Solutions

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College of Engineering and Computer Science

Mechanical Engineering Department

Mechanical Engineering 483

Alternative Energy Engineering II

Spring 2010 Number: 17724 Instructor: Larry Caretto

Solutions to April 12 Homework Problems

1. Hodge, page 146, problem 2. A contractor must select a roof covering material from the two diffuse (

=

) roof coverings whose spectral characteristics are presented in Figure

P6.2. (This figure shows that Coating A has a constant emissivity of 0.8 while material B has

= 0.6 for

≤ 4

m and

= 0.2 for

≥ 4

m.)

(a) Which of the materials would result in a lower roof temperature?

Coating A would result in a lower roof temperature. It would emit more solar radiation compared to the amount it would absorb, relative to Coating B. We can check this qualitative argument by computing the average emissivity and absorptivity of Coating B for solar and terrestrial radiation.

(Coating A will have an average value of both

and

equal to 0.8 for all wavelengths.) For solar radiation (T = 5800 K) the average emissivity, which equals the average absorptivity, is found as shown below. Here the solar radiation is assumed to be that of a black body with a temperature of 5800 K (

T = 23,200

 m·K for

= 4

 m) and the values from the blackbody radiation function table on page 118 of Hodge are used to find the fraction of radiation between the different wavelengths.

solar

 

solar

0 .

6 [

f

0 .

6

(

T

0 .

9898

23200

0

m

0 .

2

K

1

)

0 /

0 ]

0 .

2

9898

1

f

(

T

0 .

5959

23200

m

K

)

A similar calculation for terrestrial radiation at a temperature of 300K (

 m) gives

T = 1,200

 m·K for

= 4

terrestial

  

0 .

6 [

f

0 .

6

(

T

0 .

0021

1200

0

m

K

0 .

2

)

1

0 ]

0 .

2

0 .

021

1

f

(

0 .

2008

T

1200

m

K

)

For a simple model, the roof temperature can be found from an radiation energy balance between the incoming solar radiation absorbed,

solar

G solar

, and

terrestial

T

4

roof

T

4

sky

, the radiation emitted, This gives the following equation for T roof

for each material.

solar

G solar

  

terrestial

T

4

roof

T

4

sky

T

4

roof

T

4

sky

solar

G solar terrestial

Taking the difference between the roof temperatures for the two materials gives

T

4

roof

,

B

solar

,

B

 

terrestial

,

B

T

4

roof

,

A

T

4

sky

solar

,

B

G solar

 

terrestial

,

B

solar

,

A

 

terrestial

,

A

G solar

0

0

.

.

5959

2008

T

4

sky

0 .

8

0 .

8

solar

,

A

G solar

 

terrestial

,

A

G solar

1 .

97

G

solar

This shows that coating B gives the higher temperature.

(b) Which is preferred for summer use?

Coating A, which would result in a lower roof temperature, would be preferred for summer use.

Engineering Building Room 1333

E-mail: [email protected]

Mail Code

8348

Phone: 818.677.6448

Fax: 818.677.7062

April 12 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 2

(c) Which is preferred for winter use?

Coating B, which would result in a higher roof temperature, would be preferred for winter use.

(d) Sketch a spectral distribution that would be ideal for summer.

For summer we would like a low roof temperature. We would obtain this with a spectral distribution that resulted in a low value of

 solar

/

 terrestial

. This would require a distribution, in which the value of

was small at small wavelengths and large at higher wavelengths, just the opposite of Coating B.

(e) Sketch a spectral distribution that would be ideal for winter.

For winter we would like a high roof temperature. We would obtain this with a spectral distribution that resulted in a high value of

 of

 solar

/

 terrestial

. This would require a distribution, in which the value

was large at small wavelengths and small at higher wavelengths, similar to Coating B.

2. Hodge, page 146, problem 3. Two special coatings are available for use on an absorber plate for a flat-plate solar collector. The coatings are diffuse (

=

) and are characterized by the spectral distributions in Figure P6.3. (This figure shows Coating A with a spectral emissivity of 0.85 for

≤ 4

m and

= 0.05 for

≥ 4

m. Coating B in this figure has a spectral emissivity of 0.05 for

≤ 4

m and

= 0.85 for

≥ 4

m.)

