Water ﬂows steadily at 3 m/s through the vertical offset pipe bend shown in Figure 7.29. The pipe diameter is 10 cm, the upstream gage pressure is 275 kPa, and the pressure drop through the bend is estimated to be 50 kPa. Find the forces that must be exerted by each ﬂange on the pipe bend to keep

Question

Water ﬂows steadily at 3 m/s through the vertical offset pipe bend shown in Figure 7.29. The pipe diameter is 10 cm, the upstream gage pressure is 275 kPa, and the pressure drop through the bend is estimated to be 50 kPa. Find the forces that must be exerted by each ﬂange on the pipe bend to keep it in place.

Accepted Answer

We are asked to determine the forces applied to an offset pipe bend by two ﬂanges under the given conditions. Since the Reynolds number here may be calculated to be Re = 2.7 × 10⁵, the ﬂow is turbulent, and we may assume uniform properties at the ports. Choosing the mixed CV shown in Figure 7.29A with a decal surface at each ﬂange, and noting that the density and areas are constant, a steady ﬂow mass balance gives [latex]\dot M = ρA\bar V_1 = ρA\bar V_2,\ or\ \bar V_1 =\bar V_2[/latex]. Since we anticipate terms in the momentum balance in both the z and y directions, we will use the vector form of the steady ﬂow momentum balance as given by Eq. 7.19a: [latex]\int_{CS}[/latex] (ρu)(u • n)dS = F_B + F_S. To evaluate the body force we write F_B = − (W_pipe + W_water)k to account for the weight of the pipe and the water inside the pipe, since these are both inside the CV. To ﬁnish the momentum balance, we will evaluate the momentum transport at the inlet and exit, the surface force due to the pressure at these locations, the surface force applied by the air to the decal surface on the outside of the bend, and the reaction forces R₁ and R₂ applied to the CV by the ﬂanges. The forces R₁ and R₂ are the requested forces applied to the pipe bend by each ﬂange. On the inlet we have u = [latex]\bar V_1[/latex]j, n = − j, and u • n= −[latex]\bar V_1[/latex], so that the momentum transport vector there is −ρ [latex]\bar V^2_1A_1[/latex]j. On the exit, u = [latex]\bar V_2[/latex]j, n = j, and u • n = [latex]\bar V_2[/latex], and the momentum transport is +ρ[latex]\bar V^2_2A_2[/latex]j. The net momentum transport is (− ρ [latex]\bar V^2_1A_1[/latex]j + ρ [latex]\bar V^2_2A_2[/latex]j) = 0, since A₁ = A₂ = A and [latex]\bar V_1[/latex] = [latex]\bar V_2[/latex]. The surface force on the inlet and exit is due to the pressure at each location. Thus these terms give (−p₁(−j)A₁) + (−p₂(j)A₂) = (p₁ − p₂)Aj. The force of the air on the outside of the elbow will be accounted for by using gage pressure in these terms. Thus the momentum balance for the offset pipe bend gives

0 = − (W_pipe + W_water)k + [latex](p_{1_{gage} }− p_{2_{gage}})Aj+R_1 +R_2[/latex]

Solving for the desired reaction forces we have

R₁ + R₂ = [latex](p_{2_{gage} }− p_{1_{gage}})[/latex]Aj + (W_pipe + W_water)k

This is a vector equation whose components are

R_1x+ R_2x = 0 (A)

R_1y+ R_2y = [latex](p_{2_{gage} }− p_{1_{gage}})[/latex]A (B)

R_1z+ R_2z= W_pipe + W_water (C)

Equation (A) shows that if the ﬂanges apply a force in the x direction, these components must cancel. Since there is no reason to assume that these forces exist, we will assume that R_1X = R_2X = 0. Frictional pressure losses ensure that [latex]p_{1_{gage} }\gt p_{2_{gage}}[/latex], thus we see by (B) that the ﬂanges must apply a reaction force to the left. Equation (C) shows that the weight of the pipe and its contents is supported by the ﬂanges. We will assume that due to symmetry each ﬂange supports half the total weight, i.e.,

R_1z = R_2z= [latex]\frac{1}{2}[/latex](W_pipe+ W_water) (D)

Note that (B) does not allow us to determine the y component of each reaction force separately, only their sum. Figure 7.29B shows the various forces acting on the CV. To determine the y component of each reaction force, we will use Eq. 7.40 to apply a steady angular momentum balance to this CV. In this problem, the shaft torque term is zero, and the exterior torque is due to the ambient pressure (see Eq. 7.39g). Also, since the torque T_solid applied by solid supports to the control volume is created by the unknown reaction forces R₁ and R₂, we can write the torque as T_solid = r₁ × R₁ + r₂ × R₂, where r₁and r₂ are the moment arms of each reaction force. With these substitutions Eq. 7.40 becomes

