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Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)

Abstract— In this experiment we used a bomb calorimeter to accurately determine the calorific value from the weight of the liquid fuel used and the radiation correction in which it is calculated from the rates of change of temperature of the water before igniting the fuel sample and after the attainment of the maximum temperature. By carefully controlling the pressure and contents of the bomb, and by using samples such as Kerosene, Diesel, and biodiesel with known values to calibrate, we were able to calculate the value of kerosene, diesel and biodiesel reasonably close to the literature value of each sample, for kerosene (8365.2008 cal/g), diesel (10874.76 cal/g) and for biodiesel (8786.80688 cal/g).The calorific value (CV) of a specific type of fuel helps us measure and describe the energy that is produced by a given type of fuel. The bomb calorimeter is a device that burns a fuel sample and transfers the heat into a known mass of water. Most of the original error can be traced back to uncertainty in the quality of the fits of the fore- and after drift, as the original masses of sample and length of fuse wire both contribute only minimally to the final error. Nevertheless, we received a fairly accurate measurement with a good precision.

Index Terms—bomb calorimeter, calorific value, radiation Correction

Ogot, Krishna May L. Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09157101950, (e-mail: ogotlegarde@gmail.com).

Percil, Queenie Rose I. Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09168206602, (e-mail: inniedc14@gmail.com).

Compuesto, Chenny. Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09192758373, (e-mail: bischeakohahaha@gmail.com).

Eazyl D. Salazar, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09267880602, (e-mail: eazylsalazar@gmail.com).

Alwyn Wren C. Cuesta, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, 09063988292., (e-mail: alwyn_wren@yahoo.com).

Sarah May M. Pamaran, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09478483660, (e-mail: sarahmhay62@gmail.com)

I. INTRODUCTION

Calorimetry is a fundamental test of great significance to anyone concerned with the production or utilization of solid or liquid fuels. One of the most important tests in the evaluation of materials which are burned, as fuels, is the determination of the heat of combustion, or calorific value. These measurements can be made in the Bomb Calorimeter Set for Testing Calorific Value of Fuels (TBCF). The Bomb Calorimeter is a classic device used to determine the heating or calorific value of solid and liquid fuel samples at constant volume. Basically, this device burns a fuel sample and transfers the heat into a known mass of water. From the weight of the fuel sample and temperature rise of the water,

the calorific value can be calculated. The calorific value obtained in a bomb calorimeter test represents the gross heat of combustion per unit mass of fuel sample. This is the heat

produced when the sample burns, plus the heat given up when the newly formed water vapor condenses and cools to the temperature of the bomb. Determining calorific values is profoundly important; fuels are one of the biggest commodities in the world, and their calorific value. The Bomb Calorimeter study is carried out to gain a better understanding of the working principles behind the bomb calorimeter and also to find out the gross calorific values of different types of liquid fuel.

Liquid fuels are combustible or energy-generating molecules that can be harnessed to create mechanical energy, usually producing kinetic energy; they also must take the shape of their container. It is the fumes of liquid fuels that are flammable instead of the fluid.

II. DISCUSSION

Heat released in a chemical reaction can be determined experimentally by using an bomb (adiabatic) calorimeter. The reaction must proceed without any side reactions and sufficiently fast that the heat exchange with the surroundings would be negligible. The heat of combustion can be most measured conveniently using an adiabatic bomb calorimeter. In this, the combustion reaction occurs in a closed container under constant volume. The bomb is immersed in a weighted quantity or particular volume of water and surrounded by an adiabatic shield that serves as a heat insulator.

Continuous stirring ensures that heat is distributed evenly in the calorimeter. An adiabatic bomb calorimeter comprises of the bomb and the water bath which are in direct thermal contact. In this experiment, the heat of combustion of three different liquid fuels will be determined using this calorimeter. The heat of combustion is directly related to important quantities such as the internal energy and enthalpy of a chemical reaction.

III. MATERIALSAND APPARATUS

• Bomb Calorimeter Set for Testing Calorific Value of Fuels, TBCF.

