Lecture 3 - Fundamentals of internal flows (KARIMI) Flashcards
What is internal flow convection?
Convective heat transfer from fluid flow inside a conduit
What happens with velocity boundary layer at the beginning of the pipe/channel?
Boundary layer forms. The flow is divided into two distinctive regions: velocity boundary layer containing strong viscous effects and effectively inviscid core.
What happens to the BL during the entrance region?
The boundary layer grows and the inviscid core shrinks. The velocity profile is constantly changing.
Can you draw the diagram of velocity gradients and BL for internal flows?
YES OR NO
What happens in the fully developed region?
The BL fills the total volume of the pipe. The velocity profile remains unchanged.
What is HYDRONAMICALLY FULLY DEVELOPED FLOW?
u is not a function of x and the velocity only varies transversely.
For what value of Re_D is the pipe flow laminar and what is another name for this value?
Critical Reynolds Number, Re_D </= 2300
What value is flow fully turbulent?
Re_D > 10000
What does the length of the entrance region depend on?
The state of the flow (laminar or turbulent and Re)
Expression for laminar flow relating pipe distance to Re?
x_fd,h/D = 0.05 Re_D, where x_fd,h is the minimum length of the pipe downstream of that the flow is hydrodynamically fully developed (or length of the entrance region).
Range of distance/diameter for turbulent flows?
10</= 60
Can you derive the expression for Um?
Yes or no.
M dot = rho.u_m.Ac
M dot = integral (overAc) of rho.u(r,x).dAc
Equate for u_m to get an expression relating u_m, r and u(r,x)
Why is nu = 0?
du/dx + dv(nu)/dy = 0 for conservation of mass. Flow is fully developed so du,dx, dv,dy=0. From continuity: dv/dy=0 and therefore v=0. This makes sense because if the velocity in the y direction wasn’t zero, fluid would be flowing through the pipe wall. In the fully developed region, velocity field has only one axial component which remains unchanged along the pipe.
Expressions for Um and U(r)/Um?
Um = m dot / rho.Ac = -(R^(2)/8mu)(dP/dx) U(r)/Um=2[1-(r/R)^2] where U(r) = 1/4mu(dP/dx)(r^2-R^2)
Expression for shear stress in relation to Um?
tau = 8.mu.Um/D