ch03.pdf

Chapter 3

1. A turbine turbi ne develops 15,500 hp with with a decrease in head of 37 ft and a rotational speed of 106 RPM. RPM. What type of turbine is best suited for this application? RPM :=

ω :=

2⋅π min

106 106 ⋅ RPM RP M

ρ :=

62.4 ⋅

lb 3

g := 32.174 ⋅

ft

sec

Power gen := 1550 15500. 0.0 0 ⋅ hp

⎛ Power gen ⎞ ⎝ ρ ⎠ Ns := ω ⋅

ft

∆Z :=

2

37.0 ⋅ ft ft

0.5

Ns = 3.328

5

(This is the dimensionless version.)

(g ⋅ ∆Z) 4

⎛ Power gen ⎞ hp ⎠ ω Ncus := ⋅ ⎝ RPM

5

⎛ ∆Z ⎞ ⎝ ft ⎠

0.5

Ncus = 144.617 (This is the customary version.)

4

For a specific speed (customary) of 145, the Kaplan turbine is best suited.

2. A Francis turbine is used in an installation to generate generate power. The power output is 115,000 hp hp under a head of of 480 ft. The outer radius of the rotor is 6.75 ft; the inner radius is 5.75 ft. The rotational speed is 180 RPM, RPM, and the blade height is 1.5 ft. The turbine efficiency is 0.95 percent. The absolute velocity velocity leaving the rotor is in the radial direction. Determine Determine the angle the absolute absolute velocity entering the rotor makes with the radial direction.

V

0.732 1.5 ft m

Vr

α

5.52 mft 1 =6.75 r 1r =

r 2 = 5.75 ft 2⋅π

RPM :=

ρ :=

min

lb

180 180 ⋅ RPM RPM

η :=

0.95

3

sec

U1 := r 1 ⋅ ω

U1

Power out := 115000 115000 ⋅ hp

π ⋅ r 1 ⋅ height

Area rea

=

ft

g := 32.174 ⋅

ft

ω :=

Area := 2 ⋅

62.4 ⋅

=

2

r 1 := 6.75 6.75 ⋅ ft height := 1.5 ⋅ ft

-1

38.7 38.781m 81m ⋅ s

Power in :=

Power out

η

5

Power in = 1.2 1.211 × 10

2

63.6 63.617 17 ⋅ ft

With the power extracted and the available change in head known, the flow rate can be calculated. Power in

Q :=

ρ ⋅ 480 ⋅ ft ⋅

3

Q = 2.223 × 10

lbf

⋅

ft

3

sec

Q

=

5

9.977 × 10

⋅ gpm

lb The radial component of the absolute inlet velocity then becomes Q

Vm1 :=

Area

ft

Vm1 = 34.941 ⋅

sec

Since the exit absolute velocity is radial, Vu 2 = 0 and Power in = mdot dot ⋅ U1 ⋅ Vu1

Vu1 :=

Power in

Vu1 = 121.379 ⋅

ft

sec ρ ⋅ Q ⋅ U1 From the inlet velocity triangle, Vm1 = V 1 cos (α) and Vu1 = V 1 sin(α) so that

α :=

45 ⋅ de deg

Initial guess on

α

Given Vm1 Vu1

α :=

=

cos( α ) sin( α )

Find ( α )

α=

73.941 73.941 ⋅ deg

⋅ hp

3. A hydroelectric facility operates with an elevation difference of 50 m with a flow rate of 500 m 3/s. If the rotational speed is 90 RPM, determine the most suitable type of turbine and estimate the power output of the arrangement. RPM :=

2⋅

π

min

ρ :=

1000 ⋅

kg 3

m

g := 9.806 ⋅

m

sec

2

3

ω :=

90.0 ⋅ RPM

∆Z :=

50.0 ⋅ m

Q := 500 ⋅

m

sec

A reasonable efficiency for virtually any hydropower device is 0.90. The power output, neglecting major and minor losses, can be estimated as

η :=

0.90

Power gen :=


η ⋅ ρ ⋅ Q ⋅ ∆Z ⋅ g

Power gen = 2.206

×

5

10

⋅ kW

0.5

5

Ns = 1.919


Ncus = 83.388

(This is the customary version.)

( g ⋅ ∆Z) 4


5

⎛ ∆Z ⎞ ⎝ ft ⎠

0.5

4

For a specific speed (customary) of 83, either the Francis or Kaplan turbine is best suited. The estimate of an efficiency of 0.90 is congruent with Figure 3.9.

