auvw ubx byb problem: material derivative auvw ubx byb

ì

Problem: material derivative

For positive x and b, fluid particles accelerate in the positive x direction. Even though this flow is steady, there is still a non-zero acceleration field.

CHAPTER 4159

location in the flow (Fig. P4–8C). Is this a Lagrangian or anEulerian measurement? Explain.

4–9C A tiny neutrally buoyant electronic pressure probe isreleased into the inlet pipe of a water pump and transmits2000 pressure readings per second as it passes through thepump. Is this a Lagrangian or an Eulerian measurement?Explain.

4–10C A weather balloon is launched into the atmosphereby meteorologists. When the balloon reaches an altitudewhere it is neutrally buoyant, it transmits information aboutweather conditions to monitoring stations on the ground (Fig.P4–10C). Is this a Lagrangian or an Eulerian measurement?Explain.

4–13C Define a steady flow field in the Eulerian referenceframe. In such a steady flow, is it possible for a fluid particleto experience a nonzero acceleration?

4–14C List at least three other names for the material deriv-ative, and write a brief explanation about why each name isappropriate.

4–15 Consider steady, incompressible, two-dimensionalflow through a converging duct (Fig. P4–15). A simpleapproximate velocity field for this flow is

where U0 is the horizontal speed at x ! 0. Note that thisequation ignores viscous effects along the walls but is a rea-sonable approximation throughout the majority of the flowfield. Calculate the material acceleration for fluid particlespassing through this duct. Give your answer in two ways:(1) as acceleration components ax and ay and (2) as accelera-tion vector a

→.

V→

! (u, v) ! (U0 " bx) i→

# byj→

159CHAPTER 4

Helium-filledweather balloon

Transmittinginstrumentation

FIGURE P4–10C

4–11C A Pitot-static probe can often be seen protrudingfrom the underside of an airplane (Fig. P4–11C). As the air-plane flies, the probe measures relative wind speed. Is this aLagrangian or an Eulerian measurement? Explain.

Probe

FIGURE P4–11C

4–12C Is the Eulerian method of fluid flow analysis moresimilar to study of a system or a control volume? Explain.

yx

U0

FIGURE P4–15

4–16 Converging duct flow is modeled by the steady,two-dimensional velocity field of Prob. 4–15. The pressurefield is given by

where P0 is the pressure at x ! 0. Generate an expression forthe rate of change of pressure following a fluid particle.

4–17 A steady, incompressible, two-dimensional velocityfield is given by the following components in the xy-plane:

Calculate the acceleration field (find expressions for accelera-tion components ax and ay), and calculate the acceleration atthe point (x, y) ! (#2, 3). Answers: ax ! #9.233, ay ! 14.37


u ! 0.20 " 1.3x " 0.85y v ! #0.50 " 0.95x # 1.3y

u ! 1.1 " 2.8x " 0.65y v ! 0.98 # 2.1x # 2.8y

P ! P0 #r

2 B2U0 bx " b2(x 2 " y 2)R

cen72367_ch04.qxd 12/1/04 6:20 PM 59

Chapter 4 Fluid Kinematics

4-7 PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

4-17 Solution We are to calculate the material acceleration for a given velocity field. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The velocity field is

� � � �0,V u v U bx i byj � �& & &

(1)

The acceleration field components are obtained from its definition (the material acceleration) in Cartesian coordinates,

� � � �

� � � � � �

0

0

0 0 0

0 0 +0

x

y

u u u ua u v w U bx b by

t x y zv v v v

a u v w U bx by bt x y z

w w w w � � � � � � � �w w w ww w w w

� � � � � � � �w w w w

(2)

where the unsteady terms are zero since this is a steady flow, and the terms with w are zero since the flow is two-dimensional. Eq. 2 simplifies to

Material acceleration components: � � 20 x ya b U bx a b y � (3)

In terms of a vector,

Material acceleration vector: � � 20a b U bx i b yj � �

& && (4)

Discussion For positive x and b, fluid particles accelerate in the positive x direction. Even though this flow is steady, there is still a non-zero acceleration field.

4-18 Solution For a given pressure and velocity field, we are to calculate the rate of change of pressure following a fluid particle.

Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane.

Analysis The pressure field is

Pressure field: � �2 2 20 02

2P P U bx b x y

U ª º � � �¬ ¼ (1)

By definition, the material derivative, when applied to pressure, produces the rate of change of pressure following a fluid particle. Using Eq. 1 and the velocity components from the previous problem,

DP PDt t

w

w,Steady

P P Pu v w

x y zw w w

� � �w w w,

� �� Two-dimensional

2 20 0 U bx U b b x by b yU U U � � � � � �

(2)

where the unsteady term is zero since this is a steady flow, and the term with w is zero since the flow is two-dimensional. Eq. 2 simplifies to the following rate of change of pressure following a fluid particle:

� �2 2 3 2 20 02DP

U b U b x b y xDt

U ª º � � � �¬ ¼ (3)

Discussion The material derivative can be applied to any flow property, scalar or vector. Here we apply it to the pressure, a scalar quantity.




� � � �0,V u v U bx i byj � �& & &

(1)


� � � �

� � � � � �

0

0

0 0 0

0 0 +0

x

y


t x y zv v v v


w w w w � � � � � � � �w w w ww w w w

� � � � � � � �w w w w

(2)





& && (4)






2P P U bx b x y

U ª º � � �¬ ¼ (1)


DP PDt t

w

w,Steady

P P Pu v w

x y zw w w

� � �w w w,


2 20 0 U bx U b b x by b yU U U � � � � � �

(2)


� �2 2 3 2 20 02DP

U b U b x b y xDt

U ª º � � � �¬ ¼ (3)


CHAPTER 4159

location in the flow (Fig. P4–8C). Is this a Lagrangian or anEulerian measurement? Explain.

4–9C A tiny neutrally buoyant electronic pressure probe isreleased into the inlet pipe of a water pump and transmits2000 pressure readings per second as it passes through thepump. Is this a Lagrangian or an Eulerian measurement?Explain.

4–10C A weather balloon is launched into the atmosphereby meteorologists. When the balloon reaches an altitudewhere it is neutrally buoyant, it transmits information aboutweather conditions to monitoring stations on the ground (Fig.P4–10C). Is this a Lagrangian or an Eulerian measurement?Explain.

