Chapter 5

1

Chapter 5 Mass, Bernoullil, and Energy Equations

5-1 Introduction

5-2 Conservation of Mass

5-3 Mechanical Energy and Efficiency

5-4 Bernoulli Equation

5-5 General Energy Equation

Fluid Mechanics-- Chapter 5 1

5-6 Energy Analysis of Steady Flows

5-1 Introduction• Conservation laws for: mass, momentum, energy, etc.• Conservation laws were originally applied to a closed

system or just a system.y j y• In fluid mechanics, the conservation laws were extended

to regions in space call control volume.• Conservation of mass:

inoutcvsys 0 mmtd

mdtd

md&& −+==


• Conservation of momentum: (in Chapter 6)• Conservation of energy:

inoutcvsys 0 EEtd

Edtd

Ed && −+==

2

5-2 Conservation of Mass• Conservation of mass principle is one of the most

fundamental principles in nature.• Mass, like energy, is a conserved property, and it

t b t d d t d d icannot be created or destroyed during a process.• For closed systems mass

conservation is implicit since the mass of the system remains constant during a process.

• For control volumes, mass can


cross the boundaries which means that we must keep track of the amount of mass entering and leaving the control volume.

∫∫ =⋅=cc A nA

AdVAdVm ρρrr

&

• Integral in can be replaced with average values of ρ and Vn

Averaged Velocity & Volume Flow Rate

∫= n AdVV 1avg

m&

• For many flows variation of ρ is very small:

• Volume flow rate is given by:

∫cA n

cAavg

cA n AVAdVQc

avg=== ∫V&

cAVm avgρ=&

• Note: many textbooks use Qinstead of for volume flow rate

• Mass and volume flow rates are related by


V&& ρ=m

V& V& cAVavg=V&

3

Conservation of Mass Principle0or;0 inout

cvinout

cvsys =−+−+ΔΔ

==Δ

Δ∑∑∑∑ mm

tdmdmm

tm

tm

&&&&

⎭⎬⎫

⎩⎨⎧

Δ−

⎭⎬⎫

⎩⎨⎧

Δ=

⎭⎬⎫

⎩⎨⎧

Δ ttt duringcvtheleavingmassTotal

duringcvtheenteringmassTotal

duringcvthewithinmassinchangeNet

∫∫

∫∫

⋅+=

↓↓↓===Φ

⋅+=Φ

sys

cscv

sys

11|||

AdVdddm

m

AdVdtd

dtd

d

rr

rr

ρρ

φφ

ρφρφ

V

V


∫∫ +cscv

AdVdtdtd

ρρ V

∫∑∑∫ ⋅=−=csinoutcv

cv ; AdVmmdtd

dtd

md rr&& ρρ V

• For a control volume of arbitrary shape:

• General (integral) form of the continuity equation:

Continuity Equation

∫∫ ⋅+=cscv

sys AdVdtd

dtd

md rrρρ V

• If the mass of a system is constant:

For uniform flow (i.e. constant velocity):

∫∫ −=⋅cvcs

Vdtd

dAdV ρρrr

∫∑ −=⋅cv

Vdtd

dAV ρρrr


Steady, 1-D flow in a conduit:

cs td

111222cs

0 AVAVAV ρρρ −==⋅∑rr

4

• Ex. 5-1:A garden hose attached with a nozzle is used to fill a 10-gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit. If it

Continuity Equation (Cont.)

takes 50 s to fill the bucket with water, determine a) the volume and mass flow rates of water through the hose, and b) the average velocity of water at the nozzle exit.


• Ex. 1:A jet of water discharges into an open tank, and water leaves the tank through an orifice in the bottom at a rate of 0.003 m3/s. If the cross-sectional area of the jet is 0.0025 m2 where the velocity of


water is 7 m/s, at what rate is water accumulating in (or evacuating from) the tank?


5

• Ex. 2:Water flows steadily into a tank through pipes 1 and 2 and discharges at a steady rate out of the tank through pipes 3 and 4. The mean velocity of inflow and outflow in pipes 1, 2, and 3 is 50 ft/s and the hypothetical outflow velocity in pipe 4 varies linearly


ft/s, and the hypothetical outflow velocity in pipe 4 varies linearly from zero at wall to a maximum at the center of the pipe. What are the mass rate of flow discharge from pipe 4, and what is the maximum velocity in pipe 4?