4. A Pelton wheel, located at an elevation of 200 m, is used to extract power from a reservoir at an elevation of 800 m. A penstock 1500-m long and 0.8 m in diameter connects the reservoir with the nozzle of the Pelton wheel. Minor losses are K = 3 and C = 1500. The Pelton wheel is 3 m in diameter, and the nozzle exit is 0.2 m in diameter. Determine the maximum power output of the Pelton wheel. Is a Pelton wheel the appropriate turbine type for this application? ORIGIN

≡

Set origin for counters to 1 from the default value of 0.

1

Input the pipe geometry: Diameter in mm D :=

Length in m

⎛ 0.8 ⎞ ⋅ m ⎝ 0.8 ⎠

L :=

Roughness in mm:

⎛ 750 ⎞ ⋅ m ⎝ 750 ⎠

ε :=

⎛ 0.046 ⎞ ⋅ mm ⎝ 0.046 ⎠

(commercial pipe)

Input the system boundary (initial and end) conditions: Pressures in Pa

Elevations in m:

⎛ Pa ⎞ ⎛ 0 ⎞ := ⋅ Pa ⎝ P b ⎠ ⎝ 0 ⎠

⎛ Za ⎞ ⎛ 800 ⎞ := ⋅m ⎝ Z b ⎠ ⎝ 200 ⎠

Dnoz := 0.20 ⋅ m

Input the loss coefficients: Kvalue :=

⎛ D1 ⎞

4

Kvalue

⎝ Dnoz ⎠

K factor K :=

=

This accounts for the nozzle exit kinetic energy in terms of the pipe fluid velocity.

256

Equivalent length

⎛ 3 ⎞ ⎝ Kvalue ⎠

C :=

Number of pipes

⎛ 1500 ⎞ ⎝ 0 ⎠

N := length( D)

Input the fluid properties: Density in kg/m3

ρ :=

1000 ⋅

Kinematic Viscosity in m 2 /s 2

kg

ν :=

3

−6 m

1.14 ⋅ 10

⋅

m Input the flow rate in cms:

sec

No turbine or pump prior to penstock:

3

Q := 1 ⋅

m

Ws := 0 ⋅ newton ⋅

Initial guess on flow rate.

sec

Define constants and adjust units for consistency: m m ⋅ kg g := 9.806 ⋅ g c := 1 ⋅ sec

2

m kg

2

newton ⋅ sec

Define the functions for Reynolds number, fully-rough friction factor, and friction factor: Re( q, d ) :=

4 ⋅ q

π ⋅ d ⋅ ν

f T( d , ε ) :=

0.3086

⎡⎛ ε ⎞ 1.11⎤ log ⎣⎝ 3.7 ⋅ d ⎠ ⎦

2

f( q , d , ε ) :=

0.3086

⎡ log

ε ⎞ + ⎛ ⎝ 3.7 ⋅ d ⎠

6.9

⎣ Re( q, d )

64

⎤

1.11

if Re( q, d )

2

> 2300

⎦

otherwise

Re( q, d )

The generalized energy equation is: Given gc

Ws ⋅

g

=

P b

−

N

Pa

⋅ gc +

ρ⋅g

Z b

−

Za

∑

+

⎡8

2

⎢ 2 i = 1 ⎣π

Q

⋅

( i) D

⋅

4

⋅g

⎛

(

L

i

)⋅

f Q, D , ε i

⎝

i

D

+

i

K + C i

⎞⎤ ⎠⎥⎦

⋅ f D , ε i T( i i)

q := Find( Q)

=

q

3

3.166 m

-1

⋅s

=

q

5 liter 1.9 × 10 ⋅ min

Additional output of useful quantities: i := 1 .. N

V( q , D) :=

4 ⋅ q 2

π⋅D D

i

(

=

V q, D

)=

(

Re q , D

i

0.8 m

6.299

0.8

6.299

-1

m⋅ s

(

)=

)=

(

f q,D ,εi

i

i

f T D

i

, ε i) =

4.42·106

0.011

0.011

4.42·106

0.011

0.011

The jet velocity is: q

V jet := 0.25 ⋅

V jet

2

-1

=

100.786 m ⋅ s

π ⋅ Dnoz

The kinetic energy available for the Pelton wheel is 0.5 V jet 2 ; the available power is the available kinetic energy times flow rate. 2

