Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016) 5 6 1 Percil, Queenie , Salazar, Eazyl Compuesto, Chenny , Cuesta, Alwyn2, Ogot, Krishna May3 , Pamaran, Sarah May4,
Determination of Heat of Combustion of Liquid Fuels Using Bomb Calorimeter 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.
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.
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.
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:
[email protected]). Percil, Queenie Rose I. Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09168206602, (e-mail:
[email protected]). Compuesto, Chenny. Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09192758373, (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]). Alwyn Wren C. Cuesta, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, 09063988292., (e-mail:
[email protected]). Sarah May M. Pamaran, Chemical Engineering Department, Technological Institute of the Philippines/ College of Engineering and Architecture, Manila, Philippines, 09478483660, (e-mail:
[email protected])
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. MATERIALS AND APPARATUS
I.
• 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
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,
1
Determination of Heat of Combustion of Liquid Fuels Using Bomb Calorimeter IV. EXPERIMENTAL SET-UP
Fig. 4 Attachment of the ignition wire P. W. Atkins and J. de Paula, Physical Chemistry (7th ed.)
Fig. 1 Bomb Calorimeter Set Up
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. Fig. 2 Actual Bomb Calorimeter Set Up
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.
Fig. 2 Proper placement of the sample, crucible, and ignition wire
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.
P. W. Atkins and J. de Paula, Physical Chemistry (7th ed.)
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
2
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.
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.
6. Fill the calorimeter vessel by first taring the empty vessel, then add 3000 ml of water.
VI. PROCESS FLOW DIAGRAM
7. Introduce the bomb calorimeter inside the calorimeter vessel. Handle the bomb carefully during this operation so that the sample will not be disturbed.
Figure 1
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.
VII. DATA AND RESULTS Constant values: Pressure = 30 atm Fuse wire length = 10.5 cm Weight of water being heated = 3011.333152 g Room temperature = 27~29 OC
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.
Compound Trial Weight of Sample (g) Max. Temp. (C) Time at Max. (min) Equilibrium
17. Remove all unburned pieces of fuse wire from the bomb electrodes.
3
Diesel 1
2 2.20
33.4609375
35.2578125
10.275
8.478
32.9531250
34.6250000
Determination of Heat of Combustion of Liquid Fuels Using Bomb Calorimeter Temp. (C) Time at Equilibrium (min) Ave. Radiation Correction Ave. Temp. rise (C) Corrected Temp. Rise (C)
40.000
40.000
0.121632795 5.0315
15,517.79964
Calorific (cal/g)
7,053.545291
Value
Compound Trial Weight of Sample (g) Max. Temp. (C) Time at Max. (min) Equilibrium Temp. (C) Time at Equilibrium (min) Ave. Radiation Correction Ave. Temp. rise (C)
21,683.42176 7670.117354
5.1531328
Heat Absorbed by Water (cal)
Compound Trial Weight of Sample (g) Max. Temp. (C) Time at Max. (min) Equilibrium Temp. (C) Time at Equilibrium (min) Ave. Radiation Correction Ave. Temp. rise (C) Corrected Temp. Rise (C) Heat Absorbed by Water (cal) Calorific Value (cal/g)
7.2006054
Corrected Temp. Rise (C) Heat Absorbed by Water (cal) Calorific Value (cal/g)
1
34.6484375 8.783 33.8984375 40.000
Kerosene 2 2.597
Calculated CVs of liquid fuel samples
3
36.867187 5 30.140
35.507812 5 7.800
36.710937 5 42.500
34.546875 0 40.000
9201.159175
35.4453125 40.000
Diesel
10,874.76
35.1384
Kerosene