• Ex. 3:A 10-cm jet of water issues from a 1-m diameter tank. Assume that the velocity in the jet is m/s. How long will it take for the water surface into the tank to drop from h0 = 2 m to hf =0.5 m?

hg2

• Ex. 4:


Methane escape through a small (10-7 m2) hole in a 10-m3 tank. The methane escapes so slowly that the temperature in the tank remains constant at 23oC. The mass flow rate of methane through the hole is given by , where p is the pressure in the tank. Calculate the time required for the absolute pressure in the tank to decrease from 500 to 400 kPa.

TRApm 66.0=&

6

• Mechanical energy can be defined as the form of energy that can be converted to mechanical work completely and directly by an ideal mechanical device such as an ideal turbine

5-3 Mechanical Energy and Efficiency

device such as an ideal turbine• Flow P/ρ, kinetic V 2/ 2, and potential gz energy are the

forms of mechanical energy emech= P/ ρ + V 2/ 2 + gz• Mechanical energy change of a fluid during

incompressible flow becomes

( )2

12

212 zzgVVPPe −+−

+−

=Δ


• In the absence of loses, Δemech represents the work supplied to the fluid (Δemech > 0) or extracted from the fluid (Δemech < 0).

( )12mech 2zzge ++=Δ

ρ

5-3 Mechanical Energy & Efficiency (Cont.)

• Transfer of emech is usually accomplished by a rotating shaft: shaft work

• Pump, fan, propulsion: receives shaft work (e.g., from l t i t ) d t f it t th fl idan electric motor) and transfers it to the fluid as

mechanical energy• Turbine: converts emech of a fluid to shaft work• In the absence of irreversibilities (e.g., friction),

mechanical efficiency of a device or process can be defined as


• If ηmech < 100%, losses have occurred during conversion

inmech,

lossmech,

inmech,

outmech,mech 1

EE

EE

−==η

7


• In fluid systems, we are usually interested in increasing the pressure, velocity, and/or elevation of a fluid

WE &&Δ

outelec,generator

outshaft,motor

eturbine,

turbine

fluidmech,

outshaft,turbine

pump

upump,

inshaft,

fluidmech,pump

WW

WW

EW

WW

WE

&

&

&

&

&

&

&

&

&&

==

=Δ

=

=Δ

=

ηη

η

η


• Overall efficiency:

inshaft,generator

inelec,motor WW

ηη

genturbinegenturbinemotorpumpmotorpump ; ηηηηηη == −−


• Ex. 5:The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a location where the depth of the water is 50 m as shown in the picture below. Water is p pto be supplied at a rate of 5000 kg/s. If the electric power generated is measured to be 1862 kW and the generator efficiency is 95%, determine (a) the overall efficiency of the turbine-generator, (b) the mechanical efficiency of the turbine, and (c) the shaft power supplied by the turbine to the generator.


8


• Ex. 5-4:The motion of a steel ball in a hemispherical bowl of radius h, as shown in the picture below, is to be analyzed. The ball is initially held at the highest location at point A, and then it is released. g p ,Obtain relations for the conservation of energy of the ball for the case of frictionless and actual motion.



• HW. 5-20:At a certain location, wind is blowing steadily at 8 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 50-m-diameter blades at g pthat location. Also, determine the actual electric power generation assuming an overall efficiency of 30 percent. Take the air density to be 1.25 kg/m3.

Wind

8 m/s50 m

Wind

turbine


9

5-4 Bernoulli Equation

ss

WAPPAP

maF

θΔ−ΔΔ+−Δ

=∑sin)(or

• Acceleration of a fluid element:

Hence,

s

s

agsP

AsgWaAs

ρθρ

ρρ

−=+ΔΔ

⇒

ΔΔ=ΔΔΔ=

sin

Q

zP ∂∂


(Euler’s equation)saszg

sP ρρ −=

∂∂

+∂∂

tV

sVV

tdVda

tdtVsd

sVVdtsVV

s ∂∂

+∂∂

==

∂∂

+∂∂

=⇒=

Thus,

),(Q

5-4 Bernoulli Equation (Cont.)• Integrating Newton’s second law for particle motion along

a pathline provides a relationship between the change in kinetic energy and the work done on the particle.I t ti E l ’ ti l thli i t d

sazgPs

ρρ −=+∂∂ )(

• Integrating Euler’s equation along a pathline in a steady flow of an incompressible fluid yields an equivalent relationship called the Bernoulli equation.