4–13C Define a steady flow field in the Eulerian referenceframe. In such a steady flow, is it possible for a fluid particleto experience a nonzero acceleration?

4–14C List at least three other names for the material deriv-ative, and write a brief explanation about why each name isappropriate.

4–15 Consider steady, incompressible, two-dimensionalflow through a converging duct (Fig. P4–15). A simpleapproximate velocity field for this flow is

where U0 is the horizontal speed at x ! 0. Note that thisequation ignores viscous effects along the walls but is a rea-sonable approximation throughout the majority of the flowfield. Calculate the material acceleration for fluid particlespassing through this duct. Give your answer in two ways:(1) as acceleration components ax and ay and (2) as accelera-tion vector a

→.

V→

! (u, v) ! (U0 " bx) i→

# byj→

159CHAPTER 4

Helium-filledweather balloon

Transmittinginstrumentation

FIGURE P4–10C

4–11C A Pitot-static probe can often be seen protrudingfrom the underside of an airplane (Fig. P4–11C). As the air-plane flies, the probe measures relative wind speed. Is this aLagrangian or an Eulerian measurement? Explain.

Probe

FIGURE P4–11C

4–12C Is the Eulerian method of fluid flow analysis moresimilar to study of a system or a control volume? Explain.

yx

U0

FIGURE P4–15

4–16 Converging duct flow is modeled by the steady,two-dimensional velocity field of Prob. 4–15. The pressurefield is given by

where P0 is the pressure at x ! 0. Generate an expression forthe rate of change of pressure following a fluid particle.


Calculate the acceleration field (find expressions for accelera-tion components ax and ay), and calculate the acceleration atthe point (x, y) ! (#2, 3). Answers: ax ! #9.233, ay ! 14.37


u ! 0.20 " 1.3x " 0.85y v ! #0.50 " 0.95x # 1.3y

u ! 1.1 " 2.8x " 0.65y v ! 0.98 # 2.1x # 2.8y

P ! P0 #r

2 B2U0 bx " b2(x 2 " y 2)R

cen72367_ch04.qxd 12/1/04 6:20 PM Page 159 Chapter 4 Fluid Kinematics



� � � �0,V u v U bx i byj � �& & &

(1)


� � � �

� � � � � �

0

0

0 0 0

0 0 +0

x

y


t x y zv v v v


w w w w � � � � � � � �w w w ww w w w

� � � � � � � �w w w w

(2)





& && (4)






2P P U bx b x y

U ª º � � �¬ ¼ (1)


DP PDt t

w

w,Steady

P P Pu v w

x y zw w w

� � �w w w,


2 20 0 U bx U b b x by b yU U U � � � � � �

(2)


� �2 2 3 2 20 02DP

U b U b x b y xDt

U ª º � � � �¬ ¼ (3)





� � � �0,V u v U bx i byj � �& & &

(1)


� � � �

� � � � � �

0

0

0 0 0

0 0 +0

x

y


t x y zv v v v


w w w w � � � � � � � �w w w ww w w w

� � � � � � � �w w w w

(2)





& && (4)






2P P U bx b x y

U ª º � � �¬ ¼ (1)


DP PDt t

w

w,Steady

P P Pu v w

x y zw w w

� � �w w w,


2 20 0 U bx U b b x by b yU U U � � � � � �

(2)


� �2 2 3 2 20 02DP

U b U b x b y xDt

U ª º � � � �¬ ¼ (3)


ì

Problem: Couette Flow

Since vorticity is non-zero, yes this flow is rotational; it is negative, implying that particles rotate in the clockwise direction; it is constant at every location in this flow.

4–68 A cylindrical tank of water rotates about its verticalaxis (Fig. P4–67). A PIV system is used to measure the vor-ticity field of the flow. The measured value of vorticity in thez-direction is !55.4 rad/s and is constant to within "0.5 per-cent everywhere that it is measured. Calculate the angularspeed of rotation of the tank in rpm. Is the tank rotatingclockwise or counterclockwise about the vertical axis?

4–69 A cylindrical tank of radius rrim # 0.35 m rotates aboutits vertical axis (Fig. P4–67). The tank is partially filled withoil. The speed of the rim is 2.6 m/s in the counterclockwisedirection (looking from the top), and the tank has been spin-ning long enough to be in solid-body rotation. For any fluidparticle in the tank, calculate the magnitude of the componentof vorticity in the vertical z-direction. Answer: 15.0 rad/s

4–70C Explain the relationship between vorticity and rota-tionality.

4–71 Consider a two-dimensional, incompressible flowfield in which an initially square fluid particle moves anddeforms. The fluid particle dimension is a at time t and isaligned with the x- and y-axes as sketched in Fig. P4–71. Atsome later time, the particle is still aligned with the x- and y-axes, but has deformed into a rectangle of horizontal length2a. What is the vertical length of the rectangular fluid particleat this later time?

166FLUID MECHANICS

4–74 Consider fully developed Couette flow—flow betweentwo infinite parallel plates separated by distance h, with thetop plate moving and the bottom plate stationary as illustratedin Fig. P4–74. The flow is steady, incompressible, and two-dimensional in the xy-plane. The velocity field is given by

Is this flow rotational or irrotational? If it is rotational, calcu-late the vorticity component in the z-direction. Do fluid parti-cles in this flow rotate clockwise or counterclockwise?Answers: yes, !V/h , clockwise

V→

# (u , v) # V y

h i

→$ 0 j

→

y

x

a

a

FIGURE P4–71

x

yh u =V

yh

V

FIGURE P4–74

4–75 For the Couette flow of Fig. P4–74, calculate the lin-ear strain rates in the x- and y-directions, and calculate theshear strain rate exy.

4–76 Combine your results from Prob. 4–75 to form thetwo-dimensional strain rate tensor eij,

Are the x- and y-axes principal axes?

Reynolds Transport Theorem4–77C True or false: For each statement, choose whetherthe statement is true or false and discuss your answer briefly.(a) The Reynolds transport theorem is useful for transform-ing conservation equations from their naturally occurringcontrol volume forms to their system forms.(b) The Reynolds transport theorem is applicable only tonondeforming control volumes.(c) The Reynolds transport theorem can be applied to bothsteady and unsteady flow fields.(d) The Reynolds transport theorem can be applied to bothscalar and vector quantities.

