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Unformatted text preview: Squaw E 3wa “no Ahﬂmézn mﬁ . 32596 > ocm :x.. 9: E 95E 9: :0 958 $28 9: L8 $063.56 06658 2 = N
n_
N> a / N 3390
< Evaluating the surface integral in both the x and y directions, we have i
l
. [LS Uv\vp(V ~ 11) dA = (U2 COS 6)(P2U2A2) + (le‘PrutAi)
} H wtv  n) dA : out s'm oozing/a2) The accumulation term is Zero in both equations, since, for the problem considered,
ﬂow is steady. The complete momentum expressions in the x and y directions are Bx + PrAi — P2142 C05 9 = (02 COS BXPEUZAZ) + ”1(‘PIU1AO
and BJ + PZAZ sin 6 n W = (7112 sin 9)(p—_,U_,A2)
Solving for the unknown force components BX and By, we have BX Z ngzA: cos 6 e vfple  PIA1 + PZA2 cos 6
and
By = nog’pgAz sin 6 ~ PZA2 sin 6 + W Recall that we were to evaluate the force exerted on the pipe rather than that on the
ﬂuid. The force sought is the reaction to B and has components equal in magnitude and
opposite in sense to BI and By. The components of the reaction force, R. exerted on the pipe are
RX = —v§pZA2 cos 6 + vfplAl + PIA1 ~ PEA1 cos 6
and
RV = v§p2AZ sin 6 + P2142 sin 6 — W
. Some simpliﬁcation in form may be achieved if the ﬂow is steady. Applying equation (43), we have ‘31le} = 91112152 : in
Where Flt is the mass ﬂow rate.
The final solution for the components of R may now be written as
R,r = Iii(v1— U2 cos 6) + PlAl — PEA: cos 6
y: rhv2 sin 6 + PZA2 sin 6 — W The control volume shown in Figure 5.4 for which the above solution was obtained rep
resents only possible Choice. Another is depicted in Figure 5.5. This control volume is Figure 5.5 Control
volume including fluid ._A 4‘:
mlu pipe. Figure 5.3 Flow in a reducing pipe bend, Chosen in this manner is designated in Figure 5.4, showing the external forces imposed
upon it. The external forces imposed on the ﬂuid include the pressure forces at sections
(1) and (2), the body force due to the weight of fiuid in the control volume, and the forces
due to pressure and shear stress, Pm and 7W, exerted on the ﬂuid by the pipe wall The re— sultant force on the ﬂuid (due to P“, and 7;”) by the pipe will be designated B, and its 3t and
y Components symbolized as B3 and B), respectively. Figure 5.4 Control volume deﬁned by pipe surface. Considering the x and y—directional component equations, (5521) and (S5b), of the overall momentum balance, the external forces acting on the ﬂuid in the control volume
are Ea=Pﬁp+mﬁwe+a and Eﬁ=ammeew+e Each component of the unknown force, B, is assumed to have a positive sense. The actual signs for these components, when a solution is obtained, will indicate Whether or not this
assumption is correct. EXAMPLE 4.5. Water with a density of 998 kg/rn3 (62.3 lb/ft3) enters a 50 mm
(1.969—in.) pipe ﬁtting horizontally, as shown in Fig. 4.9, at a steady velocity of 1.0 m/s
(3.28 NS) and a gauge pressure of lOO kN/tnZ (14.48 lbf/in?) It leaves the ﬁtting hori~
zontally, at the same elevation, at an angle of 45° with the entrance direction, The
diameter at the outlet is 20 mm (0.787 in). Assuming the ﬂuid density is constant, the
kinetic energy and momentum correction factors at both entrance and exit are unity,
and the friction loss in the ﬁtting is negligible, calculate (a) the gauge pressure at the exit of the ﬁtting and (b) the forces in the x and y directions excited by the ﬁtting on the
ﬂuid. l
t Solution_ :K
(0) Va 2 1.0 m/S. From Eq. (4.12),   D. 3 50 2 by vzva— =l.0— :.25' b (Db) <20) 6 ms
pazioom/m2 The outlet pressure pb is found from Eq. (4.62). Since Z0 2 Zb and hf may be neglected.
Eq, (4.62) becomes FIGURE 4.9 Flow through reducing ﬁtting.
Viewed from the top,
Example 4.5. from which 99805.252 — 1.01)
— a 100
Pb p 2 1,000 x 2
: 100 718.99 = 81.01 kN/mz (11.75 lbf/in.3) (Z7) The forces acting on the ﬂuid are found by combining Eggs. (4.42) and (4.431.
For the x direction, since F8 2 O for horizontal ﬂow, this gives m<ﬁb VIM: — 18111711») : pa Smut _ pbsbac + PW, X (.463) where S“ and 5b,; are the projected areas of Sn and Sb on planes normal to the initial ﬂow direction. (Regall that pressurep is a scalar quantity.) Since the ﬂow enters in the x
direction. V4,); 2 Va and TE ,, A 2
SW = 5a : Z0.050 : 0100196~ m From Fig. 49 . Vb} : Vb cos 6 : 6.25 cos 45= = 4.42 m/s
Also sh". 2 Sb sin 0 = E00202 sin 45: = 0.000222 m3
From Eq. (4.11) :‘i m : VapSa :10 x 998 x 0.001964 21,960 kg/s Substituting in Eq. (4‘63) and solving for me assuming 1% = [3b = 1, give Fm. : 196(442 —1,0)—100,000 >< 0.001964 + 81.010 >< 0.000222
‘ _6.7 196.4 1 18.0f 17l.7N( 38.61bf) Similariy for the y direction, V3,). : O and 35.}. = 0 , and V0,). 2 17b sin e 2 4142 m/s 5),... : 5;, cos 6 2 0.000222 m2 Hence Fuk)‘ 2 VIN/527%; ‘ 19a V23) — paSa.) + prb,_\'
:1.96{4.42 — 0) — 0 8 1011 >< 0.000222 x 1.000 H 8.66 +1798 = 26.64 N(5.991bf) ...
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 Fall '07
 Poehl

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