Along a pathline:


( ) { { {constant/,or0

energykinetic

221

energypotential

energyFlow

221 =++=++ VzgPVzgP

sdd ρρρ

For a steady flow:

sddVVzgP

sdd

tρρ −=+⇒=

∂∂ )(0Q

10

Along the direction normal to the streamline:

For an irrotational flow:

5-4 Bernoulli Equation (Cont.)

rVaazgP

r nn

2

and)( −=−=+∂∂ ρρ

For an irrotational flow:

• Bernoulli equation for the steady, irrotational flow of an incompressible, inviscid fluid:

h V i th d f th fl id d t l it tconstant2

21 =++ VzgP ρρ

( ) 0221 =++⇒−= VzgP

rdd

rV

rdVd ρρ


where V is the speed of the fluid and not a velocity component

• Limitations on the use of the Bernoullil equation:Steady, incompressible, frictionless flow along a streamlineNo shaft workNo heat transfer

• Along a streamline:

HGL & EGL

{ {{ { constant

2headtotal

headelevation

velocity

2

pressure

==++ Hzg

Vg

Pρ

• Hydraulic grade line (HGL):• Energy grade line (EGL):

headhead

zgP +ρ

zgVgP ++ 22ρ


11

• For stationary bodies, the EGL and HGL coinside with the free surface of the liquid,, and can be represented by the elevation head of the free surface (z)

• THE EGL is always a distance V 2/2 above the HGL

HGL & EGL

• THE EGL is always a distance V 2/2g above the HGL

• For an idealized flow, EGL remains constant

• At the pipe exit, the HGL coinsides with the pipe outlet (Pgage = 0)

• The mechanical energy loss causes


gythe EGL and HGL to slope downward in the direction of flow

• The gage pressure of a fluid is zero at locations where the HGL intersects the fluid

Application of Bernoulli Equation

212

221

22

121

1 zzVpVp =+=+ whereρρ

• Stagnation tube:

p2: stagnation pressure

lgV 21 =

• Pitot tube:

22

221

212

121

1 zVpzVp γργρ ++=++


where

)(2 212 hhgVV −==

zph +=γ

12

Application of Bernoulli Equation (Cont.)

• Ex. 6:A mercury-kerosene manometer is connected to the Pitot tube as shown. If the deflection on the manometer is 7 in.,the deflection on the manometer is 7 in., what is the kerosene velocity in the pipe? (The specific gravity of the kerosene and mercury are 0.81 and 13.55)

• Ex. 7:A differential pressure gage is connected across the taps of a Pitot tube. When


across the taps of a Pitot tube. When this Pitot tube is used in a wind tunnel test, the gage indicates a Δp of 730 Pa. What is the air velocity in the tunnel? The pressure and temperature in the tunnel are 98 kPa absolute and 20oC.


• Ex. 8:A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap. A tap near the bottom of the tank is now opened, and water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet.


13


• Ex. 5-7:During a trip to the beach (Patm = 101.3 kPa), a car runs out of gasoline, and it becomes necessary to siphon gas out of another car as shown in the picture. To start the siphon it is necessary to insert one siphon end in the full gas tank, fill the hose with gasoline via suction, and then place the other end in a gas can below the level of the gas tank. The difference in pressure between point 1 and point 2 causes the liquid to flow from the higher to the lower elevation. Point 2 is located 0.75 m below point 1 in this case and point 3 is located 2 m above point 1


this case, and point 3 is located 2 m above point 1. The siphon diameter is 4 mm, and frictional losses in the siphon are to be disregarded. Determine (a) the minimum time to withdraw 4 L of gasoline from the tank to the can, and (b) the pressure at point 3. The density of gasoline is 750 kg/m3.