4–78 Consider the general form of the Reynolds transporttheorem (RTT) given by

dBsys

dt#

ddt

!CV

rb dV $ !CS

rbV→

r % n→ dA

eij # ¢exx exy

eyx eyy ≤

4–72 Consider a two-dimensional, compressible flow fieldin which an initially square fluid particle moves and deforms.The fluid particle dimension is a at time t and is aligned withthe x- and y-axes as sketched in Fig. P4–71. At some latertime, the particle is still aligned with the x- and y-axes buthas deformed into a rectangle of horizontal length 1.06a andvertical length 0.931a. (The particle’s dimension in the z-direction does not change since the flow is two-dimensional.)By what percentage has the density of the fluid particleincreased or decreased?

4–73 Consider the following steady, three-dimensional veloc-ity field:

Calculate the vorticity vector as a function of space (x, y, z).

# (3.0 $ 2.0x ! y) i→

$ (2.0x ! 2.0y) j→

$ (0.5xy)k→

V→

# (u , v, w)

cen72367_ch04.qxd 12/1/04 6:20 PM Page 166



4-78 Solution For a given velocity field we are to calculate the vorticity. Analysis The velocity field is

� � � � � � � �, , 3.0 2.0 2.0 2.0 0.5V u v w x y i x y j xy k � � � � �&& & &

(1)

In Cartesian coordinates, the vorticity vector is

Vorticity vector in Cartesian coordinates: w v u w v ui j k

y z z x x y§ · § ·w w w w w w§ · � � � � �¨ ¸ ¨ ¸¨ ¸w w w w w w© ¹© ¹ © ¹

&& & &] (2)

We substitute the velocity components u = 3.0 + 2.0x – y, v = 2.0x – 2.0y, and w = 0.5xy from Eq. 1 into Eq. 2 to obtain

Vorticity vector: � � � � � �� 0.5 0 0 0.5 2.0 1 0.5 0.5 3.0x i y j k x i y j k] � � � � � � � �& && & & & &

(3)

Discussion The vorticity is non-zero implying that this flow field is rotational.

4-79 Solution We are to determine if the flow is rotational, and if so calculate the z-component of vorticity. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The velocity field is given by

Velocity field, Couette flow: � �, 0yV u v V i j

h§ · �¨ ¸© ¹

& & & (1)

If the vorticity is non-zero, the flow is rotational. So, we calculate the z-component of vorticity,

z-component of vorticity: 0zv u V Vx y h h

] w w � � �w w

(2)

Since vorticity is non-zero, yes this flow is rotational. Furthermore, the vorticity is negative, implying that particles rotate in the clockwise direction. Discussion The vorticity is constant at every location in this flow.



4-80 Solution For the given velocity field for Couette flow, we are to calculate the two-dimensional linear strain rates and the shear strain rate. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The linear strain rates in the x direction and in the y direction are

Linear strain rates: xx yyu vx y

H Hw w w w

0 0 (1)

The shear strain rate in the x-y plane is

Shear strain rate: 1 1 02 2 2xy

u v V Vy x h h

H§ ·w w § · � � ¨ ¸ ¨ ¸w w © ¹© ¹

(2)

Fluid particles in this flow have non-zero shear strain rate. Discussion Since the linear strain rates are zero, fluid particles deform (shear), but do not stretch in either the horizontal or vertical directions.

4-81 Solution For the Couette flow velocity field we are to form the 2-D strain rate tensor and determine if the x and y axes are principal axes. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The two-dimensional strain rate tensor, Hij, is

2-D strain rate tensor: xx xyij

yx yy

H HH

H H§ ·

¨ ¸© ¹

(1)

We use the linear strain rates and the shear strain rate from the previous problem to generate the tensor,

2-D strain rate tensor: 0

2

02

xx xyij

yx yy

Vh

Vh

H HH

H H

§ ·¨ ¸§ ·¨ ¸ ¨ ¸¨ ¸© ¹ ¨ ¸© ¹

(2)

Note that by symmetry Hyx = Hxy. If the x and y axes were principal axes, the diagonals of Hij would be non-zero, and the off-diagonals would be zero. Here we have the opposite case, so the x and y axes are not principal axes. Discussion The principal axes can be calculated using tensor algebra.

Since the linear strain rates are zero, fluid particles deform (shear), but do not stretch in either the horizontal or vertical directions.



4-80 Solution For the given velocity field for Couette flow, we are to calculate the two-dimensional linear strain rates and the shear strain rate. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The linear strain rates in the x direction and in the y direction are


H Hw w w w

0 0 (1)


Shear strain rate: 1 1 02 2 2xy

u v V Vy x h h

H§ ·w w § · � � ¨ ¸ ¨ ¸w w © ¹© ¹

(2)


4-81 Solution For the Couette flow velocity field we are to form the 2-D strain rate tensor and determine if the x and y axes are principal axes. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The two-dimensional strain rate tensor, Hij, is


yx yy

H HH

H H§ ·

¨ ¸© ¹

(1)


2-D strain rate tensor: 0

2

02

xx xyij

yx yy

Vh

Vh

H HH

H H

§ ·¨ ¸§ ·¨ ¸ ¨ ¸¨ ¸© ¹ ¨ ¸© ¹

(2)


If the x and y axes were principal axes, the diagonals would be non-zero, and the off-diagonals would be zero. Here we have the opposite case, so the x and y axes are not principal axes

ì

Problem: Poiseuille Flow

This flow is rotational: in the lower half of the flow (y < h/2) the vorticity is negative (note that dP/dx is negative) and particles rotate in the clockwise direction in the lower half of the flow; particles rotate in the counterclockwise direction in the upper half of the flow. The vorticity varies linearly across the channel.

Since the linear strain rates are zero, fluid particles deform (shear), but do not stretch in either the horizontal or vertical directions.

If the x and y axes were principal axes, the diagonals would be non-zero, and the off-diagonals would be zero. Here we have the opposite case, so the x and y axes are not principal axes

CHAPTER 4167

where V→

r is the velocity of the fluid relative to the controlsurface. Let Bsys be the mass m of a system of fluid particles.We know that for a system, dm/dt ! 0 since no mass canenter or leave the system by definition. Use the given equa-tion to derive the equation of conservation of mass for a con-trol volume.

