Sample Doc

Technological Institute of the Philippines – ManilaUnit Operations Laboratory I, 1st Semester 2015-2016

Abstract— This experiment provides insight on the heat losses of fluid flow in pipes and the effect of insulation on the overall heat transfer coefficient. Two types of energy transfer is studied in this experiment; the two are called convection, the transfer of heat through the movement of fluids, and conduction, is the movement of heat from one solid to another one that has different temperature when they are touching each other. This experiment also involves the study of the different materials that make up thermal insulators for piping systems.

Index Terms— Heat transfer is the transfer of energy from one body to another as a result of a temperature gradient or a change in phase, Heat transfer coefficient is a quantitative characteristic of convective heat transfer between a fluid medium and surface flowed over by the fluid, Thermal insulators are materials used to inhibit the transmission of heat between bodies, Laggings are thermal insulators used mostly in equipment that contains fluids that needs its temperature conserved.

1. OBJECTIVE

The objective of this experiment is to determine the heat losses and the overall heat transfer coefficient from bare and lagged pipes.

2. INTENDED LEARNING OUTCOMES

The students shall be able to:

2.1 explain principles of combined convection and conduction using bare and lagged pipes.

2.2 apply the appropriate standards and tables in the calculation of heat losses to improve the system efficiency.

Charles Bonn Kirby F. Mayo, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09054086463 (e-mail: [email protected]).

Zeny N. Naranjo, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09351947136 (e-mail: [email protected]).

Gliezel Panopio, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 0916678120, (e-mail: [email protected]).

Eazyl D. Salazar, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09267880602, (e-mail: [email protected]).

Klinneth Joy P. Samillano, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and

Architecture, Manila, Philippines,09277514721., (e-mail: [email protected]).

3. DISCUSSION

A good pipe covering, in addition to being a good insulator, should be fireproof, waterproof, vermin proof, odorless, and light in weight. It should also be mechanically strong and should suffer no loss of insulating value due to time. The only logical method for testing commercial pipe coverings is to mount these coverings on pipe of the size for which they were intended. Extensive tests of commercial coverings have been made by various investigators, and two general methods for heat measurement have been used. For steam-pipe coverings, the most natural method is to fill the covered pipe with steam, to measure the heat content of the steam entering and leaving the test section, and to condense and weigh the steam. A dead-end pipe is ordinarily used, the test pipe itself acting as the steam condenser.

Piping of left bare can lose heat due to temperature difference between pipe surface temperature and ambient temperature. The methods of measurement and calculations for estimation of heat losses and heat gain in piping systems and insulation thickness are described. Measurements of fluid temperature and pipe surface temperatures are necessary for the calculations.

In many practical situations the surface temperatures are not known, but there is a fluid on both sides of the solid surfaces. Considering the plane wall in Fig. 7.1a, with a hot fluid at temperature T 1 on the inside surface and a cold fluid at T 4 on the outside surface. The convective coefficient on the outside is ho and hi on the inside.

The heat-transfer rate is given as

q=hi A (T 1−T2 )=k A A∆ x A

(T 2−T 3 )=ho A ( T3−T4 )

Expressing1

hi A ,∆ x A , 11

ho A as resistances and

combining the equations

q=(T 1−T 4)

1h i A +

∆ x A

k A A + 1ho A

=(T 1−T 4)

Σ R

1

Heat Losses from Bare and Lagged PipesCharles Bonn Kirby F. Mayo, Zeny N. Naranjo, Gliezel M. Panopio, Eazyl D. Salazar , Klinneth Joy P. Samillano

Experiment No. 4Flow Through Particle Layer

The overall heat transfer by combined conduction and convection is often expressed in terms of an overall heat transfer coefficient U defined by

q=UA ∆ T overall

Where ∆ T overall= T 1−T 4 and U is

U= 11hi

+∆ x A

k A+ 1

ho

Fig. 7.1 Heat flow with convective boundaries: (a) plane wall, (b) cylindrical wall

A more important application is heat transfer from a fluid outside a cylinder through a metal wall, to a fluid inside the tube. In Fig. 7.1b such a case is shown. Using the same procedure as before, the overall heat transfer rate through the cylinder is

q=(T 1−T 4)

1h i A +

ro−r i

k A A +1

ho A

=(T1−T 4)

Σ R

Where A represents2πLr , the inside area of the metal tube. Althe log mean area of the metal tube; and Aothe outside area.