• Ex. 5-8:A piezometer and a Pitot tube are tapped into a horizontal water pipe, as shown in the picture, to measure static and stagnation (static + dynamic) pressures. For the indicated water column

fheights, determine the velocity at the center of the pipe


14


• Ex. 5-10:Derive the Bernoulli equation when the compressibility effects are not negligible for an ideal gas undergoing (a) an isothermal process and (b) an isentropic process

• HW. 5-40:In a hydroelectric power plant, water enters the turbine nozzles at 800 kPa absolute with a low velocity. If the nozzle outlets are exposed to atmospheric pressure of 100 kPa, determine the maximum velocity to which water can be accelerated by the nozzle before striking the turbine blades


g

• HW. 5-41:A pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3 kPa, determine the speed of the aircraft


• Tornado:Can be modeled as a forced vortex at the center surrounded by a free vortexAt point 3:p

At point 2:

In region 2→3:3

232

13

221 zVpzVp γργρ ++=++

CzzppV ==≈≈ 3033 ;;0

CzzrVV ==== 22max2 ;ω


⇒

In region 1→2:

⇒

2222

122

2221 ωρρωρρ rzgprzgp −+=−+

2212

max0 VVpp ρρ +−=

221

0 Vpp ρ−=

15


• Pressure variation for an incompressible, irrotational, low-viscosity flow near curved boundaries:

22

02

021

02

21

⎞⎛

++=++ zVpzVp γργρ

20

2

02

021

0

20

2

02

0

0

11

112

⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−=

⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−=⇒

AA

VV

VppC

AA

VV

gVhhC

p

p

ρor



• Irrotational flow past a circular cylinder


• Pressure distribution of an irrotational flow on a cylinder

16

5-5 General Energy Equation• One of the most fundamental laws in nature is the 1st

law of thermodynamics, which is also known as the conservation of energy principle

• It states that energy can be neither created nor destroyed during a process; it can only change forms

• Falling rock, picks up speed as PE is converted to KE


as PE is converted to KE

• If air resistance is neglected,PE + KE = constant

5-5 General Energy Equation (Cont.)

• The energy content of a closed system can be changed by two mechanisms: heat transfer Q

d k fand work transfer W.• Conservation of energy for a

closed system can be expressed in rate form as

• Net rate of heat transfer to thetd

EdWQ sys

innet,innet, =+ &&


• Net rate of heat transfer to the system:

• Net power input to the system:outininnet, QQQ &&& −=

outininnet, WWW &&& −=

17

Derivation of the Energy Equation• The 1st law of thermodynamics:

E : energy of a system; Qi : heat transfer to the system;

ininsys

sysininsys WQtd

EdEEEEWQE pku

&& +=++=+=Δ

Esys: energy of a system; Qin: heat transfer to the system;Wout: work done by the system on its surroundingEu: internal energy; Ek: kinetic energy; Ep: potential energy

energy internalspecific :2

;

2

syssys

sys

WWWWW

ezgeVe

eeeem

Ee

upk

pku

&&&&&

==

++==


forces viscous theby done work

pressure theby done workflow

system;flow the to applied etc.) pump, (turbine, workshaft

:

;:

:

viscous

A

ins,

inother,viscousins,innet,

W

AVPAdVPW

W

WWWWW

Af

f

&

rrrr&

&

∑∫ ⋅==

+−−=

Derivation of the Energy Equation• Control-volume of energy conservation:

∫∫ ⋅+=Φ

cscv

sys

|||

AdVdtd

dtd

d rrρφρφ V

• Control-volume approach to the 1st law of thermodynamics:

∫∫ ⋅+=

↓↓↓===Φ

cscv

sys

|||

AdVedetd

dtd

Ed

eeE

rrρρ

φφ

V


y

innet,incscvWQAdVede

tdd

tdEd sys &&

rr+=⋅+= ∫∫ ρρ V

∫∫ ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+++⎟⎟

⎠

⎞⎜⎜⎝

⎛++=+⇒

cs

2

cv

2

innet,in 22AdVezgVdezgV

tddWQ uu

rr&& ρρ V

18

• Flow Work ( ):The work done by pressure forces as the system moves through space

Derivation of the Energy Equation

∫∑ ⋅=⋅= AdVPAVPWrrrr

&

fW&

i.e.