4–79 Consider the general form of the Reynolds transporttheorem (RTT) given by Prob. 4–78. Let Bsys be the linearmomentum mV

→of a system of fluid particles. We know that

for a system, Newton’s second law is

Use the equation of Prob. 4–78 and this equation to derivethe equation of conservation of linear momentum for a con-trol volume.

4–80 Consider the general form of the Reynolds transporttheorem (RTT) given in Prob. 4–78. Let Bsys be the angularmomentum H

→! r→ " mV

→of a system of fluid particles,

where r→ is the moment arm. We know that for a system, con-servation of angular momentum can be expressed as

where # M→

is the net moment applied to the system. Use theequation given in Prob. 4–78 and this equation to derive theequation of conservation of angular momentum for a controlvolume.

4–81 Reduce the following expression as far as possible:

(Hint: Use the one-dimensional Leibnitz theorem.) Answer:

Review Problems4–82 Consider fully developed two-dimensional Poiseuilleflow—flow between two infinite parallel plates separated bydistance h, with both the top plate and bottom plate station-ary, and a forced pressure gradient dP/dx driving the flow asillustrated in Fig. P4–82. (dP/dx is constant and negative.)

Be$ B2t2

$ Ae$ A2t2

F(t) !ddt

!x!Bt

x!At

e$ 2x2

dx

aM→

!ddt

H→

sys

a F→

! ma→

! mdV

→

dt!

ddt

(mV→

)sys

The flow is steady, incompressible, and two-dimensional inthe xy-plane. The velocity components are given by

where m is the fluid’s viscosity. Is this flow rotational or irro-tational? If it is rotational, calculate the vorticity componentin the z-direction. Do fluid particles in this flow rotate clock-wise or counterclockwise?

4–83 For the two-dimensional Poiseuille flow of Prob. 4–82,calculate the linear strain rates in the x- and y-directions, andcalculate the shear strain rate exy.

4–84 Combine your results from Prob. 4–83 to form thetwo-dimensional strain rate tensor eij in the xy-plane,


4–85 Consider the two-dimensional Poiseuille flow ofProb. 4–82. The fluid between the plates is water

at 40%C. Let the gap height h ! 1.6 mm and the pressure gra-dient dP/dx ! $ 230 N/m3. Calculate and plot seven path-lines from t ! 0 to t ! 10 s. The fluid particles are releasedat x ! 0 and at y ! 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, and 1.4 mm.

4–86 Consider the two-dimensional Poiseuille flow ofProb. 4–82. The fluid between the plates is

water at 40%C. Let the gap height h ! 1.6 mm and the pres-sure gradient dP/dx ! $ 230 N/m3. Calculate and plot sevenstreaklines generated from a dye rake that introduces dyestreaks at x ! 0 and at y ! 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, and 1.4 mm (Fig. P4–86). The dye is introduced from t ! 0 to t ! 10 s, and the streaklines are to be plotted at t ! 10 s.

eij ! ¢exx exy

eyx eyy ≤

u !1

2m dPdx

(y 2 $ hy) v ! 0

167CHAPTER 4

x

y

u (y)

h

FIGURE P4–82

4–87 Repeat Prob. 4–86 except that the dye is intro-duced from t ! 0 to t ! 10 s, and the streaklines

are to be plotted at t ! 12 s instead of 10 s.

4–88 Compare the results of Probs. 4–86 and 4–87and comment about the linear strain rate in the

x-direction.


x

y

u (y)Dye rake

h

FIGURE P4–86

cen72367_ch04.qxd 12/1/04 6:20 PM Page 167

CHAPTER 4167

where V→

r is the velocity of the fluid relative to the controlsurface. Let Bsys be the mass m of a system of fluid particles.We know that for a system, dm/dt ! 0 since no mass canenter or leave the system by definition. Use the given equa-tion to derive the equation of conservation of mass for a con-trol volume.

4–79 Consider the general form of the Reynolds transporttheorem (RTT) given by Prob. 4–78. Let Bsys be the linearmomentum mV

→of a system of fluid particles. We know that

for a system, Newton’s second law is

Use the equation of Prob. 4–78 and this equation to derivethe equation of conservation of linear momentum for a con-trol volume.

4–80 Consider the general form of the Reynolds transporttheorem (RTT) given in Prob. 4–78. Let Bsys be the angularmomentum H

→! r→ " mV

→of a system of fluid particles,

where r→ is the moment arm. We know that for a system, con-servation of angular momentum can be expressed as

where # M→

is the net moment applied to the system. Use theequation given in Prob. 4–78 and this equation to derive theequation of conservation of angular momentum for a controlvolume.

4–81 Reduce the following expression as far as possible:

(Hint: Use the one-dimensional Leibnitz theorem.) Answer:

Review Problems4–82 Consider fully developed two-dimensional Poiseuilleflow—flow between two infinite parallel plates separated bydistance h, with both the top plate and bottom plate station-ary, and a forced pressure gradient dP/dx driving the flow asillustrated in Fig. P4–82. (dP/dx is constant and negative.)

Be$ B2t2

$ Ae$ A2t2

F(t) !ddt

!x!Bt

x!At

e$ 2x2

dx

aM→

!ddt

H→

sys

a F→

! ma→

! mdV

→

dt!

ddt

(mV→

)sys

The flow is steady, incompressible, and two-dimensional inthe xy-plane. The velocity components are given by

where m is the fluid’s viscosity. Is this flow rotational or irro-tational? If it is rotational, calculate the vorticity componentin the z-direction. Do fluid particles in this flow rotate clock-wise or counterclockwise?

4–83 For the two-dimensional Poiseuille flow of Prob. 4–82,calculate the linear strain rates in the x- and y-directions, andcalculate the shear strain rate exy.

4–84 Combine your results from Prob. 4–83 to form thetwo-dimensional strain rate tensor eij in the xy-plane,



at 40%C. Let the gap height h ! 1.6 mm and the pressure gra-dient dP/dx ! $ 230 N/m3. Calculate and plot seven path-lines from t ! 0 to t ! 10 s. The fluid particles are releasedat x ! 0 and at y ! 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, and 1.4 mm.