The overall heat transfer coefficient U for the cylinder may be based on the inside area Aior the outside area Ao of the tube. Hence,

q=U i A i (T1−T 4 )=U o Ao (T 1−T 4 )=( T1−T 4 )

Σ R

U i=1

1hi

+(ro−ri) A i

k A Al+

A i

ho Ao

Uo=1

Ao

Ai hi+

(ro−r i) Ao

k A A l+ 1

ho

4. RESOURCES Equipment/Apparatus:

Bare and Lagged Pipe Assembly Non-contact Temperature Sensor Laptop

5. PROCEDURE

1. With the assistance of the laboratory technician, set the main pressure of steam to 5 bar. Fire the boiler.

2. Direct the steam to the bare and lagged pipe assembly by opening and closing the corresponding valves in the steam header.

3. With the drain valve open wide, open steam valve a small amount, and allow steam to blow through long enough to purge apparatus of all air; then close the drain valve. Read the steam temperature inside the pipes and measure the outside temperature using non-contact temperature sensor.

4. Calculate the heat losses q and the overall heat transfer coefficients Ui and Uo.

5. Compute the efficiency using this formula:

E =

6. DATA AND RESULTS

Test of Deposited Layers

Gravel, Trial 1

Q(cm3/min) h1(mm) h2(mm) S(mm) R(mm-1)

0 28.00 28.00 74.00100 29.20 26.90 74.00 17012.10200 29.70 26.40 74.00 12204.33300 30.40 25.40 74.00 12327.61400 31.10 25.20 74.00 10909.93500 31.70 24.70 74.00 10355.19600 32.45 24.00 74.00 10416.83700 33.30 23.30 74.00 10566.52800 34.20 22.50 74.00 10817.48

Average R: 11085.41

Gravel, Trial 2


0 28.00 28.00 74.00100 29.20 26.90 74.00 17012.10200 29.80 26.40 74.00 12574.17300 30.45 25.90 74.00 11218.12400 31.05 25.25 74.00 10725.02500 31.80 24.60 74.00 10651.05600 32.60 23.95 74.00 10663.38700 33.50 23.10 74.00 10989.18

2

(Heat lost from bare pipe) – (Heat lost from lagged pipe)Heat lost from bare pipeE = X100%

Heat saved by insulationHeat lost from bare pipe X100%


800 34.20 22.50 74.00 10817.48Average R: 11091.20

Gravel, Trial 3


0 28.00 28.00 74.00100 29.30 26.90 74.00 17751.76200 29.90 26.40 74.00 12943.99300 30.50 25.80 74.00 11587.95400 31.20 25.30 74.00 10909.43500 31.90 24.60 74.00 10798.99600 32.60 24.00 74.00 10601.74700 33.45 23.20 74.00 10830.69800 34.25 22.45 74.00 10909.93

Average R: 11226.10

Test of Fluidized Beds

Gravel, Trial 1

Q(cm3/min) h1(mm) h2(mm) S(mm)

0 28.00 28.00 74.00100 28.60 27.50 74.00200 28.80 27.30 74.00300 29.00 27.10 74.00400 29.30 26.90 74.00500 29.80 26.70 74.00600 30.00 26.40 74.00700 30.10 26.25 74.00800 30.40 26.00 74.00

Speed of the fluid in the space between the particles:

Gravel, Trial 2


0 28.00 28.00 74.00100 28.70 27.95 74.00200 28.85 27.30 74.00300 29.10 27.10 74.00400 29.35 26.90 74.00500 29.60 26.65 74.00600 29.90 26.50 74.00700 30.15 26.30 74.00800 30.45 26.00 74.00


