Saturday, February 27, 2010
Chapter 2 - Problem 2.F2
This problem illustrates the similarity of two distrinct geometries when the flow cross section area is very small.
Sunday, February 21, 2010
Chapter 2 - Problem 2.E2
Laminar flow trough a slit. Again, we derive a parabolic velocity profile.
The specific shear stress profile can be determined by using the following boundary condition:
The following example is based on Problem 2.B1, assuming B = 2R
We start by establishing the momentum balance:
The specific shear stress profile can be determined by using the following boundary condition: The following example is based on Problem 2.B1, assuming B = 2R
Chapter 2 - Problem 2.D2
For a descendeing liquid film.
a) The equation that represents the velocity profile is derived as follows. We start by deriving the shear stress profile:
Assuming newtonian fluid:
a) The equation that represents the velocity profile is derived as follows. We start by deriving the shear stress profile:
Assuming newtonian fluid:
The following figures show the velocity profile for different inclination angles:
The other way to derive the velocity profile:
Sunday, January 10, 2010
Chapter 9 - Problem 9.B1
The heat loss reported in the book is 10 time bigger, I believe this is a typo.
Chapter 8 - Problem 8.E2
Thermal conductivity of pure liquids. Beware of the weird energy units, ergs, only physicists and chemists use them.
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