4–86 Consider the two-dimensional Poiseuille flow ofProb. 4–82. The fluid between the plates is

water at 40%C. Let the gap height h ! 1.6 mm and the pres-sure gradient dP/dx ! $ 230 N/m3. Calculate and plot sevenstreaklines generated from a dye rake that introduces dyestreaks at x ! 0 and at y ! 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, and 1.4 mm (Fig. P4–86). The dye is introduced from t ! 0 to t ! 10 s, and the streaklines are to be plotted at t ! 10 s.

eij ! ¢exx exy

eyx eyy ≤

u !1

2m dPdx

(y 2 $ hy) v ! 0

167CHAPTER 4

x

y

u (y)

h

FIGURE P4–82

4–87 Repeat Prob. 4–86 except that the dye is intro-duced from t ! 0 to t ! 10 s, and the streaklines

are to be plotted at t ! 12 s instead of 10 s.

4–88 Compare the results of Probs. 4–86 and 4–87and comment about the linear strain rate in the

x-direction.


x

y

u (y)Dye rake

h

FIGURE P4–86

cen72367_ch04.qxd 12/1/04 6:20 PM Page 167



4-99 Solution We are to determine if the flow is rotational, and if so calculate the z-component of vorticity. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The velocity components are given by

Velocity components, 2-D Poiseuille flow: � �21 02

dPu y hy v

dxP � (1)


z-component of vorticity:

� � � �1 10 2 22 2z

v u dP dPy h y h

x y dx dx]

P Pw w

� � � � �w w

(2)

Since vorticity is non-zero, this flow is rotational. Furthermore, in the lower half of the flow (y < h/2) the vorticity is negative (note that dP/dx is negative). Thus, particles rotate in the clockwise direction in the lower half of the flow. Similarly, particles rotate in the counterclockwise direction in the upper half of the flow. Discussion The vorticity varies linearly across the channel.

4-100 Solution For the given velocity field for 2-D Poiseuille flow, we are to calculate the two-dimensional linear strain rates and the shear strain rate. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The linear strain rates in the x direction and in the y direction are


H Hw w w w

0 0 (1)


Shear strain rate:

� � � �1 1 1 12 0 22 2 2 4xy

u v dP dPy h y h

y x dx dxH

P P§ · § ·w w

� � � �¨ ¸ ¨ ¸w w© ¹ © ¹ (2)






dPu y hy v

dxP � (1)



� � � �1 10 2 22 2z

v u dP dPy h y h

x y dx dx]

P Pw w

� � � � �w w

(2)




H Hw w w w

0 0 (1)


Shear strain rate:

� � � �1 1 1 12 0 22 2 2 4xy

u v dP dPy h y h

y x dx dxH

P P§ · § ·w w

� � � �¨ ¸ ¨ ¸w w© ¹ © ¹ (2)






dPu y hy v

dxP � (1)



� � � �1 10 2 22 2z

v u dP dPy h y h

x y dx dx]

P Pw w

� � � � �w w

(2)




H Hw w w w

0 0 (1)


Shear strain rate:

� � � �1 1 1 12 0 22 2 2 4xy

u v dP dPy h y h

y x dx dxH

P P§ · § ·w w

� � � �¨ ¸ ¨ ¸w w© ¹ © ¹ (2)




4-101 Solution For the 2-D Poiseuille flow velocity field we are to form the 2-D strain rate tensor and determine if the x and y axes are principal axes. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Analysis The two-dimensional strain rate tensor, Hij, in the x-y plane,


yx yy

H HH

H H§ ·

¨ ¸© ¹

(1)


� �

� �

10 24

1 2 04

xx xyij

yx yy

dPy h

dxdP

y hdx

H H PH

H HP

§ ·�¨ ¸§ · ¨ ¸ ¨ ¸ ¨ ¸© ¹ �¨ ¸© ¹

(2)


4-102

Solution For a given velocity field we are to plot several pathlines for fluid particles released from various locations and over a specified time period. Assumptions 1 The flow is steady. 2 The flow is incompressible. 3 The flow is two-dimensional in the x-y plane. Properties For water at 40oC, P = 6.53u10-4 kg/m�s. Analysis Since the flow is steady, pathlines, streamlines, and streaklines are all straight horizontal lines. We simply need to integrate velocity component u with respect to time over the specified time period. The horizontal velocity component is

� �212

dPu y hy

dxP � (1)

We integrate as follows:

� �

� ��

end

start

10 s 2start 0

2

102

1 10 s2

t

t

dPx x udt y hy dt

dx

dPx y hy

dx

P

P

§ · � � �¨ ¸

© ¹

�

³ ³ (2)

We substitute the given values of y and the values of P and dP/dx into Eq. 2 to calculate the ending x position of each pathline. We plot the pathlines in Fig. 1. Discussion Streaklines introduced at the same locations and developed over the same time period would look identical to the pathlines of Fig. 1.

0

0.5

1

1.5

0 0.5 1 1.5x (m)

y (mm)

FIGURE 1 Pathlines for the given velocity field at t = 12 s. Note that the vertical scale is greatly expanded for clarity (x is in m, but y is in mm).

ì

Problem: Mass and Energy

CHAPTER 5231

entire air contained in a residence should be replaced by fresh outdoor air every hour. If the ventilation requirement of a 2.7-m-high, 200-m2 residence is to be met entirely by a fan, determine the flow capacity in L/min of the fan that needs to be installed. Also determine the minimum diameter of the duct if the average air velocity is not to exceed 5 m/s.

5–15 Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle. Answers: (a) 0.265 kg/s, (b) 23.2 cm2

5–16 Air at 408C flow steadily through the pipe shown in Fig. P5–16. If P1 5 50 kPa (gage), P2 5 10 kPa (gage), D 5 3d, Patm > 100 kPa, the average velocity at section 2 is V2530 m/s, and air temperature remains nearly constant, determine the average speed at section 1.

1.05 kg/m3 1.20 kg/m3

FIGURE P5–17

5–17 A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air is 1.20 kg/m3 at the inlet and 1.05 kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the hair dryer.

5–25 Electric power is to be generated by installing a hydraulic turbine–generator at a site 110 m below the free sur-face of a large water reservoir that can supply water steadily at a rate of 900 kg/s. If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.