Gravel, Trial 3


0 28.00 28.00 74.00100 28.70 27.80 74.00200 28.90 27.30 74.00300 29.15 27.10 74.00400 29.45 26.90 74.00500 29.70 26.65 74.00600 29.90 26,45 74.00700 30.20 26.30 74.00800 30.40 26.00 74.00


7. CALCULATIONS

Test of Deposited Layers

R= Ax ∆ PnL x vL

where:

A=400πmm2

nL=1x10-3Pa∙s

Trial 1

@Q=100cm3/min

R=400 π mm2 (29.20−26.90 )mm (9.81 kg

m ∙mm∙ s2 )

(1 x10−3 Pa ∙ s)(100 cm3

min)( 1min

60 s)(10 mm

1 cm)

3

R=17,012.10mm-1

@Q=200cm3/min

R=400 π mm2 (29.70−26. 40 ) mm(9.81 kg

m∙ mm∙ s2 )

(1x 10−3 Pa ∙ s)(2 00 cm3

min)(1min

60 s)( 10 mm

1 cm)3

R=12,204.33mm-1

@Q=300cm3/min

R=400 π mm2 (30.40−25 .40 )mm (9.81 kg

m ∙mm∙ s2 )

(1 x10−3 Pa ∙ s)(3 00 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=12,237.61mm-1

@Q=400cm3/min

R=400 π mm2 (31.10−25 .20 ) mm(9.81 kg

m∙ mm∙ s2 )

(1 x10−3 Pa∙ s)(4 00 cm3

min)( 1 min

60 s)( 10 mm

1cm)

3

R=10,909.93mm-1

@Q=500cm3/min

3


R=400 π mm2 (31.70−24.70 ) mm(9.81 kg

m∙ mm∙ s2 )

(1x 10−3 Pa∙ s)(500 cm3

min)(1min

60 s)( 10 mm

1 cm)3

R=10,355.19mm-1

@Q=600cm3/min

R=400 π mm2 (32.45−24.00 )mm (9.81 kg

m∙ mm∙ s2 )

(1 x 10−3 Pa ∙ s)(6 00 cm3

min)( 1min

60 s)(10 mm

1 cm)

3

R=10,416.83mm-1

@Q=700cm3/min

R=400 π mm2 (33.30−23.30 )mm (9.81 kg

m ∙mm∙ s2 )

(1 x 10−3 Pa ∙ s)(7 00 cm3

min)( 1min

60 s)(10 mm

1 cm)

3

R=10,566.52mm-1

@Q=800cm3/min

R=400 π mm2 (34.20−22.50 )mm (9.81 kg

m ∙mm∙ s2 )

(1 x 10−3 Pa ∙ s)(8 00 cm3

min)( 1min

60 s)(10 mm

1 cm)

3

R=10,566.52mm-1

Suspected outlier is R=17,1012.10mm-1

Q=17012.10−12327.6118012.10−10355.19

=0.7037

@N=8 and 95% confidence level; R=17012.10mm-1 is rejected

Average R= 12,204.33+12,327.61+10909.93+10355.19+10416.83+10566.52+10817.48

7

Average R=11085.41mm-1

Trial 2

@Q=100cm3/min

R=400 π mm2 (29.20−26.90 )mm (9.81 kg

m ∙mm∙ s2 )

(1 x10−3 Pa ∙ s)(100 cm3

min)( 1min

60 s)(10 mm

1 cm)

3

R=17,012.10mm-1

@Q=200cm3/min

R=400 π mm2 (29. 8 0−26. 40 )mm(9.81 kg

m∙ mm ∙s2 )

(1 x10−3 Pa ∙ s)(200 cm3

min)(1 min

60 s)(10 mm

1cm)

3

R=12,574.17mm-1

@Q=300cm3/min

R=400 π mm2 (30.45−25 .9 0 ) mm(9.81 kg

m∙ mm ∙s2 )