• Energy Equation:

1111111,

2222222,

APVAVPW

APVAVPW

f

f

−=⋅=

=⋅=rr

&

rr&

∫∑ ⋅=⋅=cs

cs

AdVPAVPWf


∫∫

∑∫

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛++++⎟⎟

⎠

⎞⎜⎜⎝

⎛++=

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛++++⎟⎟

⎠

⎞⎜⎜⎝

⎛++=+

cs

2

cv

2

cs

2

cv

2

,

22

22

AdVezgVPdezgVtd

d

AVezgVPdezgVtd

dWQ

uu

uuinsin

rr

rr&&

ρρ

ρ

ρρ

ρ

V

V

• For steady flow:

5-6 Energy Analysis of Steady Flows

∑∑

⎞⎛⎞⎛

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛++=⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+++=+

22

cs

2

cs

2

ins,in 22

VVP

AVhzgVAVezgVPWQ u

rrrr&& ρρ

ρ

h: specific enthalpy of the fluid = eu+P/ρ

• Steady, incompressible flow through a pipe:

∫∫ ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛++=⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛+++=

cs

2

cs

2

22AdVhzgVAdVezgVP

u

rrrrρρ

ρ

mPezgVWQ u &&&2

11,1

21

1ins,in ρα ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++++


mVAVVAdVV

AVAVAdVm

mPezgVu

&rr

rr&

&

222

where

2

222

avg

22,2

22

2

αραρ

ρρρ

ρα

==⋅

==⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

∫

∫

19

• Steady, incompressible flow through a pump/turbine system:

5-6 Energy Analysis of Steady Flows (Cont.)

pumpbyfluidtheondonework

fluidthebyturbineondonework

:

:ins, ttp

W

WWWW&

&&&& −=

• Mechanical energy can be converted to thermal energy through viscous friction between fluid particles (this process is irreversible)

pumpbyfluidtheondonework:pW

⎟⎟⎠

⎞⎜⎜⎝

⎛−−++++=+++⇒

gmQee

ggmWPz

gV

gmWPz

gV

uutp

&

&

&

&

&

&in

1,2,2

2

22

21

1

21

1 )(122 γ

αγ

α

&


• Head loss (hL) is defines as:

head turbinehead pump

flow inviscid for

::

0

22headtotal

22

22

2

headtotal

11

21

1

t

p

L

Ltp

hhh

hhPzg

VhPzg

V=

++++=+++44 344 2144 344 21γ

αγ

α

gmQee

gh in

uuL &−−= )(1

1,2,

• Ex. 5-11:Show that during steady and incompressible flow of a fluid in an adiabatic flow section (a) the temperature remains constant and there is no head loss when friction is ignored and (b) the temperature increases and some head loss occurs when frictional


temperature increases and some head loss occurs when frictional effects are considered. Discuss if it is possible for the fluid temperature to decrease during such flow.


20

• Ex. 5-12:The pump of a water distribution system is powered by a 15-kW electric motor whose efficiency is 90 percent, as shown in the picture The water flow rate through the


picture. The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the pressures at the inlet and outlet of the pump are measured to be 100 kPa and 300 kPa (absolute), respectively, determine (a) the mechanical efficiency of


( ) ythe pump and (b) the temperature rise of water as it flows through the pump due to the mechanical inefficiency.

• Ex. 5-13:In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is generated as shown in the picture. The total irreversible head loss in the piping system from point 1 to point 2 (excluding the turbine unit) is


piping system from point 1 to point 2 (excluding the turbine unit) is determined to be 35 m. If the overall efficiency of the turbine–generator is 80 percent, estimate the electric power output.