(a) Of the irradiation incident on the plate is G = 1000 W/m

2

, what is the radiant energy absorbed per m

2

for each surface coating?

We can find the mean absorptivity as was found in the previous problem. At the wavelength boundary of

= 4

 m, the value of

T = (4

 m)(5800 K) = 23,200

 m·K. For the two coatings,

solar

,

A

 

solar

,

A

0 .

85 [

0 .

85

f

(

T

0 .

9898

23200

0

m

0 .

05

K

1

)

0 ]

0 .

05

0 .

9898

1

f

(

T

0 .

8418

23200

m

K

)

solar

,

B

 

solar

,

B

0 .

05 [

0 .

05

f

(

T

0 .

9898

23200

0

m

0 .

85

1

K

)

0 ]

0 .

85

0 .

9898

1

f

(

T

0 .

05817

23200

m

K

)

For a G = 1000 W/m 2 the total energy absorbed is

841.8 W/m

2

for coating A

, and

58.17 W/m

2

for coating B

.

(b) What coating would you select for the absorber plate? Explain.

Select coating A; it will absorb more solar energy. It will emit less terrestrial tadiant energy.

3. Hodge, page 147, problem 5. An opaque solar collector surface is 3 m by 1 m and is maintained at 425 K. The surface is exposed to solar radiation with G = 800 W/m

2 surface is diffuse and its solar absorptivity varies as follows:

for 0.5

m



< 1.0

m;

= 0.5 for 1

m



< 2

m;

= 0.3 for



> 2.0

. The

= 0 for

< 0.5

m;

= 0.8 m. Determine the absorbed solar radiation, the emissive power, and the net radiation heat transfer from the surface.

Here we have a more detailed spectrum for the emissivity. We have to determine the fraction of the radiation in each wavelength band. For solar radiation at T = 5800 K we will have

T = (0.5

 m)(5800 K) = 2,900

 m·K for

= 0.5

 m;

T = (1

 m)(5800 K) = 5,800

 m·K for

= 1

 m, and

T = (2

 m)(5800 K) = 11,600

 m·K for

= 2

 m. The average properties for solar radiation are

solar

0 .

5

f

0 [

(

T f

0

(

T

2 , 900

m

K

)

10 , 600

0 .

8

0 .

72

m

K

)

0 .

2505

f

(

0 ]

T

0 .

5

0 .

8

f

5 , 800

0 .

9252

(

T

m

K

0 .

72

5 , 800

)

m

0 .

3

0 .

3

1

1

K

)

f

(

T f

(

T

0 .

9252

2 , 900

10 , 600

0 .

5007

m

m

K

)

K

)

April 12 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 3

For radiation from the surface at T = 425 K we will have

T = (0.5

 m)(425 K) = 212.5

 m·K for

= 0.5

 m;

T = (1

 m)( 425 K) = 425

 m·K for

= 1

 m, and

= 2

 m. The average properties for solar radiation are

T = (2

 m)(425 K) = 850

 m·K for

)

terrestial

0 .

5

0

0 .

8

f

0 [

1 .

2

f

(

T

(

T

x

10

8

850

212 .

5

m

K

)

0 ]

m

1

x

10

K

13

)

0

f

 

(

T

8 .

73

x

10

0 .

8

425

f

(

T

m

K

)

425

m

0 .

3

1

5

1 .

2

x

10

8

 

K f

1

)

(

T

f

(

T

850

212 .

5

m

K

)

m

8 .

73

x

10

5

0 .

3000

K

The absorbed solar radiation is

solar

G solar

= (0.5007)(800 W/m

2

=

400.6 W/m

2

.

The surface emissive power is

plate

T

4

plate

5 .

6704

x

10

8

W

425

K

4

= (0.5007)(800

m

2

K

4

W/m 2 =

555.0 W/m

2

.

The net surface radiation heat transfer is the difference between these two. Expressed as a net energy outflow from the plate it is 555.0 W/m

2

– 400.6 W/m

2

=

154.6 W/m

2

leaving the plate.