• Fuse wire

• Graduated Cylinder (2000mL) • Laptop (Lab VIEW)

• Analytical Balance • Funnel

Liquid Fuel Samples:

Kerosene Diesel

Biodiesel

Determination of Heat of Combustion of Liquid Fuels

Using Bomb Calorimeter

Compuesto, Chenny

1

, Cuesta, Alwyn

2

, Ogot, Krishna May

3

, Pamaran, Sarah May

4

,

Percil, Queenie

5

, Salazar, Eazyl

6

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IV. EXPERIMENTAL SET-UP

Fig. 1 Bomb Calorimeter Set Up

Fig. 2 Actual Bomb Calorimeter Set Up

Fig. 2 Proper placement of the sample, crucible, and ignition wire

P. W. Atkins and J. de Paula, Physical Chemistry (7th ed.)

Fig. 4 Attachment of the ignition wire

P. W. Atkins and J. de Paula, Physical Chemistry (7th ed.)

V. PROCEDURE

1. Prepare the fuel sample by placing it in a crucible and weighing it on a balance. Ensure that the sample of the fuel will not overflow the crucible. Note down the weight of the fuel sample and place the crucible containing the fuel gently in the loop holder.

2. The bomb head has been pre-attached with 10.5 cm long fuse mire between the two electrodes. Bend the use wire down just above the liquid fuel sample. The wire must not make contact with the fuel crucible. To attach the fuse to quick-grip electrodes, insert the ends of the wire into the eyelet at the end of each stem and push the cap downward to pinch the wire into place. No further threading or twisting is required.

3. It is not necessary to submerge the wire in a powdered sample. In fact, better combustions will usually be obtained it the loop of the fuse is set slightly above the surface. When using pelleted samples, bend the wire so that the loop bears against the top of the pellet firmly enough to keep it from sliding against the side of the capsule.

4. Care must be taken no to disturb the sample when moving the bomb head from to the calorimeter bomb. Check the sealing ring to be sure that it is in good condition and moisten it with a hit of water so that it will slide freely into the body of the calorimeter bomb, then slide the head into the bomb and push it down as far as it will go. Set the screw cap on the bomb and turn it down firmly by hand to a solid stop. When properly closed, no threads on the bomb should be exposed.

5. Oxygen for the bomb can be drawn from a standard commercial oxygen cylinder. Connect the regulator to the cylinder, keeping the 0-55 atm. in an upright position. The pressure connection to the bomb is made with a slip connector on the oxygen hose which slides over the gas inlet titling on the bomb head. Slide the connector onto the inlet valve body and push it down as far as it will go.

Close the outlet valve on the bomb head; then open or "crack" the oxygen tank valve not more than one-quarter turn. Open the filling connection control valve slowly and watch the gage as the bomb pressure rises to the desired filling pressure (30 atm); then close the control valve. The

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Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016) bomb inlet check valve will close automatically when the

oxygen supply is shut off, leaving the bomb filled to the highest pressure indicated on the 0-55 atm. Release the residual pressure in the filling hose by pushing downward on the lever attached to the relief valve. The gage should now return to zero.

6. Fill the calorimeter vessel by first taring the empty vessel, then add 3000 ml of water.

7. Introduce the bomb calorimeter inside the calorimeter vessel. Handle the bomb carefully during this operation so that the sample will not be disturbed.

8. Check the bomb for leaks before firing. If any gas leakage was observed, no matter how slight, do not fire the bomb. Instead remove it from the water bath; release the pressure and eliminate the leak before proceeding with combustion test.

9. Fill the jacket with water.

10. Put the cover on the jacket. Turn the stirrer by hand to be sure that it runs freely and start the motor. Install the Beckman thermometer; this thermometer should he immersed in eater and not close to the bomb.

11. Let the stirrer run for at least 5 minutes to reach equilibrium before starting a measured run.

12. The scanning of the temperature data is pre-set to be done once a minute. At the start of the fifth minute, fire the charge by pressing the firing button on the control unit, keeping the circuit closed for about 5 seconds.

13. The vessel temperature will start to rise within 20-30 seconds after firing. This rise will be rapid during the first few minutes; then it will become slower as the temperature approaches a stable maximum as shown by the typical rise curve. Accurate time and temperature observations must be recorded to identify certain points needed to calculate the calorific value of the sample.