(1 x 10−3 Pa ∙ s)(300 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=11,218.12mm-1

@Q=400cm3/min

R=400 π mm2 (31.05−25 .25 ) mm(9.81 kg

m∙ mm∙ s2 )

(1 x10−3 Pa ∙ s)(4 00 cm3

min)( 1 min

60 s)( 10 mm

1cm)

3

R=10,725.02mm-1

@Q=500cm3/min

R=400 π mm2 (31. 80−24.6 0 )mm(9.81 kg

m∙ mm ∙s2 )

(1 x 10−3 Pa ∙ s)(500 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=10,621.05m-1

@Q=600cm3/min

R=400 π mm2 (32.60−23.95 )mm (9.81 kg

m∙ mm∙ s2 )

(1x 10−3 Pa ∙ s)(6 00 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=10,663.38mm-1

@Q=700cm3/min

R=400 π mm2 (33.50−23.1 0 ) mm(9.81 kg

m∙ mm∙ s2 )

(1 x 10−3 Pa∙ s)(700 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=10,989.18mm-1

4


@Q=800cm3/min

R=400 π mm2 (34.20−22.50 )mm (9.81 kg

m ∙mm∙ s2 )

(1 x 10−3 Pa ∙ s)(8 00 cm3

min)( 1min

60 s)(10 mm

1 cm)

3

R=10,817.48mm-1


Q=17012.10−12574 .1717 012.10−10651.05

=0.6028


Average R= 12, 574.17+11,218.12+10725.02+10,651.05+10663.38+10,989.18+10817.48

7

Average R=11,091.20mm-1

Trial 3

@Q=100cm3/min

R=400 π mm2 (29.3 0−26.90 )mm (9.81 kg

m ∙mm∙ s2 )

(1 x10−3 Pa ∙s)(100 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=17,751.76mm-1

@Q=200cm3/min

R=400π mm2 (29.9 0−26. 40 )mm (9.81 kg

m∙ mm ∙s2 )

(1 x10−3 Pa ∙ s)(2 00 cm3

min)(1min

60 s)(10 mm

1cm)

3

R=12,943.99mm-1

@Q=300cm3/min

R=400 π mm2 (30.50−25.8 0 )mm (9.81 kg

m ∙mm∙ s2 )

(1x 10−3 Pa ∙ s)(300 cm3

min)( 1min

60 s)(10 mm

1 cm)

3

R=11,587.95mm-1

@Q=400cm3/min

R=400 π mm2 (31.20−25 .30 ) mm(9.81 kg

m∙ mm∙ s2 )

(1 x10−3 Pa ∙ s)(4 00 cm3

min)( 1 min

60 s)( 10 mm

1cm)

3

R=10,909.42mm-1

@Q=500cm3/min

R=400π mm2 (31. 90−24.60 ) mm(9.81 kg

m∙mm∙ s2 )

(1x 10−3 Pa ∙ s)(5 00 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=10,798.99m-1

@Q=600cm3/min

R=400 π mm2 (32.60−24.00 )mm (9.81 kg

m ∙mm∙ s2 )

(1x 10−3 Pa ∙ s)(6 00 cm3

min)( 1 min

60 s)(10 mm

1cm)

3

R=10,601.74mm-1

@Q=700cm3/min

R=400 π mm2 (33.35−23.2 0 ) mm(9.81 kg

m∙ mm∙ s2 )

(1 x 10−3 Pa∙ s)(700 cm3

min)( 1min

60 s)(10 mm

1cm)

3

R=10,830.69mm-1

@Q=800cm3/min

R=400 π mm2 (34.25−22.45 )mm (9.81 kg

m∙ mm∙ s2 )

(1x 10−3 Pa ∙ s)(800 cm3

min)( 1min

60 s)(10mm

1cm)

3

R=10,909.93mm-1


Q=17751 .56−12943.9917751.56−10601.74

=0. 6724


Average R= 12,943.99+11587.95+10,909.43+10,798.99+10,601.74+10,830.69+10,909.93

7

Average R=11,226.10mm-1

5


Anova: Single Factor

Groups Count Sum AverageVarianc

Column 1 777597.8

911085.4

1691181.

