Reconsider Couette flow between two parallel plates as derived in class, but with the top plate moving with a known velocity +i
and the bottom plate is moving with a known velocity of-Vi,i. Derive the velocity field in the fluid layer.
Calculate the shear stress vector that the fluid exerts onto the top and bottom plates. Assume steady flow, incompressible, 2D, fully developed, no pressure gradient in the x direction. Fluid density and viscosity can also be treated as known.
Calculate the shear stress vector that the fluid exerts onto the top and bottom plates. Assume steady flow, incompressible, 2D, fully developed, no pressure gradient in the x direction. Fluid density and viscosity can also be treated as known.
1 answer:
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Answer:
Initial radius of cylindrical specimen
ri = 15.46mm
Explanation:
Cw = [(A1 - A2)/A1] * 100% ...equation 1
Where CW = cold worked percentage gain for( 27% EL) ductile
A2 = final area of cylindrical specimen = pai * r²
Where r = final radius of cylindrical specimen
So therefore expanding equation 1 will give
A1 * Cw = (100 * A1) - (100 * A2)
Cw = 18% from attached graph
A2 = area of cylindrical specimen after cold work = pai * 14² = 615.75mm²
So therefore
18A1 = 100A1 - (100 * 615.75)
-82A1 = -61575.22
A1 = 750.92mm²
So solving for initial radius from initial area of specimen
r1 = (A1/pai)^½
r1 = (750.92/pai)^½
r1= 15.46mm
Answer:
I)E= 40.95 GPa
II)E=5.29 GPa
Explanation:
I)
Given that
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ = 72.5 GPa ,V₂=0.55
Longitudinal moduli given as ;
E= E₁V₁+E₂V₂
E= 2.41 x 0.45 + 72.5 x 0.55 GPa
E= 40.95 GPa
II)
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ =230 GPa ,V₂=0.55
Transverse moduli given as:
E=5.29 GPa