlet port is given by
\int_{inlet}^{}{(ρu)(u • n)dS}=\int_{0}^{2\pi }{}\int_{0}^{R_1}{(ρV_1k)(−V_1)r \ dr \ dθ =− ρV^2 _1 A_1k}
where A_1 = πR^2 _1. On the exit port, where A_2 = πR^2 _2, we get
\int_{exit}^{}{(ρu)(u • n)dS}=\int_{0}^{2\pi }{}\int_{0}^{R_2}{(ρV_2k)(V_2)r \ dr \ dθ =− ρV^2 _2 A_2k}
Amass balance on this CV shows that \dot M = ρA1V1 = ρA2V2. Since the density is constant and A2 > A1 for an expansion, we know from the mass balance that V1 > V2. Thus, the momentum transport into and out of this CV differ, and the net momentum transport can be written as
\left(ρV^2 _2 A_2 −ρV^2 _1 A_1\right) k= \dot M(V_2 −V_1)k
The surface force acting on the fluid in the control volume is given by
\int_{CS}^{}{\sum{dS} }=\int_{inlet}^{}{\sum{dS} }+\int_{exit}^{}{\sum{dS}}+\int_{wall \ decal}^{}{\sum{dS}}+\int_{washer \ decal}^{}{\sum{dS} }
On the inlet and exit ports we write the stress vector as \sum = −pn. On the two decal surfaces the stress consists of both pressure and shear stress. Thus on these surfaces we use \sum =-pn+\pmb{\tau} . The surface force becomes
\int_{CS}^{}{\sum{dS} }=\int_{inlet}^{}{−pn \ dS}+\int_{exit}^{}{−pn \ dS}+\int_{wall \ decal}^{}{(-pn+\pmb{\tau} )dS}+\int_{washer \ decal}^{}{(-pn+\pmb{\tau} )dS}The inlet and exit ports have areas A1 and A2. Assuming a uniform pressure on these surfaces, and noting that the effect of pressure on the round decal wall surface cancels for reasons of axisymmetry, we have
\int_{CS}^{}{\sum{dS} }=p_1 A_1 k− p_2 A_2k+\int_{wall \ decal}^{}{\pmb{\tau} dS}+\int_{washer \ decal}^{}{(-pn+\pmb{\tau} )dS}
The momentum balance therefore becomes
\left(ρV^2 _2 A_2 −ρV^2 _1 A_1\right) k=p_1 A_1 k− p_2 A_2k+\int_{wall \ decal}^{}{\pmb{\tau} dS}+\int_{washer \ decal}^{}{(-pn+\pmb{\tau} )dS} (A)
We see that the momentum transport vector is balanced by the pressure forces on the inlet and exit, the shear force on the wall decal surface, and the pressure and shear force on the washer decal surface adjacent to the shoulder of the expansion.
To evaluate the remaining integrals in (A), we will write the surface force on the wall decal surface as \int_{wall \ decal}^{}{\pmb{\tau} dS}=-\bar \tau _{wall}A_{wall}k, where \bar \tau _{wall} is the average wall shear stress, assumed to be acting on the wall in the flow direction. As in the preceding example, the negative sign is needed because the shear stress acting on the adjacent CV surface acts in the direction opposite to the wall shear stress. Now consider the washer decal surface. The effect of shear stress on this surface cancels for reasons of axisymmetry, so the surface force can be written in terms of an average pressure \bar p _{washer} as
\int_{washer \ decal}^{}{−pn\ dS=− \bar p_{washer} A_{washer} (−k) =\bar p_{washer} (A_2 − A_1)k}
Note the direction of the outward unit normal on this surface. Empirical data and computational fluid dynamic simulations suggest that the pressure on this surface is uniform and the same as that acting at the inlet. Taking \bar p _{washer} = p1, we find that the surface force on the washer decal surface is \bar p _{washer}(A2 − A1)k = p1(A2 − A1)k.
