Chapter 5 Energy in stead flow

Solution
For a stream passing through point A and Point B
2 vB
pB pA zB zA 2g g 2 g g
2 vA
because
vA 0
2 v B
z A zB
then So
2
pB p A
vB
2

In a word, three kinds of energies can be changed, and the sum is a constant along a streamline in steady, frictionless and incompressible flow.
2

Geometrical meaning：
and
w cos gAds cos gAdz
1 pA ( p dp )dA ( p dp )( A dA) gAdz Av[(v dv) v] 2
1 dpdA Agdz Adp Avdv 2

Because the segment is infinitesimal, the higher order infinitesimals may be omitted. Hence
p v gz C 2

2
p v z H 2g g
2
Application conditions：
(1) (2) (3) (4) (5)
Steadyflow Onlygravity mass force is Incompress ible Idealflow No energyis addedor removed
IE cm(T2 T1) c(T2 T1) Weight mg g
5.2 Equation for Steady Motion of an Ideal Fluid Along a Streamline
5.2.1 Differential Motion Equation for OneDimensional Steady flow of Incompressible Ideal Fluid

2
in which, pressure potential energy— p ：The work is done when the pressure brings a unit mass fluid from a position of p pressure to the position of zero pressure.

Because the ideal fluid is frictionless and steady, the momentum equation in the flow direction is
m(v2 v1 ) F t
1 pA ( p dp )dA ( p dp )( A dA) w cos Av[(v dv) v] 2

For a control volume in flow field, according to the principle of conservation of energy,
Inwardheat Net in fluentenergy Input power transferrate per unittime in Increase energy per unittime



(3) Pitot-static pressure tube The velocity of flow is usually measured by Pitotstatic pressure tube in engineering.
vreal 2

( pt ps )
ps
pt

Example 9 Analyze the principle of Venturi tube as shown in the figure.
(ቤተ መጻሕፍቲ ባይዱp A pB )

And because
p B gH0 p A g ( H 0 h)
pA pB gh
vB
2( PA PB )

v B 2 gh

Explaining (1) Total pressure at point A (Stagnation Pressure --PA )
(2) Potential Energy
PE mgz z Weight mg
(3) Pressure Energy (Pressure Head)
PH pAs p Weight gAs
(4) Internal Energy
Thermal Energy (Here) Nuclear Energy Chemical Energy Electrostatic Energy
2 2 p1 v1 LS p2 v2 ( z1 ) ( z2 ) g 2g gA g 2g
LS hf gA
2 2 p1 v1 p2 v2 ( z1 ) hf ( z2 ) g 2g g 2g
5.4 Energy Equation 5.4.1 Energy equation for control volume

Analysis
2 2 vs pa ve pa zs ( z s h1 ) Sream line surface outlet : 2g g 2 g g 2 2 vs pa ve pV zs ( z s h'max ) surface peak : Stream line 2g g 2 g g
Chapter 5
Energy in Steady Flow
5.1 Energies of A Flowing Fluid
(1) Kinetic Energy
1 2 mv KE v2 2 Weight gV 2g
1 2 mv KE v2 2 Mass m 2
1 2 1 mv ( V )v 2 KE v 2 2 2 Volume V V 2
v2 I gz —— the total net influent energy per 2 unit mass quantity of flow.
p v z H 2g g
energy P osition potential energy Kinetic per unit weight per unit weight P ressureenergy Constant per unit weight
qv A2

2 gh( ' )
[1 ( A2 A1 ) 2 ]
in which, is caused by viscosity and turbulence of fluid. Usually,

=0.98~0.99

Example 10 Determine h’max of the siphon pipe for saturation pressure pV
m[(v2 v1) F t2 t1
1 1 pA ( p dp )( A dA) ( p dp )dA w cos ( L dL)ds Av[(v dv) v] 2 2
dp v L dz dv ds g g gA
Measure the velocityof flow Measure the quantityof flow

Example 8 In 1773, Pitot used a bent tube to measure the velocity of stream in a river.

Water head Water head Water head of velocity of position of pressure T otalwater head Constant
5.2.3 Application of Bernoulli’s Equation
v2 v2 P I gz v d A gzdV I 2 t CV 2 CS
in which,
P ——input power. φ I ——inward heat-transfer rate. —— internal energy per unit mass quantity of flow.
and
vs 0
ve pa g
2 gh1
2 ve pV h'max 2g g
1 h'max ( pa pV ) h1 g
5.3 Equation for Steady Motion of a Real Fluid Along a Streamline
5.2.2 Bernoulli’s Equation
vdv gdz dp 0
v2 dp gz C 2

and for incompressible fluid, ρ=constant，then
p v gz C 2
2
So, Bernoulli’s Equation

Physical meaning：
p v gz C 2
energy P osition potential energy Kinetic per unitmass per unitmass P ressureenergy Constant per unitmass