3. Following the procedure outlined in the lecture on "Differential Equations of Fluid Flow", show that 2-D differential control volume analysis results in the following y-momentum equation: at dt oy C.V. +(² dr + + fes Start with the macroscopic momentum balance equation for the y-component. ΣE,- (ov,)dv] + [v, (pvm)dA. ΣΕ = C.S. ax dy

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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3. Following the procedure outlined in the lecture on "Differential Equations of Fluid Flow",
show that 2-D differential control volume analysis results in the following y-momentum
equation:
(+v+-+(6+²) Sex
dy
dy ax dy
Start with the macroscopic momentum balance equation for the y-component.
at
Z
ΣΕ = = 2 [{[[{(pv,)dv] + [[ v, (pvn)d₁.
C.S.
Txx
X
C.V.
x,y+Ay
P
Ay
x,y
1
Txy
at,
+Tyy
v(x,y) = (vx(x,y), vy(x,y))
Ax
Txy
x+Δx,y+Δy
Tyy
Tyx
P
x+Δx,y
Txx
Transcribed Image Text:3. Following the procedure outlined in the lecture on "Differential Equations of Fluid Flow", show that 2-D differential control volume analysis results in the following y-momentum equation: (+v+-+(6+²) Sex dy dy ax dy Start with the macroscopic momentum balance equation for the y-component. at Z ΣΕ = = 2 [{[[{(pv,)dv] + [[ v, (pvn)d₁. C.S. Txx X C.V. x,y+Ay P Ay x,y 1 Txy at, +Tyy v(x,y) = (vx(x,y), vy(x,y)) Ax Txy x+Δx,y+Δy Tyy Tyx P x+Δx,y Txx
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