[latex]T_{ ext{exterior}}=\int_{ ext{exterior}}^{}{} (r×(−p_An)dS[/latex] (7.39g)

[latex]\int_{CV}\frac{\partial}{\partial t}(r×ρu)dV +\sum{± \dot M_{port} [r_{port} ×u_{port} ]=r_G ×(−W_{CV} k)}T_{shaft} +T_{solid}+\sum{} [r_{port} ×(−p_{port} A_{port} n)]+T_{exterior}[/latex] (7.40)

[latex]\sum{± \dot M_{port} [r_{port} ×u_{port} ]=r_G ×(−W_{CV} k)+r_1 ×R_1 +r_2 ×R_2}[/latex][latex]\sum{[r_{port}×(−p_{port} A_{port} n)]+\int_{exterior}^{}{} (r×(−p_An)dS }[/latex]

Since the only forces acting on the CV are in the y and z directions (see (A)), we anticipate that the only component of the angular momentum balance of interest in this problem is the x component. This component describes the tendency of the forces acting on the CV to twist the pipe bend about the x axis.

Consider an origin O at the midpoint of the bend run (Figure 7.29A). To evaluate the transport term in the angular momentum balance, note that on the inlet [latex]\dot M_1 =ρ A\bar V_1[/latex], r₁ = −Hj+(L/2)k, u₁ = [latex]\bar V_1[/latex]j, hence r₁ × u₁ = [−Hj + (L/2)k] × [latex]\bar V_1[/latex]j = −(L/2) [latex]\bar V_1[/latex] i. On the exit [latex]\dot M_2 =ρ A\bar V_2[/latex], r₂ = Hj − (L/2)k, u₂ = [latex]\bar V_2[/latex]j, and r₂ × u₂ = [Hj − (L/2)k] × [latex]\bar V_2[/latex]j= (L/2) [latex]\bar V_2[/latex] i. Accounting for the signs of [latex]\dot M[/latex] at each port, and noting that [latex]\dot M =ρ A\bar V_1[/latex] = ρA [latex]\bar V_2[/latex], we ﬁnd
[latex]\sum{± \dot M_{port} [r_{port} ×u_{port} ]}=(-\dot M_1)\left(-\frac{L}{2} \bar V_1\mathrm{i} ight)+(\dot M_2)\left(\frac{L}{2} \bar V_2\mathrm{i} ight)=\dot ML \bar V_1\mathrm{i}[/latex]

Consider next the moment r_G× (−W_CVk) created by gravity acting on the CV. Owing to the symmetry, the force of gravity −(W_pipe + W_water)k acting on the CV may be considered to act at the point O. Since the moment arm for a force acting at this location is r_G = 0, the moment about O contributed by the body force is zero. The terms r₁ × R₁ and r₂ × R₂ are the unknowns in the angular momentum balance in this case, so these are left alone.

To evaluate the torque applied by atmospheric pressure on the exterior of the CV, note that if atmospheric pressure acted on the entire control surface (including the ports and small decal surface at each ﬂange), the resulting torque would be zero. We can account for the effects of atmospheric pressure acting on the portion of the control surface in contact with air by using gage pressure at the inlet and exit ports. Thus we evaluate this term and the corresponding term at the ports together by writing

[latex]\sum{[r_{port}×(−p_{port} A_{port} n)]+\int_{exterior}^{}{} (r×(−p_An)dS }[/latex]

[latex]=r_1 ×(−p_{1_{gage} }A_1n_1)+r_2 ×(−p_{2_{gage}} A_2n_2)[/latex]

Note that this approach assumes that the pipe wall is thin and the ﬂange areas are negligible. We have r₁ = −Hj + (L/2)k, n₁ = − j, and r₂ = Hj − (L/2)k, n₂ = j. Thus on the inlet we evaluate the cross product and get [−Hj + (L/2)k] × [−[latex]p_{1_{gage}}[/latex] A₁(−j)] = −(L/2)[latex]p_{1_{gage}}[/latex] A₁i. On the exit, the cross product gives −(L/2)[latex]p_{2_{gage}}[/latex] A₂i. Since A = A₁ = A₂, the two pressure terms together give

r₁× (−[latex]p_{1_{gage}}[/latex]A₁n₁) + r₂ × (−[latex]p_{2_{gage}}[/latex] A₂n₂) = [latex]\frac{L}{2}[/latex] ([latex]p_{2_{gage}}[/latex] + [latex]p_{1_{gage}}[/latex] )Ai