7,053.54529 1 9201.159175
8365.2008
10.0000
Biodiesel
7670.117354
8786.8068 8
12.7081
Where: TM = maximum temperature tM = time at maximum temperature TE = equilibrium temperature tE = time at equilibrium temperature
23,895.41038
11.080
Percentage Error (%)
Slope Calculation: TM = 33.4609375 OC tM = 10.275 min TE = 32.9531250 OC tE = 40.000 min
7.93516
36.1171875
True Value (cal/g)
Calculations for Diesel TRIAL 1: Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark
7.7473958
Biodiesel 2 2.827
Calorific Value (cal/g)
VIII. CALCULATIONS
0.1877717
1
Sample
Slope=
T M −T E t E −t M
Slope=
33.4609375−32.9531250 40−10.275
3
33.914062 5 13.563
36.390625 0 7.950
33.578125 0 40.000
35.953125 0 25.000
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
0.1597254 7.04088
4
Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016) Note that the room temperature during the experiment is 28 O C
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 time mark reading (min) change in temperature temperature at the fourth minute mark after the 0 (beginning) 28.0546875 - 28 = 0.0546875 first minute mark, divided by four 2 28.0781250 – 28 = 0.078125 v = change in change in temperature before the 3 (ignition) 28.0859375 – 28 = 0.0859375 ignition and the change in temperature at the 10.275 (max) 33.4609375 – 28 = 5.4609375 beginning, divided by four 11.275 (minute after max) 33.4453125 – 28 = 5.4453125 15.275 (fourth minute mark 33.3984375 – 28 = 5.3984375 Note that the room temperature during the experiment is 28 after the minute after max) O C Also note that it was ignited at 6-second mark so there is no data for the before-ignition-change-in-temperature n = 10.275 – 3.000 = 7.275 min Thus, the ignition’s change in temperature will be used for the computation of v’ 5.4453125−5.3984375 ' time mark reading (min) change in temperature v= =0.01171875 4 0 (beginning) 28.5625000 – 28 = 0.562500
0.078125−0.0546875 =0.005859375 4 ' −v + v ' RadiationCorrection=n∗v + 2
v=
Radiation Correction
¿ 7.275∗0.01171875 +
– – – –
28 28 28 28
= = = =
0.570312 5.257812 5.242187 5.171875
−0.005859375+0.01171875 2 n = 8.478 – 0.100 = 8.378 min
Radiation Correction = 0.08818359
5.2421875−5.1718750 =0.017578125 4 0.5703125−0.5625000 v= =0.001953125 4 ' −v + v ' RadiationCorrection=n∗v + 2 v'=
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 OC TRIAL 2: Assumption: The end of post period is at 40-minute mark Ignition: 0.1-minute mark (6 seconds after running)
Radiation Correction
¿ 8.378∗0.017578125+
Slope Calculation: TM = 35.2578125 OC tM = 8.478 min TE = 34.6250000 OC tE = 40.000 min
−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 OC
Where: TM = maximum temperature tM = time at maximum temperature TE = equilibrium temperature tE = time at equilibrium temperature
Slope=
28.5703125 35.2578125 35.2421875 35.1718750
0.100 (ignition) 8.478 (max) 9.478 (minute after max) 13.478 (fourth minute mark after the minute after max)
Calorific Value Calculations for Diesel: Average Radiation Correction of Diesel
T M −T E t E −t M
¿
0 . 08818359+0 . 1550820 2
= 0.121632795
35.2578125−34.6250000 Slope= 40−8.478
Average Rise in Temperature
¿
Slope = 0.0200752
5 . 375+4 .688 2
= 5.0315 OC
5
Determination of Heat of Combustion of Liquid Fuels Using Bomb Calorimeter beginning, divided by four Average Weight of Diesel Fuel
2 . 14+2 . 25 2
Note that the room temperature during the experiment is 27 O C. Also note that the beginning is 1 and not 0 because we = 2.20 g forgot to turn on the agitator before running the test. time mark reading (min) change in temperature Corrected Rise in Temperature 1 (beginning) 27.4609375 – 27 = 0.460937 ¿ RadiationCorrection + Rise∈Temperature 2 27.4609375– 27 = 0.460937 ¿ 0 .121632795+5 . 0315 3 (ignition) 27.4765625– 27 = 0.476562 = 5.1531328 OC 8.783 (max) 34.6484375 – 27 = 7.648437 9.783 (minute after max) 34.6484375 – 27 = 7.648437 Heat Absorbed by Water 13.783 (fourth minute mark 34.5625000 – 27 = 7.562500 ¿ Weight of Water being Heated∗Correction Rise∈Temperature after the minute after max)
