Problem 3.122 (Fox & McDonald, 4th edition)
A rectangular container of water undergoes constant acceleration down an incline as shown. Determine the slope of the free surface using the coordinate system shown.
At steady state, the water in the tank will behave like a rigid body no relative motion between any two points in the water water can be treated as inviscid
Water is incompressible
Euler’s equation can be used
steady-state Euler:
components:
p free surface = patm = constant
dp free surface = 0
NOTE: How is the direction of the surface determined?
Surfaces of constant pressure, including the free surface, most be normal to which is in the direction of the maximum variation of p.
Euler:
direction of free surface
Example: The tank shown is 4m long, 3m high, and 3m wide, and it is closed except for a small opening at the right end. It contains oil (SG=0.85) to a depth of 2m in a stationary situation. If the tank is uniformly accelerated to the right at a rate of 9.81m/s2, what will be the maximum pressure intensity in the tank during acceleration?
At steady state, oil will behave as a rigid body no relative motion between fluid particles fluid is then inviscid
Oil is incompressible
Euler’s equation can be used.
Steady-state Euler’s equation:
Components:
p free surface = patm = constantdp free surface = 0
but oil will touch the top of tank
Area occupied by the air remains the same since the amount of oil is the same.
p = pmax at x = 0, y = 0
pmax abs = 136113 Pa = 136.1 kPa
pmax gauge = 34788 Pa = 34.79 kPa
Euler’s Equation of Motion applied to a rotating fluid
Fluid rotates as a rigid body at steady state
Motor
Steady-state Euler:
(a conservative force is derivable from a potential function)
components:
(1)(r-component)
(2)(z-component)
Consider the r-component further
constant(3)
Example (Problem 5 from previous page)
At steady state liquid in tank behaves a rigid body -- liquid is inviscid since no relativemotion between liquid particles
liquid incompressible
Euler applicable from (3) and recognizing that V = r
Example (Problem #4 from previous page)
At steady state, water is in sold-body rotation -- no relative motion between fluid particles
--water is inviscid
--water is incompressible
--Euler’s equation is applicable
Consider a point C at the bottom of the tank on the axis of rotation
Between C and A
Now consider the z-component further
(4)
General control-volume formulation of the rate of change of an extensive property of a system
Let Pe be a general extensive property and Pi the corresponding intensive property
t = tt = t + t
CS: control surface
rate of net mass efflux through an elemental area dA
(of the control surface) in unit time
In-flow boundaryOutflow boundary
Conservation of Mass (continuity of flow)
Pe = m, Pi = m/m = 1
Recall:
--control-volume form of the continuity equation(conservation of mass)
For incompressible flow of a homogeneous fluid, = const
(if V is fixed)
volumetric flow rate (Q = VA or )
note: flow may or not be steady
For steady flow which is not incompressible,
mass flow into the CV via the CS since at an inflow boundary
mass flow out of the CV via the CS since at an outflow boundary
4.28Fluid with a 1050 kg/m3 density is flowing steadily through the rectangular box shown. Given A1 = 0.05 m2, A2 = 0.01 m2, A3 = 0.06 m2, and , determine the velocity .
Conservation of mass for the control volume:
Flow is steady
Assume flow is uniform at all exits and inlets:
flow at ‘3’ is an outflow
4.33Water flows steadily through a pipe of length L and radius R = 3 in. Calculate the uniform inlet velocity, U, if the velocity distribution across the outlet is given by:
and umax = 10 ft/sec.
Conservation of mass for the control volume:
Given:flow is steady
Water is the working fluid flow incompressible = const
4.38A section of pipe carrying water contains an expansion chamber with a free surface whaose area is 2 m2 . The inlet and outlet pipes are both 1 m2 in arewa. At a given instant, the velocity at section ‘1’ is 3 m/sec into the chamber. Water flows out at section ‘2’ at 4 m3/sec. Both flows are uniform. Find the rate of change of free surface level at the given instant. Indicate whether the level rises or falls.
Conservation of mass for the cortrol volume:
Given: water is the working fluid flow incompressible = const
( = const) (V is fixed)
Assume flow is uniform at all exits and inlets:
’3’ is an inflow boundary
liquid level falls
4.41A tank of 0.5 m3 volume contains compressed air. A valve is opened and air escapes with a velocity of 300 m/sec through an opening of 130 mm2 area. Air temperature passing through the opening is –15 C and the absolute pressure is 350 kPa. Find the rate of change of density of the air in the tank at this moment.
V = 0.5 m3
Ve = 300 m/sec
Ae = 130 mm2 = 130 x 10-6 m2
Conservation of mass for the control volume:
z 1 air can be treated as an ideal gas
(V is fixed)
kg/m3 sec
4.44A cylindrical tank, of diameter D = 50 mm, drains though an opening, d = 5 mm, in the bottom of the tank. The speed of the liquid leaving the tank is approximately , where y is the height from the tank bottom to the free surface. If the tank is initially filled with water to yo = 0.4 m, determine the water depth at t = 12 sec.
D = 50 mm = 0.05 m
d = 5 mm = 0.005 m
y0 = 0.4 m
Vi 0 (since D > d)
Conservation of mass for the control volume:
Assume flow is uniform at ‘e’
Assume = const since the working fluid is a liquid
( = const)
y = 0.134m
4.49Water flows steadily past a porous flat plate. Constant suction is applied along the porous section. The velocity profile at section cd is
Evaluate the mass flow rate across section bc.
Conservation of mass for the c.v.:(c.v. is abcd)
Given: working fluid is water flow is incompressible
Flow is steady
Let
Velocity profile development in pipes
“Sufficiently far from the pipe entrance, the boundary layer developing on the pipe wall reaches the pipe centerline and the flow becomes entirely viscous. The velocity profile shape changes slightly after the inviscid core disappears. When the profile shape no longer changes with increasing distance, x, the flow is fully developed. The distance downstream from the entrance to the location at which fully developed flow begins is called the entrance length. The actual shape of the fully developed velocity profile depends on whether the flow is laminar or turbulent. In Fig. 8.1 the profile is shown qualitatively for a laminar flow.
“For laminar flow, the entrance length, L, is a function of Reynolds number,
(8.1)
where D is pipe diameter, is average velocity, is fluid density, and is fluid viscosity. Laminar flow in a pipe may be expected only for Reynolds numbers less than 2300. Thus the entrance length for laminar pipe flow may be as long as
Fig. 8.1 Flow in the entrance region of a pipe
For fully-developed laminar flow in a pipe:
umax : velocity on the centreline, i.e., pipe axis
But
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