When all the terms have been gathered, the angular momentum balance becomes

[latex]\dot ML\bar V_1 \mathrm{i} [/latex] = [r₁ × R₁]_O + [r₂ × R₂]_O − [latex]\frac{L}{2}[/latex]([latex]p_{2_{gage}}[/latex] + [latex]p_{1_{gage}}[/latex])Ai

Solving for the torques applied by the ﬂanges about point O we ﬁnd

[r₁ × R₁]_O + [r₂ × R₂]_O = [latex]\left[\dot ML\bar V_1 \mathrm{i}+\frac{L}{2} (p_{2_{gage}} +p_{1_{gage}})A ight]\mathrm{i} [/latex]

The left-hand side of this vector equation is evaluated by writing

[r₁ × R₁]_O = [latex]\left(-Hj+\frac{L}{2}k ight) imes (R_{1y}jR_{1z}k)=\left(-HR_{1z}-\frac{L}{2}+R_{1y} ight)\mathrm{i}[/latex]

[r₂ × R₂]_O = [latex]\left(Hj-\frac{L}{2}k ight) imes (R_{2y}j+R_{2z}k)=\left(HR_{2z}+\frac{L}{2}R_{2y} ight)\mathrm{i}[/latex]

Thus the angular momentum balance becomes

[latex]\left(-HR_{1z}-\frac{L}{2}R_{1y} ight)\mathrm{i}+ \left(HR_{2z}+\frac{L}{2}R_{2y} ight)\mathrm{i}=\left[\dot ML\bar V_1 \mathrm{i}+\frac{L}{2} (p_{2_{gage}} +p_{1_{gage}})A ight]\mathrm{i}[/latex]

We see that there is only an x component to this equation, as expected. Earlier we assumed each ﬂange supports half the weight, thus we can put R_1z = R_2z. Solving for R_1y we obtain

[latex]R_{1y}= R_{2y}−2\dot M\bar V_1 −(p_{2_{gage} }+p_{1_{gage}})A[/latex] (E)

To solve for the y component of each reaction force, we can use (B) to write

[latex]R_{2y}= -R_{1y}+(p_{2_{gage} }-p_{1_{gage}})A[/latex]

Substituting the preceding expression into (E) and solving for R_1y we ﬁnd

[latex]R_{1y}= -\dot M V_1 −p_{1_{gage}}A[/latex] (F)

The remaining component is then found to be

[latex]R_{2y}= -\dot MV_1 +p_{2_{gage}}A[/latex] (G)

The forces applied by the two ﬂanges are thus

[latex]R_1 =\left[0,−\dot MV_1 − p_{1_{gage}} A,\frac{1}{2}(W_{pipe}+W_{water}) ight] [/latex] and [latex]R_2=\left[0,\dot MV_1 +p_{2_{gage}} A,\frac{1}{2}(W_{pipe}+W_{water}) ight] [/latex]

By using the data, we ﬁnd

[latex]\dot M =ρ A\bar V_1=(998\ kg/m^3)\left[\frac{\pi (0.1\ m)^2}{4} ight](3\ m/s)=23.5\ kg/s[/latex]

and with [latex]p_{1_{gage}}[/latex] = 275 kPa, we know [latex]p_{2_{gage}}[/latex] = 275 kPa − 50 kPa = 225 kPa. The y components of the reaction force are

[latex]R_{1y}= -\dot M\bar V_1 −p_{1_{gage}}A[/latex] = − (23.5 kg/s)(3 m/s)−(275,000 N/m² ) [latex]\left[\frac{\pi (0.1\ m)^2}{4} ight][/latex]

= − 2.23 kN

[latex]R_{2y}= -\dot M V_1 +p_{2_{gage}}A[/latex] = (23.5 kg/s)(3 m/s)+(225,000 N/m² ) [latex]\left[\frac{\pi (0.1\ m)^2}{4} ight][/latex]

= 1.84 kN

We are not given the weight of the pipe bend, so we will write the z components of the reactions in symbolic form as R_1z = R_2z = [latex]\frac{1}{2} [/latex](W_pipe + W_water).

The choice of an origin for the angular momentum balance is arbitrary. We could choose point G in the ﬁgure to evaluate the angular momentum balance and obtain the same results.

Question 7.17: Water ﬂows steadily at 3 m/s through the vertical offset pip...

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