5–26 Consider a river flowing toward a lake at an average speed of 4 m/s at a rate of 500 m3/s at a location 70 m above the lake surface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location. Answer: 347 MW

River 4 m/s

70 m

FIGURE P5–26

∆T = 0.048°F

Pump

FIGURE P5–24EFIGURE P5–16

P1

P2

dD

1 2

Mechanical Energy and Efficiency5–18C Define turbine efficiency, generator efficiency, and combined turbine–generator efficiency.

5–19C What is mechanical efficiency? What does a mechan-ical efficiency of 100 percent mean for a hydraulic turbine?

5–20C How is the combined pump–motor efficiency of a pump and motor system defined? Can the combined pump–motor effi-ciency be greater than either the pump or the motor efficiency?

5–21C What is mechanical energy? How does it differ from thermal energy? What are the forms of mechanical energy of a fluid stream?

5–22 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 that location. Also determine the actual electric power generation assuming an overall effi-ciency of 30 percent. Take the air density to be 1.25 kg/m3.

5–23 Reconsider Prob. 5–22. Using EES (or other) software, investigate the effect of wind velocity

and the blade span diameter on wind power generation. Let the velocity vary from 5 to 20 m/s in increments of 5 m/s, and the diameter to vary from 20 to 80 m in increments of 20 m. Tabulate the results, and discuss their significance.

5–24E A differential thermocouple with sensors at the inlet and exit of a pump indicates that the temperature of water rises 0.0488F as it flows through the pump at a rate of 1.5 ft3/s. If the shaft power input to the pump is 23 hp and the heat loss to the surrounding air is negligible, determine the mechanical efficiency of the pump. Answer: 72.4 percent

5–27 Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of

185-242_cengel_ch05.indd 231 12/21/12 2:26 PM

Chapter 5 Mass, Bernoulli, and Energy Equations

PROPRIETARY MATERIAL. © 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

5-8

5-15 Solution Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined.

Assumptions Flow through the nozzle is steady.

Properties The density of air is given to be 2.21 kg/m3 at the inlet, and 0.762 kg/m3 at the exit.

Analysis (a) The mass flow rate of air is determined from the inlet conditions to be

kg/s 0.265# kg/s 2652.0)m/s 20)(m 0.006)(kg/m 21.2( 23111 VAm U�

(b) There is only one inlet and one exit, and thus � � �m m m1 2 .

Then the exit area of the nozzle is determined to be

2cm 23.2 �o� 23

222222 m 00232.0

m/s) )(150mkg/ (0.762kg/s 0.2652

VmAVAm

UU

��

Discussion Since this is a compressible flow, we must equate mass flow rates, not volume flow rates.

5-16 Solution Air flows in a varying crosss section pipe. The speed at a specified section is to be determined. Assumptions Flow through the pipe is steady. Analysis

Applying conservation of mass for the cv shown,

³ ³ ��ww oo

cv cs

dAnVdt

0.... UU , 0.... 22211 �� VAAV UU 11

2221 .

..A

VAVU

U ,

RTPabs U ,

1

)(11 RT

P abs U , 2

)(22 RT

P abs U

)(1

)(2

1

2

abs

abs

PP

UU ,

2

2

2

1

2

4

4 ¸¹·

¨©§

Dd

D

d

AA

S

S

m/s2.44 ¸¹·

¨©§ 30.

31.

150110 2

1V

AIR V2 = 150 m/s

V1 = 20 m/s A1 = 60 cm2



5-8

5-15 Solution Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined.

Assumptions Flow through the nozzle is steady.

Properties The density of air is given to be 2.21 kg/m3 at the inlet, and 0.762 kg/m3 at the exit.

Analysis (a) The mass flow rate of air is determined from the inlet conditions to be

kg/s 0.265# kg/s 2652.0)m/s 20)(m 0.006)(kg/m 21.2( 23111 VAm U�

(b) There is only one inlet and one exit, and thus � � �m m m1 2 .

Then the exit area of the nozzle is determined to be

2cm 23.2 �o� 23

222222 m 00232.0

m/s) )(150mkg/ (0.762kg/s 0.2652

VmAVAm

UU

��

Discussion Since this is a compressible flow, we must equate mass flow rates, not volume flow rates.

5-16 Solution Air flows in a varying crosss section pipe. The speed at a specified section is to be determined. Assumptions Flow through the pipe is steady. Analysis

Applying conservation of mass for the cv shown,

³ ³ ��ww oo

cv cs

dAnVdt

0.... UU , 0.... 22211 �� VAAV UU 11

2221 .

..A

VAVU

U ,

RTPabs U ,

1

)(11 RT

P abs U , 2

)(22 RT

P abs U

)(1

)(2

1

2

abs

abs

PP

UU ,

2

2

2

1

2

4

4 ¸¹·

¨©§

Dd

D

d

AA

S

S

m/s2.44 ¸¹·

¨©§ 30.

31.

150110 2

1V

AIR V2 = 150 m/s

V1 = 20 m/s A1 = 60 cm2

CHAPTER 5231

entire air contained in a residence should be replaced by fresh outdoor air every hour. If the ventilation requirement of a 2.7-m-high, 200-m2 residence is to be met entirely by a fan, determine the flow capacity in L/min of the fan that needs to be installed. Also determine the minimum diameter of the duct if the average air velocity is not to exceed 5 m/s.

5–15 Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle. Answers: (a) 0.265 kg/s, (b) 23.2 cm2

5–16 Air at 408C flow steadily through the pipe shown in Fig. P5–16. If P1 5 50 kPa (gage), P2 5 10 kPa (gage), D 5 3d, Patm > 100 kPa, the average velocity at section 2 is V2530 m/s, and air temperature remains nearly constant, determine the average speed at section 1.

1.05 kg/m3 1.20 kg/m3

FIGURE P5–17

5–17 A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air is 1.20 kg/m3 at the inlet and 1.05 kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the hair dryer.

5–25 Electric power is to be generated by installing a hydraulic turbine–generator at a site 110 m below the free sur-face of a large water reservoir that can supply water steadily at a rate of 900 kg/s. If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.