21

• Ex. 5-14:A fan is to be selected to cool a computer case whose dimensions are 12-cm×40-cm×40-cm, as shown in the picture. Half of the volume in the case is expected to be filled with components and the other half to be air space A 5 cm diameter hole is available at


the other half to be air space. A 5-cm diameter hole is available at the back of the case for the installation of the fan that is to replace the air in the void spaces of the case once every second. Small low-power fan–motor combined units are available in the market and their efficiency is estimated to be 30 percent. Determine (a) the wattage of the fan–motor unit to be purchased and (b) the pressure difference across the fan. Take the air density to be 1.20 kg/m3.


g

• Ex. 5-15:A submersible pump with a shaft power of 5 kW and an efficiency of 72 percent is used to pump water from a lake to a pool through a constant diameter pipe as shown in the picture. The free surface

f th l i 25 b th f f f th l k If th


of the pool is 25 m above the free surface of the lake. If the irreversible head loss in the piping system is 4 m, determine the discharge rate of water and the pressure difference across the pump.


22

• Example:A steam turbine receives superheated steam at 1.4 MPa absolute and 400oC (h = 3121 kJ/kg). The steam leaves the turbine at 101 kPa absolute and 100oC (h = 2676 kJ/kg). The steam enters the turbine at 15 m/s and exits at 60 m/s The elevation difference


turbine at 15 m/s and exits at 60 m/s. The elevation difference between and exit ports is negligible. The heat lost through the turbine wall is 7600 kJ/h. Calculate the power output if the mass flow through the turbine is 0.5 kg/s.


• Example:The velocity distribution for laminar flow in a pipe is given by the equation

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

2

max 1orrVV


where ro is the radius of the pipe and r is the radial distance from the center. Determine the average velocity Vave in terms of Vmax and evaluate the kinetic-energy correction factor α

⎦⎣


23

• Example:A horizontal pipe carries cooling water for a thermal power plant from a reservoir as shown. The head loss in the pipe is given as

VDL2

)(02.0 2


Where L is the length of the pipe from the reservoir to the point in question, V is the mean velocity in the pipe, and D is the diameter of the pipe. If the pipe diameter is 20 cm and the rate of flow is 0.06 m3/s, what is the pressure in the pipe at L = 2000m?

g2


• Example:The pipe in the figure is 50 cm in diameter and carries water at a rate of 0.5 m3/s. Also, z2 = 40 m, z1 = 30 m, and p1 = 70 kPa gage. What power in Kilowatts and in horsepower must be supplied to the flow by the pump if the gage pressure at section 2 is to be 350 kPa?


flow by the pump if the gage pressure at section 2 is to be 350 kPa? (Assume hL = 3m of water and α1 = α2 = 1)


24

• Example:At the maximum rate of power generation this hydroelectric power plant takes a discharge of 141 m3/s. If the head loss through the intakes, penstock, and outlet works is 1.52 m, what is the rate of

ti ? (A 1)


power generation? (Assume α1 = α2 = 1)


• HW. 5-86:Water flows at a rate of 20 L/s through a horizontal pipe whose diameter is constant at 3 cm. The pressure drop across a valve in the pipe is measured to be 2 kPa, as shown in the picture below. D t i th i ibl h d l f th l d th f l


Determine the irreversible head loss of the valve, and the useful pumping power needed to overcome the resulting pressure drop.


25

Homework5-7, 5-9, 5-14, 5-23, 5-24, 5-41 (292 km/h), 5-44 (1.33 m/s), 5-45,5-47, 5-48, 5-54 (6.24 cm), 5-55 (4.48 ft3/s), 5-57, 5-61 (33.8 m/s), 5-76,5-85 (47.1%), 5-89 (55 kW), 5-90, 5-91 (6.76 kW, 23 m), 5-95 (0.0133 m3/s, 278 kPa), 5-104 5-107 5-113 (97 4 kPa)5-104, 5-107, 5-113 (97.4 kPa)

5-45: (a)

(b)

zgDmDD oT ⎟

⎞⎜⎛ −− 2

41

in212

21 2

l2&

&πρπρ

5-47: (a)

(b)

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

gH

gH

DDt f

22

20

gH

DDt f

22

20=

2

20

inmax

421

⎟⎟⎠

⎞⎜⎜⎝

⎛=

Dm

gh

πρ&


( ) tm

gmzgD

gDo

o

o

T =⎟⎟⎠

⎜⎜⎝

+in

4inin

241

2241

2 ln22 &

&ρ

πρπρ

ρ(b)

Chapter 6: Momentum Analysis of Flow Systems


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Chapter 5