14. Usually the temperature will reach a maximum then it will drop very slowly. But this is not always true since a low starting temperature may result in a slow continuous rise without reaching a maximum. As stated, the difference between successive readings must be noted and the readings continued until the rate of the temperature change becomes constant over a period of 5 minutes.

15. After the last temperature reading, stop the stirrer. Let the bomb stand in the calorimeter vessel for at least 3 minutes. Then remove the jacket cover and extract the bomb calorimeter. Wipe the bomb with a clean cloth.

16. Open the valve knob on the bomb head slightly to release all residual gas pressure before attempting to remove the screw cap. This release should proceed slowly over a period of not less than one minute to avoid entrainment losses. After all pressure has been released, unscrew the cap; lift the head out of the cylinder. Do not twist the head during removal. Pull it straight out to avoid sticking. Examine the interior of the bomb for soot or other evidence of incomplete combustion. If such evidence is found, the test will have to be discarded.

17. Remove all unburned pieces of fuse wire from the bomb electrodes.

18. On completion of experiment, wash all inner surfaces of the bomb and the combustion crucible with a jet of distilled water and collect the washings. Keep the bomb set dry and clean with some wiping tissue.

VI. PROCESS FLOW DIAGRAM

Figure 1

VII. DATAANDRESULTS

Constant values: Pressure = 30 atm

Fuse wire length = 10.5 cm Weight of water being heated = 3011.333152 g Room temperature = 27~29 OC Compound Diesel Trial 1 2 Weight of Sample (g) 2.20 Max. Temp. (C) 33.4609375 35.2578125 Time at Max. (min) 10.275 8.478 Equilibrium 32.9531250 34.6250000

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Temp. (C) Time at Equilibrium (min) 40.000 40.000 Ave. Radiation Correction 0.121632795 Ave. Temp. rise

(C) 5.0315 Corrected Temp. Rise (C) 5.1531328 Heat Absorbed by Water (cal) 15,517.79964 Calorific Value (cal/g) 7,053.545291 Compound Kerosene Trial 1 2 3 Weight of Sample (g) 2.597 Max. Temp. (C) 34.6484375 36.8671875 35.5078125 Time at Max. (min) 8.783 30.140 7.800 Equilibrium Temp. (C) 33.8984375 36.7109375 34.5468750 Time at Equilibrium (min) 40.000 42.500 40.000 Ave. Radiation Correction 0.1877717 Ave. Temp. rise (C) 7.7473958 Corrected Temp. Rise (C) 7.93516 Heat Absorbed by Water (cal) 23,895.41038 Calorific Value (cal/g) 9201.159175 Compound Biodiesel Trial 1 2 3 Weight of Sample (g) 2.827 Max. Temp. (C) 36.1171875 33.9140625 36.3906250 Time at Max. (min) 11.080 13.563 7.950 Equilibrium Temp. (C) 35.4453125 33.5781250 35.9531250 Time at Equilibrium (min) 40.000 40.000 25.000 Ave. Radiation Correction 0.1597254 Ave. Temp. rise (C) 7.04088 Corrected Temp. Rise (C) 7.2006054 Heat Absorbed by Water (cal) 21,683.42176 Calorific Value (cal/g) 7670.117354

Calculated CVs of liquid fuel samples Sample Calorific Value (cal/g) True Value (cal/g) Percentage Error (%) Diesel 7,053.54529 1 10,874.76 35.1384 Kerosene 9201.159175 8365.2008 10.0000 Biodiesel 7670.117354 8786.8068 8 12.7081 VIII.CALCULATIONS

Calculations for Diesel TRIAL 1:

Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark

Slope Calculation: TM = 33.4609375 OC tM = 10.275 min TE = 32.9531250 OC tE = 40.000 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

t

E

t

M

Slope=

33.4609375−32.9531250

40−10.275

Slope = 0.0170837

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the beginning, divided by four

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Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)

Note that the room temperature during the experiment is 28

OC

time mark reading (min) change in temperature

0 (beginning) 28.0546875 - 28 = 0.0546875

2 28.0781250 – 28 = 0.078125

3 (ignition) 28.0859375 – 28 = 0.0859375

10.275 (max) 33.4609375 – 28 = 5.4609375

11.275 (minute after max) 33.4453125 – 28 = 5.4453125

15.275 (fourth minute mark after the minute after max)