Power := q ⋅ ρ

⋅

V jet

4

Power = 1.608 × 10

2

⋅ kW

β :=

175 ⋅ deg

Where β is a reasonable value 2

Power out := q ⋅ ρ ⋅ V jet

⋅

1

−

cos( β )

4

Power out = 1.605 × 10

4

⋅ kW

For maximum efficiency of a Pelton wheel, the blade speed is 0.5 of the jet speed. -1

U := 0.5 ⋅ V jet

U = 50.393 m ⋅ s

Dwheel := 3.0 ⋅ m Torque :=

Power out

ω

ω :=

RPM :=

U

ω =

0.5 ⋅ Dwheel

Torque

=

5

4.778 × 10

2⋅

π

min -1

33.595 s

⋅ newton ⋅ m

ω=

320.811 ⋅ RPM

The head corresponding to V jet is 2

∆h :=

V jet

∆h =

2⋅g

517.937 m

The power specific speed for this example, using the head corresponding to V jet , is

⎛ Power out ⎞ ⎝ ρ ⎠ Ns := ω ⋅

0.5

5

Ns = 0.099


Ncus = 4.314


( g ⋅ ∆h ) 4 ω

Ncus := 2⋅

π min

⎛ Power out ⎞ hp ⎠ ⋅ ⎝

0.5

5

⎛ ∆h ⎞ ⎝ ft ⎠

4

Then the efficiency is approximately 88 percent from Figure 12.32. Power max := Power out ⋅ 0.88 Power max = 1.412

×

4

10

⋅ kW

5. A proposed design for a hydroelectric project is based on a discharge of 0.25 m /s through the penstock and turbine as illustrated in the figure. The minor losses are considered negligible. (a) Determine the power in kW that can be expected from the facility, if the turbine efficiency is 0.85 (b) What type of turbine should be installed if the desired rotational speed is 1200 RPM? 3

el = 982 m L = 350 m D = 300 mm

el = 915 m

Turbine

ORIGIN

≡

1



⎛ 0.3 ⎞ ⋅ m ⎝ 0.3 ⎠

Length in m L :=

Roughness in mm:

⎛ 175 ⎞ ⋅ m ⎝ 175 ⎠

ε :=

⎛ 0.046 ⎞ ⋅ mm ⎝ 0.046 ⎠

(commercial pipe)


⎛ Pa ⎞ ⎛ 0 ⎞ := ⋅ Pa P 0 ⎝ ⎠ b ⎝ ⎠

Elevations in m:

⎛ Za ⎞ ⎛ 982 ⎞ := ⋅m Z 915 ⎝ ⎠ b ⎝ ⎠

Input the loss coefficients: K factor K :=

⎛ .78 ⎞ ⎝ 1.0 ⎠

Equivalent length C :=

⎛ 0 ⎞ ⎝ 0 ⎠


Kinematic Viscosity in m 2 /s

Number of pipes N := length( D)

ρ :=

1000 ⋅

2

kg

ν :=

3

−6 m

1.14 ⋅ 10

⋅

m

Input the flow rate in cms:

sec

Initial guess on turbine decrease in head:

3

Q := 0.25 ⋅

m

Ws :=

Flow rate.

sec


2

−15 ⋅ newton ⋅

m kg

2

newton ⋅ sec

Define the functions for Reynolds number, fully-rough friction factor, and friction factor: 4 ⋅ q

Re( q, d ) :=

π ⋅ d ⋅ ν

⎡ log

0.3086

⎡⎛ ε ⎞ 1.11⎤ log ⎣⎝ 3.7 ⋅ d ⎠ ⎦

f( q , d , ε ) :=

f T( d , ε ) :=

0.3086

ε ⎞ + ⎛ ⎝ 3.7 ⋅ d ⎠

6.9

⎣ Re( q, d )

64 Re( q, d )

⎤

1.11

2

if Re( q, d )

2

> 2300

⎦

otherwise

The generalized energy equation is: Given Ws ⋅

gc

=

g

P b

−

Pa

ρ⋅g

N

⋅ gc +

Z b

−

Za +

∑

⎡8

⎢ 2 i = 1 ⎣π

2

Q

⋅

( i) D

4

⋅ ⋅g

⎛

(

)⋅

f Q, D , ε i

⎝

i

L

i

+

D

K + C

i

i

⎞⎤ ⎠⎥⎦

⋅ f D , ε i T( i i)