4. Determine the efficiency of a solar collector with the following design and operating data:

collector area = 3 m

2

,

absorber plate thickness = 0.6 mm, absorber plate thermal conductivity =

385 W/m·K,

absorber plate solar absorptivity = 0.95,

Transmissivity of glass covers = 0.95, inner and outer diameter of absorber tubes = 10 mm, 11 mm, overall heat loss coefficient = 5 W/m tube spacing, w = 0.15 m,

2

·K bond conductance = 385 W/m·K, heat transfer coefficient inside absorber tube = 300 W/m ambient temperature = 18 solar isolation = 700 W/m o

2

C,

, water mass flow rate = 0.25 kg/s, and inlet water temperature = 30 o

C.

2

·K,

We need to find the useful heat which depends on the absorbed solar radiaiotn, H a

, and the heatremoval factor, F

R

, (in addition to the data items). The heat-removal factor depends on the collector efficiency factor, F’, which, in turn, depends on the fin efficiency, F.

To do these calculations we first find the absorbed solar radiation, H a

, as the product of the incident radiation, H i

, the transmissivity of the glass covers,

, and the absorptivity of the collector plate,

: H a

= H i



= (700 W/m

2

)(0.95)(0.95) = 631.75 W/m

2

.

Next we do the calculations to find F, F’, and F

R

. The fin effectiveness depends on the parameter m, which is found as follows.

April 12 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 4

m

U c kt

385

W m

K

5

W

m

2

0 .

6

K mm

0 .

001

m mm

4 .

652

m

The fin efficiency depends on the product mL where L = (w

– D)/2. w = 0.15 is the tube spacing and D = 11 mm = 0.011 m is the outer tube diameter. This gives L = (0.15 m

– 0.011 m)/2 =

0.0695 m, so that mL = (4.652/m)( 0.0695 m) = 0.3233. Then F = tanh(Ml)/(mL) = tanh(0.3233) /

0.3233 = 0.9665.

We can now compute the collector efficiency factor, F’.

1

F

'

U c w

U c

1

2

LF m

2

K

D

1

C

B

1

h c

,

i

D i

5

W

0 .

15

m

5

W m

2

K

1

2

0 .

0695

m



0 .

9665

0 .

011

m

1

35

W m

K

m

2

K

300

W

1

1

0 .

010

m

0 .

8826

We can now find the heat removal factor, F

R

, which depends on the following computational group found below.

c p

U c

A c

0 .

25

kg

4 , 186

J W s m

5

W

2

kg

3

K

 

1

J

K

s

69 .

77

We can use this value in the equation for F

R

.

F

R

c p

U c

A c

1

e

U c

A c

F

c p

'

69 .

77

1

e

0 .

8826

69 .

77

0 .

8771

Now that F

R

and H a

are known we can calculate the useful heat transfer, Q u

.

Q u

A c

F

R

H a

U c

T f

,

in

T a

 

0 .

8771



631 .

75

W m

2

5

W m

2

K

30

18

K



1504

W

We can now compute the efficiency.

c

Q u

A c

H i

1 , 504

W

 

700

W m

2

71 .

6 %

April 12 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 5

5. Determine the stagnation temperature (the average absorber plate temperature when there is no useful heat) for the solar collector of problem 4.

The stagnation temperature, T max

, is given by the following equation for no useful heat transfer.

Q u

A c

H a

U c

T p

,

average

T a

0

when T p

,

average

T p

, max

This gives the general result:

T p

, max

H

U c a

T a

Using the data of problem 4 and the result found there that H a

= 631.75 W/m 2 , gives:

T p

, max

H

U c a

T a

631 .

75

W m

2

5

W m

2

K

18

o

C

T p,max

= 144 o

C

6. Using some software such as Excel or Matlab plot the efficiency of the solar collector whose data are given in problem 4 as a function of the inlet water temperature that varies from 10 o

C to 60 o

C. Create one plot with curves for mass flow rates of 0.001 kg/s, 0.01 kg/s, 0.1 kg/s, 1 kg/s and 10 kg/s. Prepare a plot of water outlet temperatures for the same cases.