¿
¿ 3011 .333152∗5 .1531328
= 15,517.79964 cal n = 8.783 – 3.000 = 5.783 min
Calorific Value of Diesel
Heat Absorbed by Water AverageWeight of Fuel 15 , 517 . 79964 ¿ 2 . 20
7.6484375−7.5625000 =0.01171875 4 0.4609375−0.4609375 v= =0.0000000 4
¿
'
v=
= 7,053.545291 cal/g
RadiationCorrection=n∗v ' + Calculations for Kerosene TRIAL 1: Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark
Radiation Correction
¿ 5.783∗0.01171875 +
Slope Calculation: TM = 34.6484375 OC tM = 8.783 min TE = 33.8984375 OC tE = 40.000 min
−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 OC
Where: TM = maximum temperature tM = time at maximum temperature TE = equilibrium temperature tE = time at equilibrium temperature
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
T −T E Slope= M t E −t M Slope=
−v + v ' 2
34.6484375−33.8984375 40−8.783
Where: TM = maximum temperature tM = time at maximum temperature TE = equilibrium temperature tE = time at equilibrium temperature
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
Slope=
T M −T E t E −t M
Slope=
36.8671875−36.7109375 42.5−30.141
Slope = 0.0126426
6
Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016) TE = equilibrium temperature tE = time at equilibrium temperature
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
Slope=
T M −T E t E −t M
Slope=
35.5078125−34.5468750 40−7.8
Slope = 0.0298428
Radiation Correction Calculation: Note that the room temperature during the experiment is 29 n = time difference between maximum temperature and O C ignition Also note that it was ignited at 6-second mark so there is no v’ = change in change in temperature at the data for the before-ignition-change-in-temperature first minute mark after the attainment of Thus, the ignition’s change in temperature will be used for maximum temperature and the change in the computation of v’ temperature at the fourth minute mark after the time mark reading (min) change in temperature first minute mark, divided by four 0 (beginning) 29.1015625 – 29 = 0.1015625 v = change in change in temperature before the ignition and the change in temperature at the 0.100 (ignition) 29.1015625 – 29 = 0.1015625 beginning, divided by four 30.140 (max) 36.8671875 – 29 = 7.8671875 Note that the room temperature during the experiment is 27 31.140 (minute after max) 36.8593750 – 29 = 7.859375 O C 35.140 (fourth minute mark 36.8203125 – 29 = 7.8203125 Also note that it was ignited at 6-second mark so there is no after the minute after max) data for the before-ignition-change-in-temperature Thus, the ignition’s change in temperature will be used for the computation of v’ n = 30.140 – 0.100 = 30.04 min time mark reading (min) change in temperature 0 (beginning) 27.2031250 – 27 = 0.203125 7.859375−7.8203125 '
v=
=0.0097656 4 0.1015625−0.1015625 v= =0.0000000 4 ' −v + v ' RadiationCorrection=n∗v + 2
Radiation Correction
¿ 30.04∗0.0097656+
0.100 (ignition) 7.14 (max) 8.14 (minute after max) 12 (fourth minute mark after the minute after max)
−0+ 0.0097656 2
27.2031250 35.5078125 35.5000000 35.3984375
– – – –
27 27 27 27
n = 7.140 – 0.100 = 7.040 min
Radiation Correction = 0.2982414
8.5000000−8.3984375 =0.0253906 4 0.2031250−0.2031250 v= =0.0000000 4 ' −v + v ' RadiationCorrection=n∗v + 2 v'=
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 OC
Radiation Correction
TRIAL 3: Assumption: The end of post period is at 40-minute mark Ignition: 0.1-minute mark (6 seconds after running)
¿ 7.040∗0.0253906+
−0+0.0253906 2
Radiation Correction = 0.1914451
Slope Calculation: TM = 35.5078125 OC tM = 7.800 min TE = 34.5468750 OC tE = 40.000 min