Ws := Find( Ws) 2

Ws = −542.897 m

-2

⋅s

Ws = −542.897 ⋅ newton ⋅

m kg


V( q , D) :=

4 ⋅ q 2

π⋅D D

i

=

(

V Q, D

)=

(

Re Q , D

i

0.3 m

3.537

0.3

3.537

-1

m⋅ s

)=

i

(

)=

f Q, D , ε i i

ρ ⋅ Ws

Power gen := 0.85 ⋅ Power

i

, ε i) =

9.307·105

0.014

0.013

9.307·105

0.014

0.013

The head decrease at the turbine can be used to find the power generated. Power := Q ⋅

(

f T D

Power = −135.724 ⋅ kW Power gen = −115.366 ⋅ kW

The power specific speed for this example is

ω :=

1200 ⋅

2⋅π min

The change in head across the device is

∆h :=

−Ws

∆h =

g

⎛ −Power gen ⎞ ⎝ ρ ⎠ Ns := ω ⋅

55.364 m

0.5

Ns = 0.515

5


( g ⋅ ∆h ) 4

ω

Ncus := 2⋅

π min

⎛ −Power gen ⎞ hp ⎠ ⋅ ⎝ 5

⎛ ∆h ⎞ ⎝ ft ⎠

0.5

Ncus = 22.383


4

Based on the customary value of the specific speed, 22, the choice for the turbine type would be a Francis turbine.

6. A hydroelectric facility extracts power from a reservoir at an elevation of 800 m discharging into a tailrace at an elevation of 200 m. A penstock 1500 m long and 0.8 m in diameter connects the reservoir to the tailrace. Minor losses are K = 3 and C = 1500. Determine the power extracted by the turbine if the flow rate is 2 m 3 /s. If a Francis turbine is to be used, specify an appropriate speed. ORIGIN

≡


1


Length in m

⎛ 0.8 ⎞ ⋅ m ⎝ 0.8 ⎠

L :=

Roughness in mm:

⎛ 750 ⎞ ⋅ m ⎝ 750 ⎠

ε :=

⎛ 0.046 ⎞ ⋅ mm ⎝ 0.046 ⎠

(commercial pipe)


Elevations in m:

⎛ Pa ⎞ ⎛ 0 ⎞ := ⋅ Pa ⎝ P b ⎠ ⎝ 0 ⎠

⎛ Za ⎞ ⎛ 800 ⎞ := ⋅m ⎝ Z b ⎠ ⎝ 200 ⎠


Equivalent length

⎛ 3 ⎞ ⎝ 0 ⎠

C :=

Number of pipes

⎛ 1500 ⎞ ⎝ 0 ⎠

N := length( D)


ρ :=

1000 ⋅


kg

ν :=

3

−6 m

1.14 ⋅ 10

⋅


sec

Initial guess on turbine change in head:

3

Q := 2.0 ⋅

m


sec


2

m kg

2

newton ⋅ sec


4 ⋅ q

π ⋅ d ⋅ ν

f T( d , ε )

:=

0.3086

⎡⎛ ε ⎞ 1.11⎤ log ⎣⎝ 3.7 ⋅ d ⎠ ⎦

2

f( q , d , ε )

:=

0.3086

⎡ log

6.9

⎣ Re( q, d )

64

ε ⎞ + ⎛ ⎝ 3.7 ⋅ d ⎠

⎤

1.11

if Re( q, d )

2

> 2300

⎦

otherwise

Re( q, d )


Ws ⋅

g

P b

=

−

Pa

ρ⋅g

⎡8

N

⋅ gc +

−

Z b

Za +

Ws := Find( Ws) 3

2

Ws = −5.559 × 10 m

∑

2

⎢ 2 i = 1 ⎣π

Q

⋅

( i) D

4

⋅ ⋅g

⎛

(

L

i

)⋅

f Q, D , ε i

⎝

i

D

i

+

K + C i

⎞⎤ ⎠⎥⎦

⋅ f D , ε i T( i i)

