IfP at 22 is 70 kNm and 5 Consider the viscous flow of wate

. IfP at 22) is 70 kN/m and 5. Consider the viscous flow of water given by the two dimensional velocity field V-yi +2xj neglecting gravity: (a) Find P at (3 ,2); (b) What is the AP due to viscosity? (c) Compute the value (N/m2) and the sign of all nine neglecting grarviy. (P due to viscosity Comp components of the stress tensor Ty at (2.2). Approx Ans: (a) P(2)~ 40 kPa; (b) AP 1.5x 10* Pa ; (e) Tyx ~4x 103 Pa

Solution

V = y^2 i + 2x j

Comparing it V = ui + vj we get

u = y^2

v = 2x

del u / del x = 0

del u / del y = 2y

del v / del x = 2

del v / del y = 0

del (u^2) / del x = del (y^4) / del x = 0

del (uv) / del y = del (2xy^2) / del y = 4xy

del (v^2) / del y = del (4x^2) / del y = 0

del (uv) / del x = del (2xy^2) / del x = 2y^2

-del p / del x = del (u^2) / del x + del (uv) / del y

-del p / del x = 4xy

-del p / del y = del (v^2) / del y + del (uv) / del x

-del p / del y = 2y^2

Partially integrating it, p = -2y^3 /3 + f(x)

Partially differentiating it with x, del p / del x = df/dx = -4xy

f(x) = -2yx^2 + C

So, p = -2y^3 /3 - 2yx^2 + C

At (2,2) p = -2*2^3 /3 - 2*2*2^2 + C = 70

C = 274/3

p = -2y^3 /3 - 2yx^2 + 274/3

a)

At (3,2) we get p = p = -2*2^3 /3 - 2*2*3^2 + 274/3

p = 50 kPa

b)

Water viscosity meu = 8.9*10^-4 Pa-s

dP due to viscosity = meu*(del2 u / del y^2 + del2 v / del x^2)

= 8.9*10^-4 *(2)

At (2,2) dP = 8.9*10^-4 *(2) = 1.78*10^-3 Pa

c)

Water viscosity meu = 8.9*10^-4 Pa-s

Tyx = meu*(del u / del y + del v / del x)

= 8.9*10^-4 *(2y + 2)

At (2,2) Tyx = 8.9*10^-4 *(2*2+2) = 5.34*10^-3 Pa

Since water is Newtonian fluid, other shear stress will be 0.