33.3984375 – 28 = 5.3984375 n = 10.275 – 3.000 = 7.275 min

v

'

=

5.4453125−5.3984375

4

=0.01171875

v =

0.078125−0.0546875

4

=0.005859375

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

7.275∗0.01171875 +

−0.005859375+0.01171875

2

Radiation Correction = 0.08818359

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 5.4609375 - 0.0859375 =

5.375 O C TRIAL 2:

Assumption: The end of post period is at 40-minute mark Ignition: 0.1-minute mark (6 seconds after running)

Slope Calculation: TM = 35.2578125 OC tM = 8.478 min TE = 34.6250000 OC tE = 40.000 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

t

E

t

M

Slope=

35.2578125−34.6250000

40−8.478

Slope = 0.0200752

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the beginning, divided by four

Note that the room temperature during the experiment is 28

OC

Also note that it was ignited at 6-second mark so there is no data for the before-ignition-change-in-temperature

Thus, the ignition’s change in temperature will be used for the computation of v’

time mark reading (min) change in temperature

0 (beginning) 28.5625000 – 28 = 0.5625000

0.100 (ignition) 28.5703125 – 28 = 0.5703125

8.478 (max) 35.2578125 – 28 = 5.2578125

9.478 (minute after max) 35.2421875 – 28 = 5.2421875

13.478 (fourth minute mark after the minute after max)

35.1718750 – 28 = 5.1718750 n = 8.478 – 0.100 = 8.378 min

v

'

=

5.2421875−5.1718750

4

=0.017578125

v =

0.5703125−0.5625000

4

=0.001953125

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

8.378∗0.017578125+

−0.00195315+0.0175781

2

Radiation Correction = 0.1550820

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 5.2578125 - 0.5703125 = 4.688 O C

Calorific Value Calculations for Diesel: Average Radiation Correction of Diesel

¿

0 . 08818359+0 . 1550820

2

= 0.121632795

Average Rise in Temperature

¿

5 . 375+4 .688

2

(6)

Average Weight of Diesel Fuel

¿

2 . 14+2 . 25

2

= 2.20 g

Corrected Rise in Temperature

¿

RadiationCorrection+Rise∈Temperature

¿

0 .121632795+5 . 0315

= 5.1531328 O C

Heat Absorbed by Water

¿

Weight of Water being Heated∗Correction Rise∈Temperature

¿

3011.333152∗5 .1531328

= 15,517.79964 cal Calorific Value of Diesel

¿

Heat Absorbed by Water

AverageWeight of Fuel

¿

15 , 517 . 79964

2 . 20

= 7,053.545291 cal/g Calculations for Kerosene TRIAL 1:

Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark

Slope Calculation: TM = 34.6484375 OC tM = 8.783 min TE = 33.8984375 OC tE = 40.000 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

t

E

t

M

Slope=

34.6484375−33.8984375

40−8.783

Slope = 0.0240254

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the

beginning, divided by four

Note that the room temperature during the experiment is 27

OC.

Also note that the beginning is 1 and not 0 because we forgot to turn on the agitator before running the test.

time mark reading (min) change in temperature

1 (beginning) 27.4609375 – 27 = 0.4609375

2 27.4609375– 27 = 0.4609375

3 (ignition) 27.4765625– 27 = 0.4765625

8.783 (max) 34.6484375 – 27 = 7.6484375

9.783 (minute after max) 34.6484375 – 27 = 7.6484375

13.783 (fourth minute mark after the minute after max)

34.5625000 – 27 = 7.5625000 n = 8.783 – 3.000 = 5.783 min

v

'

=

7.6484375−7.5625000

4

=0.01171875

v =

0.4609375−0.4609375

4

=0.0000000

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

5.783∗0.01171875 +

−0+0.01171875

2

Radiation Correction = 0.0736289

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 7.6484375 - 0.4765625 = 7.171875 CO

TRIAL 2:

Assumption: The end of post period is at 42.5-minute mark Ignition: 0.1-minute mark (6 seconds after running)

Slope Calculation: TM = 36.8671875 OC tM = 30.140 min TE = 36.7109375 OC tE = 42.500 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

t

E

t

M

Slope=

36.8671875−36.7109375

42.5−30.141

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Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the beginning, divided by four

Note that the room temperature during the experiment is 29

OC

Also note that it was ignited at 6-second mark so there is no data for the before-ignition-change-in-temperature

Thus, the ignition’s change in temperature will be used for the computation of v’

time mark reading (min) change in temperature

0 (beginning) 29.1015625 – 29 = 0.1015625

0.100 (ignition) 29.1015625 – 29 = 0.1015625

30.140 (max) 36.8671875 – 29 = 7.8671875

31.140 (minute after max) 36.8593750 – 29 = 7.859375

35.140 (fourth minute mark after the minute after max)

36.8203125 – 29 = 7.8203125 n = 30.140 – 0.100 = 30.04 min

v

'

=

7.859375−7.8203125

4

=

0.0097656

v =

0.1015625−0.1015625

4

=0.0000000

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

30.04∗0.0097656+

−0+0.0097656

2

Radiation Correction = 0.2982414

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 7.8671875 - 0.1015625 = 7.765625 O C

TRIAL 3:

Assumption: The end of post period is at 40-minute mark Ignition: 0.1-minute mark (6 seconds after running)

Slope Calculation: TM = 35.5078125 OC tM = 7.800 min TE = 34.5468750 OC tE = 40.000 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

t

E

t

M

Slope=

35.5078125−34.5468750

40−7.8

Slope = 0.0298428

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the beginning, divided by four

Note that the room temperature during the experiment is 27

OC

Also note that it was ignited at 6-second mark so there is no data for the before-ignition-change-in-temperature

Thus, the ignition’s change in temperature will be used for the computation of v’

time mark reading (min) change in temperature

0 (beginning) 27.2031250 – 27 = 0.2031250

0.100 (ignition) 27.2031250 – 27 = 0.2031250

7.14 (max) 35.5078125 – 27 = 8.5078125

8.14 (minute after max) 35.5000000 – 27 = 8.5000000

12 (fourth minute mark after the minute after max)

35.3984375 – 27 = 8.3984375 n = 7.140 – 0.100 = 7.040 min

v

'

=

8.5000000−8.3984375

4

=0.0253906

v =

0.2031250−0.2031250

4

=0.0000000

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

7.040∗0.0253906+

−0+0.0253906

2

Radiation Correction = 0.1914451

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 8.5078125 - 0.2031250 = 8.3046875 O C

Calorific Value Calculations for Kerosene

(8)

Average Radiation Correction of Kerosene

¿

0 . 0736289+0 . 2982414+0 . 1914451

3

= 0.1877718

Average Rise in Temperature

¿

7 . 171875+7 . 765625+8 .3046875

3

= 7.7473958 O C

Average Weight of Diesel Fuel

¿

2 . 64+2 . 50+2. 65

3

= 2.597 g

Corrected Rise in Temperature

¿

RadiationCorrection+Rise∈Temperature

¿

0 .1877718+7 . 7473958

= 7.93516 O C

Heat Absorbed by Water

¿

Weight of Water being Heated∗Correction Rise∈Temperature

¿

3011.333152∗7 . 93516

= 23,895.41038 cal Calorific Value of Diesel

¿

Heat Absorbed by Water

AverageWeight of Fuel

¿

23 , 895 . 41038

2 . 597

= 9201.159175 cal/g Calculations for Biodiesel TRIAL 1:

Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark

Slope Calculation: TM = 36.1171875 OC tM = 11.080 min TE = 35.4453125 OC tE = 40.000 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

t

E

t

M

Slope=

36.1171875−35.4453125

40−11.080

Slope = 0.0232322

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the beginning, divided by four

Note that the room temperature during the experiment is 28

OC.