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 OC
Where: TM = maximum temperature tM = time at maximum temperature
Calorific Value Calculations for Kerosene
7
= = = =
0.203125 8.507812 8.500000 8.398437
Determination of Heat of Combustion of Liquid Fuels Using Bomb Calorimeter 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
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 OC
Note that the room temperature during the experiment is 28 O C. 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 = 0.375000 2 28.3906250 – 28 = 0.390625 3 (ignition) 28.3984375– 28 = 0.398437 11.080 (max) 36.1171875 – 28 = 8.117187 12.080 (minute after max) 36.1171875 – 28 = 8.117187 16.080 (fourth minute mark 36.0234375 – 28 = 8.023437 after the minute after max)
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 OC Heat Absorbed by Water
¿ Weight of Water being Heated∗Correction Rise∈Temperature n = 11.080 – 3.000 = 8.080 min ¿ 3011 .333152∗7 . 93516
= 23,895.41038 cal
8.1171875−8.0234375 =0.0234375 4 0.3906250−0.3750000 v= =0.0039063 4 −v + v ' RadiationCorrection=n∗v ' + 2 v'=
Calorific Value of Diesel
Heat Absorbed by Water AverageWeight of Fuel 23 , 895 . 41038 ¿ 2 . 597 ¿
Radiation Correction
= 9201.159175 cal/g
¿ 8.080∗0.0234375+ Calculations for Biodiesel TRIAL 1: Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark
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 OC
Slope Calculation: TM = 36.1171875 OC tM = 11.080 min TE = 35.4453125 OC tE = 40.000 min
TRIAL 2: Assumption: The end of post period is at 40-minute mark Ignition: 3-minute mark
Where: TM = maximum temperature tM = time at maximum temperature TE = equilibrium temperature tE = time at equilibrium temperature
Slope= Slope=
−0.0039063+0.0234375 2
Slope Calculation: TM = 33.9140625 OC tM = 13.563 min TE = 33.5781250 OC tE = 40.000 min
T M −T E t E −t M
Where: TM = maximum temperature tM = time at maximum temperature TE = equilibrium temperature tE = time at equilibrium temperature
36.1171875−35.4453125 40−11.080
Slope = 0.0232322
Slope=
Radiation Correction Calculation: n = time difference between maximum temperature and ignition
8
T M −T E t E −t M
Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)
Slope=
33.9140625−33.5781250 40−13.563
TM = maximum temperature tM = time at maximum temperature TE = equilibrium temperature tE = time at equilibrium temperature
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
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 Note that the room temperature during the experiment is 28 O first minute mark after the attainment of C. maximum temperature and the change in Also note that the beginning is 1 and not 0 because we temperature at the fourth minute mark after the forgot to turn on the agitator before running the test. first minute mark, divided by four time mark reading (min) change in temperature v = change in change in temperature before the 1 (beginning) 28.6250000 – 28 = 0.6250000 ignition and the change in temperature at the 2 28.7187500 – 28 = 0.7187500 beginning, divided by four 3 (ignition) 28.7265625– 28 = 0.7265625 13.563 (max) 33.9140625 – 28 = 5.9140625 Note that the room temperature during the experiment is 28 14.563 (minute after max) 33.9062500 – 28 = 5.9062500 O C 18.563 (fourth minute mark 33.8593750 – 28 = 5.8593750 Also note that it was ignited at 6-second mark so there is no after the minute after max) 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 = 0.835937
n = 13.563 – 3.000 = 10.563 min
5.9062500−5.8593750 =0.01171875 4 0.7187500−0.6250000 v= =0.0234375 4 ' −v + v ' RadiationCorrection=n∗v + 2 v'=
0.100 (ignition) 7.95 (max) 8.95 (minute after max) 12.95 (fourth minute mark after the minute after max)
28.8281250 36.3906250 36.3750000 36.2968750
– – – –
28 28 28 28
= = = =
Radiation Correction