The following VBA function was used to solve this problem:

Option Explicit

Const PI as double = 3.14159265358979

Function solarCollector(incidentRadiation As Double, transmissivity As Double, _

absorptivity As Double, centerLineDistance As Double, _

tubeOD As Double, tubeID As Double, collectorU As Double, _

plateK As Double, plateThickness As Double, _

bondConductance As Double, tubeInnerH As Double, _

ambientT As Double, waterInT As Double, mDotWater As Double, _

collectorArea As Double, cpWater As Double) As Variant

Dim absorbedRadiation As Double

Dim lengthBetweenTubes As Double

Dim finM As Double

Dim finML As Double

Dim finEffectiveness As Double

Dim collectorEfficiencyFactor As Double

Dim heatRemovalFactor As Double

Dim usefulHeat As Double

Dim waterOutT As Double

Dim plotParameter As Double

Dim collectorEfficiency As Double

Dim x(1 To 4) As Variant

absorbedRadiation = incidentRadiation * transmissivity * absorptivity

lengthBetweenTubes = (centerLineDistance - tubeOD) / 2

finM = Sqr(collectorU / (plateK * plateThickness))

finML = finM * lengthBetweenTubes

finEffectiveness = Application.WorksheetFunction.Tanh(finML) / finML

collectorEfficiencyFactor = (1 / collectorU) / (centerLineDistance * (1 / (collectorU _

* (2 * lengthBetweenTubes * finEffectiveness + tubeOD)) _

+ 1 / bondConductance + 1 / (tubeInnerH * PI * tubeID)))

heatRemovalFactor = mDotWater * cpWater / (collectorU * collectorArea) _

* (1 - Exp(-(collectorU * collectorArea _

* collectorEfficiencyFactor / (mDotWater * cpWater))))

usefulHeat = collectorArea * heatRemovalFactor * (absorbedRadiation - collectorU _

* (waterInT - ambientT))

collectorEfficiency = usefulHeat / (incidentRadiation * collectorArea)

waterOutT = waterInT + usefulHeat / (mDotWater * cpWater)

plotParameter = (waterInT - ambientT) / incidentRadiation

April 12 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 6

x(1) = collectorEfficiency

x(2) = usefulHeat

x(3) = waterOutT

x(4) = plotParameter

solarCollector = Application.WorksheetFunction.Transpose(x)

End Function

The plotted results are shown below.

Collector Efficiency vs . Inlet Temperature and Mass Flow Rate

90%

80%

70%

60%

50%

40%

mdot = 10 kg/s mdot = 0.1 kg/s mdot = 0.01 kg/s mdot = 0.001 kg/s

30%

20%

10%

0%

-0.02

-0.01

0 0.01

0.02

0.03

0.04

0.05

0.06

(T in

- T ambient

)/H incident

An evaluation of the equations in the efficiency calculation show that as the mass flow rate of water decreases the parameter

U c

A c

F

'

c p

increases. As this parameter increases the ratio

F

R

/F’ decreases. Since F’ is fixed by the collector design, the heat removal factor, F

R

decreases as the water flow rate decreases. This decrease in the heat removal factor decreases the collector efficiency and consequently decreases the useful heat as the mass flow rate decreases.

However, as the mass flow rate

increases

, the temperature rise is negligible and the efficiency does not increase significantly for further increases in mass flow rate. At this point, the temperature rise is nearly zero.

The figure on the next page shows the temperature increase of the water. The lowest mass flow rate leads to a temperature that is higher than the boiling point of water and is not feasible.

However, it is included in the plot to show the trend of temperature rise with mass flow rate.

The combination of the two figures shows that a reduced mass flow rate can produce a higher exit temperature, but this higher exit temperature causes a significant decrease in efficiency.

April 12 homework solutions ME 483, L. S. Caretto, Spring 2010 Page 7

140

Temperature Rise vs . Inlet Temperature and Mass Flow Rate

120

100

80

60

40

20

0

-0.02

-0.01

0

mdot = 0.001 kg/s mdot = 0.01 kg/s mdot = 0.1 kg/s mdot = 10 kg/s

0.01

0.02

0.03

(T in

- T ambient

)/H incident

0.04

0.05

0.06

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