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

Question

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.

Accepted Answer

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

[latex]\int_{CS}^{}{(ρu)(u • n)dS=F_B +F_S }[/latex] (7.19a)

[latex]\int_{CS}^{}{(ρu)(u • n)dS}=\int_{CV}^{}{ρfdV}+\int_{CS}^{}{\sum{} dS }[/latex] (7.19b)

[latex]\int_{CS}^{}{(ρu)(u • n)dS}=\int_{CS}^{}{\sum{} dS }[/latex]

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

[latex]\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}[/latex]

where [latex]A_1 = πR^2 _1[/latex]. On the exit port, where [latex]A_2 = πR^2 _2[/latex], we get

[latex]\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}[/latex]

Amass balance on this CV shows that [latex]\dot M[/latex] = 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

[latex]\left(ρV^2 _2 A_2 −ρV^2 _1 A_1 ight) k= \dot M(V_2 −V_1)k [/latex]

The surface force acting on the ﬂuid in the control volume is given by

[latex]\int_{CS}^{}{\sum{dS} }=\int_{inlet}^{}{\sum{dS} }+\int_{exit}^{}{\sum{dS}}+\int_{wall \ decal}^{}{\sum{dS}}+\int_{washer \ decal}^{}{\sum{dS} }[/latex]

On the inlet and exit ports we write the stress vector as [latex]\sum [/latex]= −pn. On the two decal surfaces the stress consists of both pressure and shear stress. Thus on these surfaces we use [latex]\sum =-pn+\pmb{ au} [/latex]. The surface force becomes

[latex]\int_{CS}^{}{\sum{dS} }=\int_{inlet}^{}{−pn \ dS}+\int_{exit}^{}{−pn \ dS}+\int_{wall \ decal}^{}{(-pn+\pmb{ au} )dS}+\int_{washer \ decal}^{}{(-pn+\pmb{ au} )dS}[/latex]

The inlet and exit ports have areas A₁and 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

[latex]\int_{CS}^{}{\sum{dS} }=p_1 A_1 k− p_2 A_2k+\int_{wall \ decal}^{}{\pmb{ au} dS}+\int_{washer \ decal}^{}{(-pn+\pmb{ au} )dS}[/latex]

The momentum balance therefore becomes

[latex]\left(ρV^2 _2 A_2 −ρV^2 _1 A_1 ight) k=p_1 A_1 k− p_2 A_2k+\int_{wall \ decal}^{}{\pmb{ au} dS}+\int_{washer \ decal}^{}{(-pn+\pmb{ au} )dS}[/latex] (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 [latex]\int_{wall \ decal}^{}{\pmb{ au} dS}=-\bar au _{wall}A_{wall}k[/latex], where [latex]\bar au _{wall}[/latex] 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 [latex]\bar p _{washer}[/latex] as

[latex]\int_{washer \ decal}^{}{−pn\ dS=− \bar p_{washer} A_{washer} (−k) =\bar p_{washer} (A_2 − A_1)k}[/latex]

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 [latex]\bar p _{washer}[/latex] = p₁, we ﬁnd that the surface force on the washer decal surface is [latex]\bar p _{washer}[/latex](A₂ − A₁)k = p₁(A₂ − A₁)k.

The completed momentum balance is

[latex]\left(ρV^2 _2 A_2 −ρV^2 _1 A_1 ight) k=p_1 A_1 k− p_2 A_2k-\bar au _{wall}A_{wall}k+p_1(A_2− A_1)k[/latex]

Solving for the pressure change we have

[latex](p_2 − p_1 )A_2=-\left(ρV^2 _2 A_2 −ρV^2 _1 A_1 ight) k-\bar au _{wall}A_{wall}[/latex] (B)

which can also be written by using [latex]\dot M[/latex] = ρA₁V₁ = ρA₂V₂ (i.e., V₂ = V₁(A₁/A₂)) as

[latex]\frac{p_2 − p_1}{\frac{1}{2} ho V^2_1}=2\frac{A_1}{A_2}\left(1-\frac{A_1}{A_2} ight)-\frac{\bar au _{wall}A_{wall}}{\frac{1}{2} ho V^2_1A_2} [/latex] (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

[latex]\frac{p_2 − p_1}{\frac{1}{2} ho V^2_1}=-\frac{\bar au _{wall}A_{wall}}{\frac{1}{2} ho V^2_1A_2} [/latex] or [latex]p_1-p_2=\frac{\bar au _{wall}A_{wall}}{A_2} [/latex]

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

[latex]\frac{p_2 − p_1}{\frac{1}{2} ho V^2_1}=2\frac{A_1}{A_2}\left(1-\frac{A_1}{A_2} ight)[/latex] (D)

This formula is plotted in Figure 7.16C. It is easy to show that (D) predicts a maximum pressure increase of (p₂ − p₁ )/[latex]\frac{1}{2}ρV^2 _1=\frac{1}{2}[/latex]
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 [latex]p_2 − p_1 =[\frac{1}{2} ρ(\bar V^2 _1 − \bar V^2 _2 )]−\Delta p_F[/latex] , where [latex]\Delta p_F[/latex] is the frictional pressure loss. This loss, which is due to the effects of turbulence and recirculation, is given by Eq. 3.22 as [latex]\Delta p_F=K_E\frac{1}{2}ρ \bar V^2 _1[/latex], where the K_Eis the expansion loss coefficient. Combining these two equations we ﬁnd:

[latex]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[/latex] (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

[latex]\begin{aligned}p_2-p_1 &=2 \frac{A_1}{A_2}\left(1-\frac{A_1}{A_2} ight)\left(\frac{1}{2} ho V_1^2 ight) \&=\left[2 \frac{A_1}{A_2}-2\left(\frac{A_1}{A_2} ight)^2 ight]\left(\frac{1}{2} ho V_1^2 ight)+\left(\frac{1}{2} ho V_1^2-\frac{1}{2} ho V_2^2 ight)-\left(\frac{1}{2} ho V_1^2-\frac{1}{2} ho V_2^2 ight) \&=\left(\frac{1}{2} ho V_1^2-\frac{1}{2} ho V_2^2 ight)-\left[1-2 \frac{A_1}{A_2}+2\left(\frac{A_1}{A_2} ight)^2-\frac{V_2^2}{V_1^2} ight]\left(\frac{1}{2} ho V_1^2 ight)\end{aligned}[/latex]

After using the fact that [latex]V^2 _2 /V^2 _1 = (A_1/A_2)^2[/latex] from the mass balance and simplifying, we have

[latex]p_2-p_1=\left(\frac{1}{2} ho V_1^2-\frac{1}{2} ho V_2^2 ight)-\left[1-2 \frac{A_1}{A_2}+\left(\frac{A_1}{A_2} ight)^2 ight]\left(\frac{1}{2} ho V_1^2 ight)[/latex]

Since the ﬂow is approximately uniform, V₁ = [latex]\bar V_1[/latex] and V₂ = [latex]\bar V_2[/latex]. Thus we can write K_E = [1−2(A₁/A₂) + (A₁/A₂)²] and conclude that the expansion loss coefficient predicted by the momentum balance is

[latex]K_E=\left(1-\frac{A_1}{A_2} ight)^2[/latex] (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.

Question 7.8: The steady turbulent ﬂow of a constant density ﬂuid through ...

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