Also note that the beginning is 1 and not 0 because we forgot to turn on the agitator before running the test.

time mark reading (min) change in temperature

1 (beginning) 28.3750000 – 28 = 0.3750000

2 28.3906250 – 28 = 0.3906250

3 (ignition) 28.3984375– 28 = 0.3984375

11.080 (max) 36.1171875 – 28 = 8.1171875

12.080 (minute after max) 36.1171875 – 28 = 8.1171875

16.080 (fourth minute mark

after the minute after max) 36.0234375 – 28 = 8.0234375

n = 11.080 – 3.000 = 8.080 min

v

'

=

8.1171875−8.0234375

4

=0.0234375

v =

0.3906250−0.3750000

4

=0.0039063

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

8.080∗0.0234375+

−0.0039063+0.0234375

2

Radiation Correction = 0.1991406

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 8.1171875 – 0.3984375 = 7.71875 O C

TRIAL 2:

Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark

Slope Calculation: TM = 33.9140625 OC tM = 13.563 min TE = 33.5781250 OC tE = 40.000 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

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Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)

Slope=

33.9140625−33.5781250

40−13.563

Slope = 0.0127071

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the beginning, divided by four

Note that the room temperature during the experiment is 28

OC.

Also note that the beginning is 1 and not 0 because we forgot to turn on the agitator before running the test.

time mark reading (min) change in temperature

1 (beginning) 28.6250000 – 28 = 0.6250000

2 28.7187500 – 28 = 0.7187500

3 (ignition) 28.7265625– 28 = 0.7265625

13.563 (max) 33.9140625 – 28 = 5.9140625

14.563 (minute after max) 33.9062500 – 28 = 5.9062500

18.563 (fourth minute mark after the minute after max)

33.8593750 – 28 = 5.8593750 n = 13.563 – 3.000 = 10.563 min

v

'

=

5.9062500−5.8593750

4

=0.01171875

v =

0.7187500−0.6250000

4

=0.0234375

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

10.563∗0.01171875+

−0.0234375+0.01171875

2

Radiation Correction = 0.1179258

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 5.9140625 – 0.07265625 = 5.8414063 CO

TRIAL 3:

Assumption: The end of post period is at 25-minute mark Ignition: 0.1-minute mark (6 seconds after running)

Slope Calculation: TM = 36.3906250 OC tM = 7.950 min TE = 35.9531250 OC tE = 25.000 min Where: TM = maximum temperature

tM = time at maximum temperature

TE = equilibrium temperature

tE = time at equilibrium temperature

Slope=

T

M

T

E

t

E

t

M

Slope=

36.3906250−35.9531250

25−7.95

Slope = 0.0256598

Radiation Correction Calculation:

n = time difference between maximum temperature and ignition

v’ = change in change in temperature at the first minute mark after the attainment of maximum temperature and the change in temperature at the fourth minute mark after the first minute mark, divided by four

v = change in change in temperature before the ignition and the change in temperature at the beginning, divided by four

Note that the room temperature during the experiment is 28

OC

Also note that it was ignited at 6-second mark so there is no data for the before-ignition-change-in-temperature

Thus, the ignition’s change in temperature will be used for the computation of v’

time mark reading (min) change in temperature

0 (beginning) 28.8359375 – 28 = 0.8359375

0.100 (ignition) 28.8281250 – 28 = 0.8281250

7.95 (max) 36.3906250 – 28 = 8.3906250

8.95 (minute after max) 36.3750000 – 28 = 8.3750000

12.95 (fourth minute mark after the minute after max)

36.2968750 – 28 = 8.2968750 n = 7.95 – 0.100 = 7.850 min

v

'

=

8.3750000−8.2968750

4

=0.0195313

v =

0.8281250−0.8359375

4

=0.0019531

RadiationCorrection=n∗v

'

+

v +v '

2

Radiation Correction

¿

7.850∗0.0195313+

−0.0019531+0.0195313

2

Radiation Correction = 0.16210981

Rise in Temperature during Test = change in temperature at maximum – change in temperature at ignition

Rise in Temperature during Test = 8.3906250 - 0.8281250 = 7.5625 O C

(10)

Calorific Value Calculations for Biodiesel: Average Radiation Correction of Biodiesel

