Fo=-Sprido σ
= ροΐας - ερξατ - ορατ = = Dt →since the volume of the differential element is arbitrary to get this, the inside must be zero for any o 0 = 20[ds +
Momentum Conservation.
d Prove: pxa = Sp → DX
Using Reynold's T.T.
a(px)
0=20 + 2 = RHS
[+ (px)] α φ = RHS d
2(px) + x(pv)+pvvx] d = RHS
de + X +0 7X]d 6 = RHS
x(+pv))=0 = DX 70 م
Dt Now I can say that at =
2 δρότατα do Using Reynold's T.T. on defovde = F + Fs
St de + Spy PV (Vn) do = F + Fs 0=20
And for a C.V.. dt Substituting 2(pv) + ατ*=*
= Steady, =0, and neglect Fisc
*2 pådo dp*=*
Mass conservation is always
used
Steady, p=const
FIND D
includes some viscosity effects
2 Mass: Steady
Assume-0
Momentum Spin) do = +Sendo = - +2t pvde Neglecting F, and viscous forces compared to pressure forces
b/2 JU(2)1.dy 6/2 + Vel (1)dx +
6/2 - U+ Sve(x)dx + Sob1210 uly)dy=0 Know ve (x) now
374 14 surface normal
are orthogonal
b/2 Spui (ut. (-x)) 2.dy + 5^ [ui+ve (x)]] (ui+ve (x)j). (2).dx 0 -612 + puly) (uly)î•î) 1.dy = Fp
-pu²½i+fp(ui+ve (x)) Ve (x)dx + pity) i dy = Fo Fp=-pido P = ・ST・(-1) 1.dy-ff-Poj) j (2) dx - - √(-PoT). ↑ (1)dy - - SCP+0P)j-jdx = 0
P and SPod=0 -[Pondo + Sopή] 014 =>-pu²½î+ps" (ui+ve (x)î) ve (x)dx + pou² (y)îdy = - (2↑+Fyj) = b