−0.0234375+0.01171875 n = 7.95 – 0.100 = 7.850 min 2 Radiation Correction = 0.1179258 8.3750000−8.2968750 v'= =0.0195313 4 Rise in Temperature during Test = change in 0.8281250−0.8359375 temperature at maximum – change in v= =0.0019531 4 temperature at ignition −v + v ' Rise in Temperature during Test = 5.9140625 – RadiationCorrection=n∗v ' + 0.07265625 = 5.8414063 OC 2 ¿ 10.563∗0.01171875 +
TRIAL 3: Assumption: The end of post period is at 25-minute mark Ignition: 0.1-minute mark (6 seconds after running)
Radiation Correction
Slope Calculation: TM = 36.3906250 OC tM = 7.950 min TE = 35.9531250 OC tE = 25.000 min
Radiation Correction = 0.16210981
¿ 7.850∗0.0195313+
−0.0019531+0.0195313 2
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 OC
Where:
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0.828125 8.390625 8.375000 8.296875
Determination of Heat of Combustion of Liquid Fuels Using Bomb Calorimeter
¿ experimental−¿ theoretical l X 100 ¿ theoretical
Calorific Value Calculations for Biodiesel:
%error=l
Average Radiation Correction of Biodiesel
True Calorific value of KEROSENE:8786.80688 cal/g
0 . 1991406+0 . 1179258+0 . 16210981 ¿ 3
= 0.1597254
%error=l
7670.117354−8786.80688 l X 100 = 8786.80688
12.7081 %
Average Rise in Temperature
¿
7 . 71875+5 . 8414063+7 .5625 3
= 7.04088 OC Average Weight of Diesel Fuel
¿
2 . 85+2 .81+2 . 82 3
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:
= 2.827 g Corrected Rise in Temperature
¿ RadiationCorrection + Rise∈Temperature ¿ 0 .1597254 +7 . 04088
= 7.2006054 OC Heat Absorbed by Water
Sample
Calorific Value (cal/g)
True Value (cal/g)
Percentage Error (%)
Diesel
10,874.76
35.1384
Kerosene
7,053.54529 1 9201.159175
8365.2008
10.0000
Biodiesel
7670.117354
8786.8068 8
12.7081
¿ 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
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.
% Error of DIESEL (TRIAL 1) %error=l
¿ experimental−¿ theoretical l X 100 ¿ theoretical
True Calorific value of Diesel:10874.76 cal/g %error=l
7053.545291−10874.76 l X 100 = 10874.76
35.1384 % % Error of KEROSENE (TRIAL 1) %error=l
¿ experimental−¿ theoretical l X 100 ¿ theoretical
X. HAZARDS AND COUNTERMEASURES 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.
True Calorific value of KEROSENE:8365.2008 cal/g %error=l
9201.159175−8365.2008 l X 100 = 8365.2008
10.0000 % % Error of BIODIESEL (TRIAL 1)
XI.
WASTE DISPOSAL
Properly segregate or provide a secured bin for the rags, cloths, and tissues used to wipe the crucible and the liquid fuel spills.
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Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)
XII. APPENDIX
Oxygen gas tank for pressure filling
Bomb Calorimeter Set Up
Post-laboratory group picture
DataStudio displaying the measured temperature inside the calorimeter
Post-laboratory group picture
From left to right: diesel, kerosene, and biodiesel
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Determination of Heat of Combustion of Liquid Fuels Using Bomb Calorimeter XIII. AUTHORS
Krishna May Ogot Currently resides in Taguig City. She took up Chemical Engineering Technology in Technological University of the PhilippinesTaguig 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 PhilippinesTaguig 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 TUPTaguig 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.
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 2 nd 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, extracurricular 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.
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
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Technological Institute of the Philippines - Manila (Chemical Engineering Calculations II, 2nd Semester, 2015-2016)
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