¿

0 . 1991406+0 . 1179258+0 . 16210981

3

= 0.1597254

Average Rise in Temperature

¿

7 . 71875+5 . 8414063+7 .5625

3

= 7.04088 O C

Average Weight of Diesel Fuel

¿

2 . 85+2 .81+2 . 82

3

= 2.827 g

Corrected Rise in Temperature

¿

RadiationCorrection+Rise∈Temperature

¿

0 .1597254 +7 . 04088

= 7.2006054 O C

Heat Absorbed by Water

¿

Weight of Water being Heated∗Correction Rise∈Temperature

¿

3011.333152∗7 .2006054

= 21,683.42176 cal :

Calorific Value of Diesel

¿

Heat Absorbed by Water

AverageWeight of Fuel

¿

21 , 683 . 42176

2 . 827

= 7670.117354 cal/g

% Error of DIESEL (TRIAL 1)

%error=l

¿

experimental−

¿

theoretical

¿

theoretical

l X 100 True Calorific value of Diesel:10874.76 cal/g

%error=l

7053.545291−10874.76

10874.76

l X 100 =

35.1384 %

% Error of KEROSENE (TRIAL 1)

%error=l

¿

experimental−

¿

theoretical

¿

theoretical

l X 100 True Calorific value of KEROSENE:8365.2008 cal/g

%error=l

9201.159175−8365.2008

8365.2008

l X 100 =

10.0000 %

% Error of BIODIESEL (TRIAL 1)

%error=l

¿

experimental−

¿

theoretical

¿

theoretical

l X 100 True Calorific value of KEROSENE:8786.80688 cal/g

%error=l

7670.117354−8786.80688

8786.80688

l X 100 =

12.7081 %

IX. CONCLUSION

In this experiment, we used Lab VIEW in conjunction with a bomb calorimeter to determine the calorific value of different types of Fuel. The experiments carried out were quite successful, and yielded valid results. The final results of the experiment are given as follows:

Sample Calorific Value (cal/g) True Value (cal/g) Percentage Error (%) Diesel 7,053.54529 1 10,874.76 35.1384 Kerosene 9201.159175 8365.2008 10.0000 Biodiesel 7670.117354 8786.8068 8 12.7081

Our calculated values of the calorific value of our known samples, though not perfect, are from bad, with respectable for diesel trial 1 (29.57%), trial 2 (45.467%), for kerosene trial 1 (1.57%), trial 2 (15.677%), and trial 3 (14.98%) , for Biodiesel Trial 1 (5.15%), trial 2 (27.59%) and trial 3 (6.48%) error from literature values. Our result is understandable and adequate. Understanding how bomb calorimeter is different from standard constant-pressure calorimetry methods is a key to realizing why bomb calorimeter is the method of choice for accurate measurement of energies and elemental analysis.

X. HAZARDSANDCOUNTERMEASURES Skin burns – refrain from touching the calorimeter immediately right after the trial was done. Wait for a few minutes for its system to cool down.

Serious facial injury – secure that the calorimeter is tightly sealed before pressurizing it to avoid injuries that the loose lid might cause.

Electrocution – check for any submerged or broken electrical wires before powering up or setting up the apparatus.

Explosion – refrain from using any materials that can induce combustion of the liquid fuel samples while performing the experiment.

XI. WASTEDISPOSAL

Properly segregate or provide a secured bin for the rags, cloths, and tissues used to wipe the crucible and the liquid fuel spills.

(11)

Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)

XII. APPENDIX

Bomb Calorimeter Set Up

DataStudio displaying the measured temperature inside the calorimeter

From left to right: diesel, kerosene, and biodiesel

Oxygen gas tank for pressure filling

Post-laboratory group picture

(12)

XIII. AUTHORS

Krishna May Ogot

Currently resides in Taguig City. She took up Chemical Engineering Technology in Technological University of the Philippines-Taguig for three years and had a Supervised Industrial Training in S.C. Johnson and Son for a span of six months. At present she is now continuing her studies in Technological Institute of the Philippines- Manila with a program of B.S. Chemical Engineering. Not much of an achievement can be said to her because she is still learning on how to become a full pledge Chemical Engineer. Aside from being a licensed chemical engineer she wants to continue her studies to masteral degree and if possible until doctorate but she knows overcoming this goals can be hard but fulfilling. However with the support from her family and with God she can make these things possible.