Column 2 7 77638.4 11091.2468581.

Column 3 773469.0

5 11226.1217548.

Source of Variation SS df MS

Between Groups163964

0 2 8198201.78569

Within Groups826387

1 18459103.

9

Total990351

1 20

8. CONCLUSION

After performing the experiment, we have concluded that, the heat losses from fluid flow in pipe systems can be reduced by covering bare pipes with laggings. Laggings inhibit the flow of heat from the fluid inside the pipes to the air around them. The laggings does this by reducing the overall heat transfer coefficient of the system. Also, the efficiency of laggings in reducing the heat loss varies on what type of material makes up the lagging.

9. QUESTION/PROBLEMS

1. What is the basic mechanisms of heat transfer? Discuss.2. What is thermal conductivity?3. A wall of furnace 0.244 m thick is constructed of material having a thermal conductivity of. The wall will be insulated on the outside with the material having an average k of 0.306 W/m-K, so the heat loss from the furnace will be equal to or less than 1830 W/m2. The inner surface temperature is 1588 K and the outer 299 K. Calculate the thickness of insulation required.4. A glass window with an area of 0.557 m2 is installed in the wooden outside wall of a room. The wall dimensions are 2.44 x 10 x 3.05 m. the wood has a k of 0.1505 W/m-K and is 25.4 mm thick. The glass is 3.18 mm thick and has a k of 0.692. The inside room temperature is 299.9 K and the outside air temperature is 266.5 K. The convection coefficient hi on the inside wall of the glass and the wood is estimated as 8.5 W/m2-K; the outside ho is also estimated as 8.5 for both surfaces. Calculate the heat loss through the wooden wall, through the glass, and the total.5. Water at an average of 70 °F is flowing in a 2-in steel pipe, schedule 40. Steam at 220 °F is condensing on the outside of the pipe. The convective coefficient of the water inside the pipe is h = 500 Btu/hr-ft2-°F and the condensing steam coefficient on the outside is h = 1500.a. Calculate the heat loss per unit length of 1 m pipe using resistances.b. Repeat, using the overall Ui based on the inside area Ai.c. Repeat, using Uo.6. It is desired to predict the heat transfer coefficient for air being blown past an apple lying on a screen with large openings. The air velocity is 0.61 m/s at 101.32 kPa pressure and 316.5 K. The surface of the apple is at 277.6 K and its average diameter is 114 mm. Assume that it is a sphere.

10.ANSWERS

1. Heat can be transferred in three modes: conduction, convection and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. It can take place in solids, liquids, or gases. The rate of heat conduction through a medium depends on the geometry of the medium, its thickness, and the material of the medium, as well as the temperature difference across the medium. Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. The faster the fluid motion, the greater the convection heat transfer. Radiation is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules. Unlike conduction and convection, the transfer of energy by radiation does not require the presence of an intervening medium. In fact, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum.

6


2. The thermal conductivity of a material can be defined as the rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference. It is a measure of the ability of the material to conduct heat. A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator.3. Given: T1 = 1588 K

T3 = 299 KkA = 1.3 W/m-KΔxA = 0.244 mkB = 0.306 W/m-Kq/A = 1830 W/m2

Required: ΔxB

Solution: qA

=kA

∆xA(T1- T2 ) =

kB

∆xB(T2- T3 )

1830 Wm2 =

1.3 Wm-K

0.244 m ( 1588-T2)