5–26 Consider a river flowing toward a lake at an average speed of 4 m/s at a rate of 500 m3/s at a location 70 m above the lake surface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location. Answer: 347 MW

River 4 m/s

70 m

FIGURE P5–26

∆T = 0.048°F

Pump

FIGURE P5–24EFIGURE P5–16

P1

P2

dD

1 2

Mechanical Energy and Efficiency5–18C Define turbine efficiency, generator efficiency, and combined turbine–generator efficiency.

5–19C What is mechanical efficiency? What does a mechan-ical efficiency of 100 percent mean for a hydraulic turbine?

5–20C How is the combined pump–motor efficiency of a pump and motor system defined? Can the combined pump–motor effi-ciency be greater than either the pump or the motor efficiency?

5–21C What is mechanical energy? How does it differ from thermal energy? What are the forms of mechanical energy of a fluid stream?

5–22 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 that location. Also determine the actual electric power generation assuming an overall effi-ciency of 30 percent. Take the air density to be 1.25 kg/m3.

5–23 Reconsider Prob. 5–22. Using EES (or other) software, investigate the effect of wind velocity

and the blade span diameter on wind power generation. Let the velocity vary from 5 to 20 m/s in increments of 5 m/s, and the diameter to vary from 20 to 80 m in increments of 20 m. Tabulate the results, and discuss their significance.

5–24E A differential thermocouple with sensors at the inlet and exit of a pump indicates that the temperature of water rises 0.0488F as it flows through the pump at a rate of 1.5 ft3/s. If the shaft power input to the pump is 23 hp and the heat loss to the surrounding air is negligible, determine the mechanical efficiency of the pump. Answer: 72.4 percent

5–27 Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of

185-242_cengel_ch05.indd 231 12/21/12 2:26 PM



5-11

5-22

Solution Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation potential, and the actual electric power generation are to be determined.

Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency of the wind turbine is independent of the wind speed.

Properties The density of air is given to be U = 1.25 kg/m3.

Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work

entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and 2/2Vm� for a given mass flow rate:

kJ/kg 032.0/sm 1000

kJ/kg 1

2

)m/s 8(

2 22

22

mech ¸¹·

¨©§

Vkee

kg/s 635,194

m) (50m/s) 8)(kg/m 25.1(

4

23

2

SS

UUD

VVAm�

kW 628 kJ/kg) 2kg/s)(0.03 635,19(mechmechmax emEW ��

The actual electric power generation is determined by multiplying the power generation potential by the efficiency,

kW 188 kW) 628)(30.0(max turbinewindelect WW �� K

Therefore, 283 kW of actual power can be generated by this wind turbine at the stated conditions.

Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.

Wind

8 m/s 50 m

Wind turbine

ì

Problem: Bernoulli Equation



5-67

5-94

Solution Underground water is pumped to a pool at a given elevation. The maximum flow rate and the pressures at the inlet and outlet of the pump are to be determined.

Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the inlet and the outlet of the pump is negligible. 3 We assume the frictional effects in piping to be negligible since the maximum flow rate is to be

determined, .0 pippingloss, mech E� 4 The effect of the kinetic energy correction factors is negligible, D = 1.

Properties We take the density of water to be 1 kg/L = 1000 kg/m3.

Analysis (a) The pump-motor draws 5-kW of power, and is 78% efficient. Then the useful mechanical (shaft) power it delivers to the fluid is

kW 9.3kW) 5)(78.0(electricmotor-pumpu pump, WW �� K

We take point 1 at the free surface of underground water, which is also taken as the reference level (z1 = 0), and point 2 at the free surface of the pool. Also, both 1 and 2 are open to the atmosphere (P1 = P2 = Patm), the velocities are negligible at both points (V1 # V2 # 0), and frictional losses in piping are disregarded. Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to

lossmech,turbine2

22

22

pump1

21

11

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U

In the absence of a turbine, pipingloss, mech pumploss, mechloss mech, EEE �� and

pumploss, mechpump upump, EWW �� .

Thus, 2 upump, gzmW �� .

Then the mass and volume flow rates of water become

kg/s 25.13kJ 1

/sm 1000

m) )(30m/s (9.81

kJ/s 9.3 22

22

u pump, ¸¸¹

·¨¨©

§

gz

Wm

��

/sm 0.0133/ 3# sm 01325.0kg/m 1000

kg/s 25.13 33U

m��V

(b) We take points 3 and 4 at the inlet and the exit of the pump, respectively, where the flow velocities are

m/s 443.34/m) (0.07

/sm 01325.0

4/ 2

3

233

3 SSDA

VVV ��

, m/s 748.64/m) (0.05

/sm 01325.0

4/ 2

3

244

4 SSDA

VVV ��

We take the pump as the control volume. Noting that z3 = z4, the energy equation for this control volume reduces to

pumploss, mechturbine4

24

44

pump3

23

33

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U o

V�

�upump,

24

23

34 2

)( WVVPP �

� �UD

Substituting,

kPa 278# ��

¸¹·

¨©§ �

�¸¸¹

·¨¨©

§

��

�

kPa 5.277kN/m )3.2948.16(

kJ 1

mkN 1

/sm 0.01325

kJ/s 9.3

m/skg 1000

kN 1

2

]m/s) (6.748)m/s 443.3()[kg/m (1000

2

32

223

34 PP

Discussion In an actual system, the flow rate of water will be less because of friction in the pipes. Also, the effect of flow velocities on the pressure change across the pump is negligible in this case (under 2%) and can be ignored.