The completed momentum balance is
\left(ρV^2 _2 A_2 −ρV^2 _1 A_1\right) k=p_1 A_1 k− p_2 A_2k-\bar \tau _{wall}A_{wall}k+p_1(A_2− A_1)k
Solving for the pressure change we have
(p_2 − p_1 )A_2=-\left(ρV^2 _2 A_2 −ρV^2 _1 A_1\right) k-\bar \tau _{wall}A_{wall} (B)
which can also be written by using \dot M = ρA1V1 = ρA2V2 (i.e., V2 = V1(A1/A2)) as
\frac{p_2 − p_1}{\frac{1}{2}\rho V^2_1}=2\frac{A_1}{A_2}\left(1-\frac{A_1}{A_2} \right)-\frac{\bar \tau _{wall}A_{wall}}{\frac{1}{2}\rho V^2_1A_2} (C)
Result (C) seems reasonable and tells us that the pressure change consists of an increase in pressure due to the flow slowing down as the area increases and a decrease in pressure due to the effects of the wall shear stress. Note that there is no expansion of the pipe as A1/A2 → 1, and we find
\frac{p_2 − p_1}{\frac{1}{2}\rho V^2_1}=-\frac{\bar \tau _{wall}A_{wall}}{\frac{1}{2}\rho V^2_1A_2} or p_1-p_2=\frac{\bar \tau _{wall}A_{wall}}{A_2}
The only pressure change in this case is a drop in pressure p1 > p2 due to the effect of friction at the wall. This agrees with (B) of Example 7.7, which considered the flow through a round pipe of constant diameter.
The flow through a sudden expansion at the high Reynolds numbers of engineering interest is accompanied by recirculation zones near the expansion plane, which persist for some distance downstream in the larger pipe, as shown in Figure 7.16A. The wall shear stress is very small and is directed opposite to the flow direction because the wall is exposed to a very low speed flow in the reverse direction. Thus the term in (C) involving the wall shear stress is actually positive but small enough to be neglected, and the pressure change for a flow through a sudden expansion is traditionally written as
\frac{p_2 − p_1}{\frac{1}{2}\rho V^2_1}=2\frac{A_1}{A_2}\left(1-\frac{A_1}{A_2} \right) (D)
This formula is plotted in Figure 7.16C. It is easy to show that (D) predicts a maximum pressure increase of (p2 − p1 )/\frac{1}{2}ρV^2 _1=\frac{1}{2}
for A1/A2 = 0.5, and no change in pressure for A1/A2 → 0 and A1/A2 → 1.
To determine the loss coefficient, recall that in our case study of Section 3.3.2, on flow through an area change, the empirical formula for the pressure change across a sudden expansion is given by Eq. 3.20 as p_2 − p_1 =[\frac{1}{2} ρ(\bar V^2 _1 − \bar V^2 _2 )]−\Delta p_F , where \Delta p_F is the frictional pressure loss. This loss, which is due to the effects of turbulence and recirculation, is given by Eq. 3.22 as \Delta p_F=K_E\frac{1}{2}ρ \bar V^2 _1, where the KE is the expansion loss coefficient. Combining these two equations we find:
p_2 − p_1 =[\frac{1}{2} ρ(\bar V^2 _1 − \bar V^2 _2 )]−K_E\frac{1}{2}ρ \bar V^2 _1 (E)
Our analysis of this problem by means of a momentum balance allows us to find the expansion loss coefficient. Rearranging (D) to put it into the form of (E), we have
After using the fact that V^2 _2 /V^2 _1 = (A_1/A_2)^2 from the mass balance and simplifying, we have
p_2-p_1=\left(\frac{1}{2} \rho V_1^2-\frac{1}{2} \rho V_2^2\right)-\left[1-2 \frac{A_1}{A_2}+\left(\frac{A_1}{A_2}\right)^2\right]\left(\frac{1}{2} \rho V_1^2\right)
Since the flow is approximately uniform, V1 = \bar V_1 and V2 = \bar V_2. Thus we can write KE = [1−2(A1/A2) + (A1/A2)2] and conclude that the expansion loss coefficient predicted by the momentum balance is
K_E=\left(1-\frac{A_1}{A_2}\right)^2 (F)
This result, which is plotted in Figure 7.16D (and appeared earlier in the case study of Section 3.3.2 as Figure 3.13), is in excellent agreement with experimental measurements.