Sarah May Manzano Pamaran

22 years old. I graduated from Technological University of the Philippines-Taguig Campus as a Chemical Technician. One of my biggest dreams is to become an Engineer, that’s why I’ve decided to continue my Bachelor’s Degree here in TIP-Manila Campus. I’m a hard working person, though sometimes I wanted to give up in this program, because we all know that Engineering Program specifically CHEMICAL ENGINEERING is not easy as the other people thought. I remember when I was studying in TUP-Taguig one of the hardest Program there is Chemical Engineering Technology, our professor always tells us that if we can’t handle being a Chem. Tech student, we have no rights to pursue BS Chemical Engineering. So, as I study here in TIP and took some major courses, it taught me how to handle problems and manage my time in terms of school, family, and friends because in Engineering Program you have no choice but to study and study and study and study. As of now one of the hardest course that I have encounter in TIP was Chemical Calculation1 and also Chemical Calculation2. I don’t know but the CheCal course is not just a course that you’ll have to calculate this using this formula, you need to analyze and understand carefully each problem because this course is very complicated. But then I’m so thankful that we have Engr. Crizaldy Tugade to teach us, he always let us understands the topic clearly.

Queenie Rose Percil

A simple Chemical Engineering student who is the eldest among my siblings. I love to sing and do a lot of physical activities such as hiking. Being the eldest among my siblings, I am entitled with a big responsibility. Aside from that, I am also a good and caring friend that you can rely on everytime. You will get wrong with with me the first time you met me because I look so snobbish but reality says that I am really approachable. One of my biggest dreams is to see my parents during my graduation. Aside from being a Chemical Engineer, I also wanted to be a Doctor of Internal Medicine. I really wanted my parents to be so proud of me. I'll prove to them that I am completely different a lot bigger than those people they are comparing to me.

Alwyn Wren Cuesta

Alwyn Wren Cuesta was born on 11 November 1997

in Quezon City, Philippines. He is a junior student at the Technological Institute of the Philippines Manila and currently taking up a bachelor’s degree in Chemical Engineering. He is an avid reader, an otaku, a gamer, an inventor and a violinist. As a chemical engineering student he has trained and still training to perform highly in different fields such as mathematics, biochemistry, particle

technology, computers, plant designs and conversions of raw materials into advanced materials. He is highly imaginative and an introvert, and his conviction and dedication are what set him apart from anybody else. As of now, he is struggling against a series of unfortunate events towards his dreams such as two of his major courses- Integration Course 1 and Chemical Engineering Calculations 2. Despite of almost collapsing from numerous numbers of projects and having only a few hours of sleep, he never gave up on Chemical Engineering because of his extreme love with it. He dreams of using his skills in Chemical Engineering to create all of his fictitious and astounding imagination in the future such asbuilding the Iron Man Armor, creating the Dragon Blade of Hiccup, and the invention of a medicine that regenerates telomeres to achieve immortality.

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.

Chenny Ibañez Compuesto

I am Chenny Ibanez Compuesto, a Chemical Engineering student. I was born on May 13, 1996 in Antipolo City. I didn’t imagine that I would take Chemical Engineering. I knew back then that ENGINEERING isn’t an easy way to be successful. After all, my high-school crush, who at first wanted to take this program but resorted to BS Math in UP Diliman, warned me that it will be full of Math, and he is right. But, because of a sudden turn of events, my dream to be a simple chemical analyst was redirected into this new path: to be a chemical engineer. (At least there are a few differences between a chemist and Ch.E., since both has board exams.) Now, I enjoy my studying here, although I experience difficulty and pressure in keeping up with school work, extra-curricular activities and varying attitudes of upperclassmen, underclassmen and batch mates. I am currently involved as a committee member in one of the organizations in my department, and I still compete in the quiz bees here in TIP and even for the first time in the National Quiz bee. I still have my aim to finish my undergraduate studies here, and soon enough, be a topnotcher, if not, a Ch.E. board passer, but for now, I’ll enjoy every single moment that I have to make here, so that when time comes, I’ll never have any regrets.

(13)

Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)

References

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