30 m

Pool

Pump

2

1



5-67

5-94








lossmech,turbine2

22

22

pump1

21

11

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U





kg/s 25.13kJ 1

/sm 1000

m) )(30m/s (9.81

kJ/s 9.3 22

22

u pump, ¸¸¹

·¨¨©

§

gz

Wm

��

/sm 0.0133/ 3# sm 01325.0kg/m 1000

kg/s 25.13 33U

m��V


m/s 443.34/m) (0.07

/sm 01325.0

4/ 2

3

233

3 SSDA

VVV ��

, m/s 748.64/m) (0.05

/sm 01325.0

4/ 2

3

244

4 SSDA

VVV ��



24

44

pump3

23

33

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U o

V�

�upump,

24

23

34 2

)( WVVPP �

� �UD

Substituting,

kPa 278# ��

¸¹·

¨©§ �

�¸¸¹

·¨¨©

§

��

�

kPa 5.277kN/m )3.2948.16(

kJ 1

mkN 1

/sm 0.01325

kJ/s 9.3

m/skg 1000

kN 1

2

]m/s) (6.748)m/s 443.3()[kg/m (1000

2

32

223

34 PP


30 m

Pool

Pump

2

1

1

2



5-67

5-94








lossmech,turbine2

22

22

pump1

21

11

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U





kg/s 25.13kJ 1

/sm 1000

m) )(30m/s (9.81

kJ/s 9.3 22

22

u pump, ¸¸¹

·¨¨©

§

gz

Wm

��

/sm 0.0133/ 3# sm 01325.0kg/m 1000

kg/s 25.13 33U

m��V


m/s 443.34/m) (0.07

/sm 01325.0

4/ 2

3

233

3 SSDA

VVV ��

, m/s 748.64/m) (0.05

/sm 01325.0

4/ 2

3

244

4 SSDA

VVV ��



24

44

pump3

23

33

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U o

V�

�upump,

24

23

34 2

)( WVVPP �

� �UD

Substituting,

kPa 278# ��

¸¹·

¨©§ �

�¸¸¹

·¨¨©

§

��

�

kPa 5.277kN/m )3.2948.16(

kJ 1

mkN 1

/sm 0.01325

kJ/s 9.3

m/skg 1000

kN 1

2

]m/s) (6.748)m/s 443.3()[kg/m (1000

2

32

223

34 PP


30 m

Pool

Pump

2

1



5-67

5-94








lossmech,turbine2

22

22

pump1

21

11

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U





kg/s 25.13kJ 1

/sm 1000

m) )(30m/s (9.81

kJ/s 9.3 22

22

u pump, ¸¸¹

·¨¨©

§

gz

Wm

��

/sm 0.0133/ 3# sm 01325.0kg/m 1000

kg/s 25.13 33U

m��V


m/s 443.34/m) (0.07

/sm 01325.0

4/ 2

3

233

3 SSDA

VVV ��

, m/s 748.64/m) (0.05

/sm 01325.0

4/ 2

3

244

4 SSDA

VVV ��



24

44

pump3

23

33

22EWgz

VPmWgz

VPm �� ¸

¸¹

·¨¨©

§�� ¸

¸¹

·¨¨©

§�� D

UD

U o

V�

�upump,

24

23

34 2

)( WVVPP �

� �UD

Substituting,

kPa 278# ��

¸¹·

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2

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2

32

223

34 PP


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Pool

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3

233

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2

32

223

34 PP


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Pool

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2

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22

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gz

Wm

��

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4/ 2

3

233

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244

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kJ 1

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2

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2

32

223

34 PP


30 m

Pool

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2

1



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22

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gz

Wm

��

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4/ 2

3

233

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kJ 1

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m/skg 1000

kN 1

2

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2

32

223

34 PP


30 m

Pool

Pump

2

1



5-67

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22

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gz

Wm

��

/sm 0.0133/ 3# sm 01325.0kg/m 1000

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4/ 2

3

233

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kJ 1

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m/skg 1000

kN 1

2

]m/s) (6.748)m/s 443.3()[kg/m (1000

2

32

223

34 PP


30 m

Pool

Pump

2

1

238BERNOULLI AND ENERGY EQUATIONS

5–94 Underground water is to be pumped by a 78 percent efficient 5-kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. Determine (a) the maximum flow rate of water and (b) the pressure difference across the pump. Assume the eleva-tion difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible.

Suppose a utility company is selling electric power for $0.06/kWh at night and is willing to pay $0.13/kWh for power produced during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir 50 m above the lake level, pumping water from the lake to the reser-voir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump–motor operates as a turbine–generator during reverse flow. Preliminary analysis shows that a water flow rate of 2 m3/s can be used in either direction, and the irreversible head loss of the piping system is 4 m. The combined pump–motor and tur-bine–generator efficiencies are expected to be 75 percent each. Assuming the system operates for 10 h each in the pump and turbine modes during a typical day, determine the potential rev-enue this pump–turbine system can generate per year.

5–98 When a system is subjected to a linear rigid body motion with constant linear acceleration a along a distance L, the modified Bernoulli Equation takes the form

aP1

r1

V 21

21 gz1b 2 a

P2

r1

V 22

21 gz2b 5 aL 1 Losses

where V1 and V2 are velocities relative to a fixed point and ‘Losses’ which represents frictional losses is zero when the frictional effects are negligible. The tank with two discharge pipes shown in Fig. P5–98 accelerates to the left at a constant linear acceleration of 3 m/s2. If volumetric flow rates from both pipes are to be identical, determine the diameter D of the inclined pipe. Disregard any frictional effects.

30 m

Pool

FIGURE P5–94

5–95 Reconsider Prob. 5–94. Determine the flow rate of water and the pressure difference across the pump if the irre-versible head loss of the piping system is 4 m.

5–96E A 73-percent efficient 12-hp pump is pumping water from a lake to a nearby pool at a rate of 1.2 ft3/s through a con-stant-diameter pipe. The free surface of the pool is 35 ft above that of the lake. Determine the irreversible head loss of the pip-ing system, in ft, and the mechanical power used to overcome it.

5–97 The demand for electric power is usually much higher during the day than it is at night, and utility companies often sell power at night at much lower prices to encourage consumers to use the available power generation capacity and to avoid building new expensive power plants that will be used only a short time during peak periods. Utilities are also willing to purchase power produced during the day from private parties at a high price.

Pump–turbine

Lake

50 m

Reservoir

FIGURE P5–97

5–99 A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diam-eter pipe at a rate of 0.04 m3/s and discharging it through a hose nozzle with an exit diameter of 5 cm. The total irrevers-ible head loss of the system is 3 m, and the position of the nozzle is 3 m above sea level. For a pump efficiency of 70 percent, determine the required shaft power input to the pump and the water discharge velocity. Answers: 39.2 kW, 20.4 m/s

A

L = 8 m

h = 3 ma = 3 m/s

d = 1 cm

BC

D

V1 V2

FIGURE P5–98

185-242_cengel_ch05.indd 238 12/17/12 10:56 AM

3

4

auvw ubx byb problem: material derivative auvw ubx byb

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