T2=1244.52 K

1830 Wm2 =

0. 306 Wm-K

∆ xB(1244.52-299 ) K

∆ xB =0.1581 m4. Given: for glass: A = 0.557 m2

Δx = 3.18 mm = 0.00318 mk = 0.692 W/m-Khi = ho = 8.5 W/m2-K

for wood: A = ((2.44 x 3.05) – 0.557) m2 = 6.885 m2

Δx = 25.4 mm = 0.0254 mk = 0.1505 W/m-Khi = ho = 8.5 W/m2-K

T1 = 299.9 KT2 = 266.5 K

Required: qwood, qglass, qtotal

Solution: q=

T 1−T 4

1h i A

+∆ x A

k o A+ 1

ho A

qglass = (299.9-266.5 ) K

[1(8.5 Wm2-K ) (0.557 m2 ) ]+[0.00318m

(0.692 Wm-K ) (0.557m ) ]+[1(8.5 W

m2-K )( 0.557 m2 ) ]qglass =77.55 W

qwood = (299.9-266.5 )K

[1(8.5 wm2-K ) (6.885 m2 ) ]+[0.0254m

(0.1505 Wm-K ) (6.885 m2 ) ]+ [1(8.5 W

m2 -K ) (6.885 m2 ) ]qwood =569.11 W

q total = qglass +qwood

q total =77.55 W+569.11 Wq total =646.66 W5. Given: T1 = 220 °F

T4 = 70 °Fhi = 500 Btu/hr-ft2-°Fho = 1500 Btu/hr-ft2-°Ffor 2-in. Schecdule 40 pipe: Do = 2.375 in

Di = 2.067 inΔx = 0.154 in

Required: a. q if L=1ftb. Ui

c. Uo

Solution: Ao =π (2.375 in )(1 ft12 in )(1 ft )=0.6218 ft2

Ai =π (2.067 in )(1 ft12 in ) (1 ft ) =0.5411 ft2

Alm =(0.6218-0.5411) ft2

ln(0.62180.5411 )

=0.5805 ft2

a. q=

T1- T4

1h i A i+

∆xk Alm

+1ho Ao

where, k = 26

Btu/hr-ft-°F @ 220 °F

q= (220-70 )℉

1

(500 Btuhr- ft2 -℉ ) (0.5411 ft2 )

+0.154 in (1 ft

12 in )(26 Btu

hr-ft-K ) (0.5805 ft 2)+1

(1500 Btuhr- ft2-℉ ) (0.6218 ft2)

q=26696.97 Btuhr

b. U i=

11h i

+∆x Ai

k Alm+

A i

ho Ao

U i=1

1

500 Btuhr- ft2-℉

+0.154 in (1 ft

12 in ) (0.5411 ft2 )

26 Btuhr-ft-℉ (0.5805 ft2 )

+0.5805 ft2

1500 Btuhr- ft2-℉

(0.5411 ft2 )

U i=314.93 Btuhr- ft2 -℉

c. Uo=1

Ao

h i A i+

∆x Ao

k A lm+1

ho

7


U i=1

0.6218 ft2

500 Btuhr- ft2-℉

( 0.5411 ft2 )+

0.154 in (1 ft12 in )( 0.6218 ft2 )

26 Btuhr-ft-℉ ( 0.5805 ft2 )

+1

1500 Btuhr- ft2-℉

U i= 286.23 Btuhr- ft2-℉

6. Given: v = 0.61 m/sP = 101.32 kPaTair = 316.5 KTapple = 277.6 KD = 114 mm = 0.114 m

Required: h

Solution: T= (316.5+277.6 ) K2

=297.05 K, @ 297.05

K, properties of air: ρ = 1.1923 kg/m3, NPr = 0.709, µ = 1.84 x 10-5 kg/m-s, k = 0.0259 W/m-K

NRe=Dvρμ

=0.114 m(0.61 m

s )(1.1923 kgm3 )

1.84x 10-5 kgm-s

=4506.12

NNu= hDk

=2.0+0.60 NRe0.5 NPr

13

h (0.114 m )

0.0259 Wm-K

=2.0+0.60 (4506.12 )0.5 (0.709 )13

h=8.61 Wm2-K

11. FLOWCHART

8

START

With the assistance of the laboratory technician, set the main pressure of steam to 5 bar. Fire the boiler

Direct the steam to the bare and lagged pipe assembly by opening and closing the corresponding valves in the steam header

With the drain valve open wide, open steam valve a small amount, and allow steam to blow through long enough to purge apparatus of all air; then close the drain valve. Read the steam temperature inside

the pipes and measure the outside temperature using non-contact temperature sensor.

