Page 144 - Analysis and Design of Energy Geostructures

116 Analysis and Design of Energy Geostructures

magnitude of the shear stresses at the surface wall, τ s . This relationship is usually
expressed through the local friction coefficient of the surface:

C f τ s ð3:48Þ
ρ f v N
2
For Newtonian fluids, that is fluids for which the viscous stresses arising from their
flow are linearly proportional to the local strain rate at every point, the surface shear
stress can be determined through Newton’s law as

τ 5 μ dv rf ;i ð3:49Þ
f dn i
With regards to the problem depicted in Fig. 3.20A, the surface shear stress can
thus be determined as

τ s 5 μ dv x ð3:50Þ
y50
f
dy
The development of the thermal boundary layer is a consequence of the presence
of a temperature difference between the surface and the free stream (cf. Fig. 3.20B).
Due to the presence of such a temperature difference, the fluid temperature is equal
to the surface temperature at the wall in the boundary layer, varies within the bound-
ary layer and remains constant at a temperature outside the boundary layer that is
usually termed the free stream temperature, T N . The thickness of the thermal
boundary layer, δ th , is usually considered to correspond to a fluid temperature of
T δ 5 ðT s 2 TÞ=ðT s 2 T N Þ 5 0:99. The significance of the thermal boundary layer
thus depends on the magnitude of difference between the temperature at the surface
wall, T s , and the temperature of the free stream, T N . This relationship is usually
expressed through the local convection heat transfer coefficient, h c , which can be
determined by considering that at the surface wall heat transfer occurs by conduction
only (i.e. Fourier’s law can be applied to the fluid) and convection governs heat trans-
fer in the fluid in motion (i.e. Newton’s law of cooling can be used). With regards to
the problem depicted in Fig. 3.20B, Fourier’s law reads

52 λ @T
_ q j y50 ð3:51Þ
cond
@y
and can be substituted into Newton’s law of cooling to yield the convection heat
transfer coefficient

2 λ @T j
h c 5 @y y50 ð3:52Þ
T s 2 T N