-2

⋅s


4 ⋅ q

V( q , D) :=

2

π⋅D D

i

(

=

V q, D

0.8 m

)=

6.299 6.299

0.8

Power extracted :=

(

Re q , D

i

-1

m⋅ s

ρ ⋅ Q ⋅ Ws

(

)=

)=

f q , D , εi

i

i

(

f T D

i

, ε i) =

4.42·106

0.011

0.011

4.42·106

0.011

0.011

Power extracted = −1.112

×

4

10

⋅ kW

The change in head across the device is

∆h :=

−Ws

∆h =

g

566.856 m

A reasonable efficiency is 0.90. Power out := 0.90 ⋅ Power extracted

4

Power out = −1.001 × 10

⋅ kW

For a Francis turbine, 40 is a reasonable value of the customary (dimensional) power specific speed. 2⋅π Ncus := 40 RPM := ω := 10 ⋅ RPMInitial guess on turbine speed (in RPM). min Given

ω

Ncus = 2⋅

ω :=

Find ( ω )

π min

ω=

⎛ −Power out ⎞ hp ⎠ ⋅ ⎝

0.5

5

⎛ ∆h ⎞ ⎝ ft ⎠ -1

441.646 s

4

ω = 4.217 ×

3

10

⋅ RPM

This is an unreasonable value for the turbine speed. Hence, for these conditions, a Francis turbine is not a good choice.

7. A Pelton wheel is used produce electricity in a hydroelectric facility. The radius of the wheel is 1.83 m, and velocity of the fluid exiting the 10-cm diameter nozzle is 102 m/sec. The exit blade angle, β, is 165 degrees. (a) Sketch the “bucket” of the Pelton wheel and the inlet and outlet velocity triangles. (b) Calculate the flow rate. (c) What is the speed in RPM for maximum power extraction? (d) If the efficiency is 0.82, estimate the output shaft power. (e) Calculate the power specific speed. Based on the power specific speed, are these conditions suitable for a Pelton turbine? Splitter- ridge

β2

Vr 2 V2 Vr 2

U

Vr 1 V1

Outlet

U Inlet RPM :=

2⋅

π

ρ :=

min

kg

g := 9.806 ⋅

3

m

Dwheel := 2 ⋅ 1.83 ⋅ m Vnoz := 102 ⋅

1000 ⋅

m sec

m sec

Dnoz := 0.10 ⋅ m

Anoz := 0.25 ⋅

Q := Vnoz ⋅ Anoz

Q = 0.801 m

β 2 :=

2 2

3

165 ⋅ deg

−3 2

π ⋅ Dnoz

Anoz = 7.854 × 10

-1

Q = 1.27 × 10

⋅s

4

⋅ gpm

For maximum power extraction, the wheel speed, U, is one-half the nozzle velocity. U := 0.5 ⋅ Vnoz

ω :=

U 0.5 ⋅ Dwheel

-1

U = 51 m ⋅ s

ω =

-1

ω=

27.869 s 2

Power max := 0.25 ⋅ ρ ⋅ Q ⋅ Vnoz Power out := 0.82 ⋅ Power max

⋅ ( 1 −

cos( β 2))

266.128 ⋅ RPM

Power max = 4.096 Power out = 3.359

×

×

3

10

3

10

⋅ kW

⋅ kW

m

For a nozzle velocity of 102 m/sec, the elevation difference, neglecting major and minor losses, must be at least 2

∆Z :=

Vnoz

∆Z =

2⋅g

ω

Ncus := 2⋅

π min

⎛ Power out ⎞ hp ⎠ ⋅ ⎝ 5

⎛ ∆Z ⎞ ⎝ ft ⎠

530.492 m

0.5

Ncus = 1.589


4

These conditions are suitable for a Pelton wheel application.

8. A 0.9 efficient Francis turbine produces 200 MW with a speed of 150 RPM. The blade height is 1.0 m, and the inlet radius of the rotor is 7.0 m. The flow exits the turbine in a radial direction. (a) If the angle between the radial direction and the absolute velocity at the inlet is 30 degrees, determine the volume flow rate. (b) What is the change in head across the turbine?