Calculate the heat losses q and the overall heat transfer coefficients Uo and Uo

Calculate the efficiency

E= ( Heat lost from bare pipe ) -(Heat lost from lagged pipe)Heat lost from bare pipe

x100

END


12. HAZARDS AND COUNTERMEASURES

Explosion - secure that the boiler’s water level will not reach the top level or the low level, as well as the water tank’s water level. Also ensure that the pressure will not exceed 5 bars. If the following were not followed, a sudden explosion might occur which may result in serious physical injury.

Skin Burns – avoid being severely burned by not touching the pipes after the boiler was turned on.

13.WASTE DISPOSAL

No waste generated in this experiment

APPENDIX

Group 2 members

Observing the boiler

Boiler Set-Up

Checking the boiler’s pressure

Boiler starts to operate

9


Infrared sensors for non-contact temperature measurement

Checking the tank’s water level

Calibrating the infrared sensors for non-contact temperature measurement

Calcium Silicate 1x2 in. thickness

Double-Sided Aluminum Foil 1 in. Fiber Glass (Blanket Type)

Fiber Glass Pre-Molded 1x2 in. thickness

10


Rockwool Pre-Molded 1x2 in. thickness

Single-Sided Aluminum Foil 1 in. Fiber Glass (Blanket Type)

Observing the pressure of each pipes

Measuring the temperature of each pipes

Checking the temperature of the bare pipe

Recorded temperature being saved digitally

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Turning the boiler off

Charles Bonn Kirby F. Mayo finished his elementary studies at the University of Santo Tomas, then finished high school at the Ramon Magsaysay High School, in his high school years, he joined many research related competition. He was part of a research team which developed a paint that could absorb nearby pollution, the research led to qualifying for the research competition nationals in SIBOL Cebu back in 2009. In the following year, as part of a different team, he helped develop a portable water quality analyzer and an air filter used for chimneys, these researches qualified for the International Exhibition for Young Inventors (IEYI) which was held in Hanoi, Vietnam, the researches won 2nd and 3rd place respectively.. After high school, he enrolled in the Pamantasan ng Lungsod ng Maynila in the Chemical Engineering Program, after 4 years, he transferred to the Technological Institute of the Philippines.

Klinneth Joy P. Samillano a modest and compassionate student who focuses in the field of Chemical Engineering. Capable of showing her willingness to learn more about everything that deals with her field of specialization. Committed to the Center of NanoBiomaterials Engineering, specifically to the Decellularization Team, with the aim of expanding the use of tissues to create biomaterials that can be use in the near future. A dedicated member of the Jr. Piche and Society Of Chemistry and Environment. She have established her communication skills and feel able to interconnect with other people.

Gliezel M. Panopio finished her primary and secondary studies at Lady Fatima Montessori School in the Province of Batangas and La Consolacion College-Pasig, respectively. After graduating high school, she continued her education to pursue a degree in Chemical Engineering at Adamson University, Manila. And after four years, she transferred to Technological Institute of the Philippines, Manila taking up the same degree.

Eazyl D. Salazar finished her elementary and secondary studies at Holy Word Academy. She was awarded as the class salutatorian and consistently part of the top three (3rd honorable mention) students during her elementary and high school years respectively. Aside from her academic awards, she was active in participating on extra-curricular activities resulting on becoming one of the representatives of the said school for its music team, and the short story writer for Junior Student Convention and National Student Convention of School of Tomorrow Philippines. She had won several awards such as consistent 6th place for her two short stories (in Filipino), and 2nd and 3rd place for the Trio and Duet Female respectively. She started her tertiary education at Adamson University under the program Chemical Engineering from year 2010 to 2012. She then continued the said program at Technological Institute of the Philippines after being in her previous school for two years.

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