V

0.732 m 1m

2⋅π

RPM :=

ω :=

ρ :=

min

U1 := r 1 ⋅

MW := 1000 ⋅ kW

=

3

30

g := 9.806 ⋅

m

150 ⋅ RPM

Power in

kg

1000 ⋅

η :=

ω

=

r 1 := 7.0 ⋅ m height := 1.0 ⋅ m

2 -1

109.956 m ⋅ s

Power out := 200 ⋅ MW

0.9

222.222 ⋅ MW Area := 2 ⋅

m sec

U1

Vr

o

5.52 1 =7 r 1r = mm

Power in :=

Power out

η 2

π ⋅ r 1 ⋅ height

Area = 43.982m

Since the absolute exit velocity is in the radial direction Power = mdot ⋅ U1 ⋅ Vu1 = Vu1 := 1 ⋅

m

ρ ⋅ A ⋅ U1 ⋅ Vm1 ⋅ Vu1 = ρ ⋅ A ⋅ U1 ⋅ Vu1 ⋅ cot( α ) ⋅ Vu1

Initial guess on Vu.

sec

Given Power in =

2

ρ ⋅ Area ⋅ U1 ⋅ cot( 30 ⋅ deg) ⋅ Vu1 -1

Vu1 := Find( Vu1)

Vu1

Vm := cot( 30 ⋅ deg) ⋅ Vu1

Vm

Q := Area ⋅ Vm

Q = 392.377 m

∆Z :=

Power in

ρ⋅g⋅Q

=

=

5.151 m ⋅ s

-1

8.921 m ⋅ s 3

∆Z =

57.755 m

-1

⋅s

6

Q = 6.219 × 10

⋅

gal min

9. A developer has constructed an elevated reservoir, as shown in the sketch, and estimates that a flow rate of 200 L/min is available on a continuous basis. (a)

Determine the power in kW that can be expected from the arrangement, if the turbine efficiency is 0.85. Discuss the implications of the results of this problem for the individual home owner in Mississippi.

(b)

L = 105 m D = 100 mm ε = 0.046 mm K=2

el 270 m

el 247 m Turbine ORIGIN

≡


1


Length in m

⎛ 0.1 ⎞ ⋅ m ⎝ 0.1 ⎠

L :=

Roughness in mm:

⎛ 5 ⎞ ⋅ m ⎝ 100 ⎠

ε :=

⎛ 0.046 ⎞ ⋅ mm ⎝ 0.046 ⎠

(commercial pipe)


Elevations in m:

⎛ Pa ⎞ ⎛ 0 ⎞ := ⋅ Pa P 0 ⎝ ⎠ ⎝ b ⎠

⎛ Za ⎞ ⎛ 270 ⎞ := ⋅m Z 247 ⎝ ⎠ ⎝ b ⎠


Equivalent length

⎛ 1.78 ⎞ ⎝ 2 ⎠

C :=

Number of pipes

⎛ 0 ⎞ ⎝ 0 ⎠

N := length( D)


ρ

:= 1000 ⋅

Kinematic Viscosity in m 2 /s

kg 3

m

2

ν :=

−6 m

1.14 ⋅ 10

⋅

sec

Input the flow rate in cms: liter

Q := 200 ⋅

Initial guess on turbine decrease in head: Ws :=

Flow rate.

min


2

−15 ⋅ newton ⋅

m kg

2

newton ⋅ sec


4 ⋅ q

π ⋅ d ⋅ ν

0.3086

⎡ log

0.3086

⎡⎛ ε ⎞ 1.11⎤ log ⎣⎝ 3.7 ⋅ d ⎠ ⎦

f( q , d , ε ) :=

f T( d , ε ) :=

ε ⎞ + ⎛ ⎝ 3.7 ⋅ d ⎠

6.9

⎣ Re( q, d )

64

⎤

1.11

2

if Re( q, d )

2

> 2300

⎦

otherwise

Re( q, d )

The generalized energy equation is: Given Ws ⋅

gc g

=

P b

−

Pa

ρ⋅g

⎡8

N

⋅ gc +

−

Z b

Za +

∑

⎢ 2 i = 1 ⎣π

2

Q

⋅

( i) D

4

⋅ ⋅g

⎛

(

)⋅

f Q, D , ε i

⎝

i

L

i

D

+

K + C i

i

⎞⎤ ⎠⎥⎦

⋅ f D , ε i T( i i)

Ws := Find( Ws) 2

Ws = −222.989 m

-2

⋅s

Ws = −222.989 ⋅ newton ⋅

m kg


V( q , D) :=

4 ⋅ q 2

π⋅D D

i

=

(

V Q, D

0.1 m

)=

0.424

0.1

(

Re Q , D

i

0.424

-1

m⋅ s

)=

i

(

)=

f Q, D , ε i i

(

f T D

i

, ε i) =

3.723·104

0.023

0.016

3.723·104

0.023

0.016

The head decrease at the turbine can be used to find the power generated. Power := Q ⋅

ρ ⋅ Ws

Power gen := 0.85 ⋅ Power

Power = −0.743 ⋅ kW Power gen = −0.632 ⋅ kW

This is not much power, less than a kW. Hence, this system would not be very useful for a home owner unless great care was taken to minimize the electrical demand.

10. A 1-m diameter penstock is 10-km long and carries water to an impulse turbine. If the turbine/generator is 83 percent efficient, what power can be extracted if the elevation difference between the reservoir and the nozzle is 650 m. The jet (nozzle) diameter is 16 cm. What should the Pelton wheel diameter for maximum efficiency be if 360 rpm is to be maintained? What is the power specific speed in EE (English engineering) units? Would this be a good application for an impulse turbine? Explain. ORIGIN

≡


1


Length in m

⎛ 1.0 ⎞ ⋅ m ⎝ 1.0 ⎠

L :=

Roughness in mm:

⎛ 5 ⎞ ⋅ km ⎝ 5 ⎠

ε :=

⎛ 0.046 ⎞ ⋅ mm ⎝ 0.046 ⎠

(commercial pipe)


Elevations in m:

⎛ Pa ⎞ ⎛ 0 ⎞ := ⋅ Pa P 0 ⎝ ⎠ b ⎝ ⎠

⎛ Za ⎞ ⎛ 650 ⎞ := ⋅m Z 0 ⎝ ⎠ b ⎝ ⎠

Dnoz := 0.16 ⋅ m

Input the loss coefficients: The problem statement contains no mention of minor losses. Minor losses are neglected. However, the large kinetic energy at the nozzle exit must be considered.

Kvalue :=

⎛ D1 ⎞

4

Kvalue

⎝ Dnoz ⎠

K factor K :=

=

This accounts for the nozzle exit kinetric energy in terms of the pipe fluid velocity.

3

1.526 × 10

Equivalent length

⎛ 0 ⎞ ⎝ Kvalue ⎠

C :=

Number of pipes

⎛ 0 ⎞ ⎝ 0 ⎠

N := length( D)


ρ :=

1000 ⋅


kg

ν :=

3

−6 m

1.14 ⋅ 10

⋅


sec

No turbine or pump prior to penstock:

3

Q := 1 ⋅

m


Initial guess on flow rate.

sec


2

m kg

2

newton ⋅ sec


4 ⋅ q

π ⋅ d ⋅ ν

f T( d , ε )

:=

0.3086

⎡⎛ ε ⎞ 1.11⎤ log ⎣⎝ 3.7 ⋅ d ⎠ ⎦

2

f( q , d , ε )

:=

0.3086

⎡ log

ε ⎞ + ⎛ ⎝ 3.7 ⋅ d ⎠

6.9

⎣ Re( q, d )

64

⎤

1.11

if Re( q, d )

2

> 2300

⎦

otherwise

Re( q, d )


Ws ⋅

=

g

−

P b

⎡8

N

Pa

⋅ gc +

ρ⋅g

Z b

−

∑

Za +

2

⎢ 2 i = 1 ⎣π

Q

⋅

( i) D

4

⋅ ⋅g

⎛

(

L

i

)⋅

f Q, D , ε i

⎝

i

D

i

+

K + C i

⎞⎤ ⎠⎥⎦

⋅ f D , ε i T( i i)

q := Find( Q)

=

q

3

2.189 m

-1

⋅s

q

=

5 liter 1.314 × 10 ⋅ min


V( q , D) :=

4 ⋅ q 2

π⋅D D

i

(

=

V q, D

)=

(

Re q , D

i

1 m

2.788

1

2.788

-1

m⋅ s

)=

i

(

)=

f q , D , εi i

(

f T D

i

, ε i) =

2.445·106

0.011

0.01

2.445·106

0.011

0.01

The jet velocity is: q

V jet := 0.25 ⋅

V jet

2

=

-1

108.897 m ⋅ s

π ⋅ Dnoz

The kinetic energy available for the Pelton wheel is 0.5 V jet 2 ; the available power is the available kinetic energy times flow rate. Power := q ⋅

η :=

ρ⋅

2 4

Power = 1.298 × 10

2

⋅ kW

Efficiency of turbine/generator.

0.83

Power out :=

V jet

η ⋅ Power

Power out = 1.078

×

4

10

⋅ kW

For maximum efficiency of a Pelton wheel, the blade speed is 0.5 of the jet speed. 2⋅π -1 U := 0.5 ⋅ V jet U = 54.448 m ⋅ s RPM := min

ω :=

360 ⋅ RPM

The Pelton wheel diameter is D pelton :=

2⋅U

ω

D pelton

=

2.889 m

Torque :=

Power out

Torque

ω

=

5

2.858 × 10

⋅ newton ⋅ m

The head corresponding to V jet is 2

∆h :=

V jet

∆h =

2⋅g

604.655 m

The power specific speed for this example, using the head corresponding to V jet , is T

⎛ Power out ⎞ ⎝ ρ ⎠ Ns := ω ⋅

0.5

5

Ns = 0.075


Ncus = 3.269


( g ⋅ ∆h ) 4 ω

Ncus := 2⋅

π min

⎛ Power out ⎞ hp ⎠ ⋅ ⎝ 5

⎛ ∆h ⎞ ⎝ ft ⎠

0.5

4

The customary power specific speed of nearly 3.3 is appropriate for a Pelton Wheel..

11. Turbines at the Conowingo Plant on the Susquehanna River each develop 54,000 bhp at 82 rpm under a head of 89 feet. What is the turbine type? Estimate the flow rate of each turbine. 2⋅π lb ft RPM := g := 32.174 ⋅ ρ := 62.4 ⋅ min

ω :=

82 ⋅ RPM

3

ft

sec

Power gen := 54000.0 ⋅ hp


∆Z :=

2

89.0 ⋅ ft

0.5

5

Ns = 1.604


Ncus = 69.706


( g ⋅ ∆Z ) 4


0.5

5

⎛ ∆Z ⎞ ⎝ ft ⎠

4

A specific speed (customary) value of 70 is at the overlap region between the Francis turbine and the Kaplan turbine. A reasonable efficiency is 0.90.

η :=

0.90

Power avail :=

Q :=

54000 ⋅ hp

Power avail

ρ ⋅ g ⋅ ∆Z

η

4

Power avail = 6 × 10 3

Q = 168.262 m

-1

⋅s

⋅ hp 6

Q = 2.667 × 10

⋅ gpm

This value of flow rate neglects major and minor losses in the system and asserts that the elevation difference is available for the turbine.

12. Many areas of the country have reservoirs/waterways with locks and dams. An interesting problem is to evaluate the potential for hydroelectric generation at a local facility with a spillway. For example: Would a hydroelectric facility be attractive at the Columbus (Mississippi) Lock and Dam on the Tennessee-Tombigbee Waterway? The web is a good source of information on many waterway components. The Columbus Lock and Dam (John C. Stennis Lock and Dam) is shown in the photograph.

The yearly average volume rate is 7094 acre-ft, and the pool average elevation difference between this dam and the next downstream dam is 27 ft. The average flow rate is Q :=

ρ :=

7094 ⋅ acre ⋅ ft 8760hr 1000 ⋅

kg 3

m Power avg :=

ρ ⋅ Q ⋅ g ⋅ 27 ⋅ ft

3

Q = 0.277 m g := 9.806 ⋅

-1

⋅s

3

Q = 4.398 × 10

⋅ gpm

m sec

2

Power avg = 22.392 ⋅ kW

With an average power available of only 22.4 kW, the installation of a turbine is not recommended.

13. If a hydroelectric facility is to produce 100,000 hp with an elevation difference of 300 feet, what is a suitable speed (in RPM) if a Kaplan turbine, operating at high efficiency, is used. RPM :=

2⋅

π

min

ρ :=

lb

62.4 ⋅

g := 32.174 ⋅

3

ft

Power gen := 100000.0 ⋅ hp

∆Z :=

ft sec

2

300.0 ⋅ ft

From Figure 3.9, a specific speed (customary) of 100 yields is a good value for a high-efficiency Kaplan turbine. Ncus := 100

ω :=

106 ⋅ RPM

Initial guess on speed.

Given

Ncus =

⎛ Power gen ⎞ hp ⎠ ω ⋅ ⎝

RPM

5

⎛ ∆Z ⎞ ⎝ ft ⎠ ω :=

Find ( ω )

0.5

ω =

4

-1

41.346 s

Major and minor losses are neglected.

ω=

394.822 ⋅ RPM

ch03.pdf

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