Experiment 3. Determination of Heating Value of Liquid Fuels

I. Objectives1. To determine the heating value of a liquid fuel specifically butanol using the

bomb calorimeter2. To compare the experimental net calorific value of butanol to its theoretical

value

II.Theory

The energy from the sun is captured and utilized though the process of photosynthesis which occurs in plants, algae and some types of bacteria. Photosynthesis converts the light energy into chemical energy and stores it in bonds of sugar. The light energy is used to convert carbon dioxide and water into simple sugar glucose (Encyclopædia Britannica, 2012). The chemical energy is then stored within the plant and can be transferred through the different levels of consumers in the food pyramid. On the other hand, the energy within the dead plants that are not consumed by organisms is released in the form of light and heat during combustion. Thus, remains of living organisms are great source of energy and these are used as fuels. Natural gas, wood, charcoal and fuel oil are some of the fuels used in households and industrial plants to produce heat.

Fuel is any substance that releases significant amounts of heat or energy when it is burned. It is a substance that reacts with oxygen to form carbon dioxide and water vapor, releasing large amounts of heat (Litwinienko, n.d.). Matinez (2015) stated that fuels can be classified according to the period of natural renovation and physical state. On the basis of period of natural renovation, fuels can be classified as fossil fuels or renewable fuels. Fossil fuels such as coal, crude oil and natural gas were formed millions of years ago by high pressure decomposition of the remains of living organisms. This kind of fuel is considered non-renewable and continually depletes due to commercial consumption. On the other hand, renewable fuels, also known as biomass, are formed in a few years basis. It is more dispersed, has less energy content, has more moisture and ash content and requires more handling effort compared to fossil fuels.

Fuels can also be classified according to physical state: solid, liquid and gas. Coal from mineral, charcoal from the combustion of wood and biomass from dung and wood are some examples of solid fuels. Solid fuels are easy to transport, convenient to store without any risk of contamination, and has low production cost. The examples of liquid fuels are crude oil derivatives (such as gasoline, diesel and fuel oil), alcohols and LPG (Liquified Petroleum Gas). Lastly, examples of gaseous fuels are natural gas, acetylene and biogas. Liquid and gaseous fuels are preferably used in industrial plants compared to solid fuels because these types of fuels are easier to handle and can be transported easily through pipelines (Preserve Articles, 2012). Liquid and gaseous fuels do not leave any residue after burning. Most importantly, these types of fuel have higher calorific values than the solid fuels. This means that for a given mass of liquid or gaseous fuel, it produces more heat. For this experiment, it focuses more on liquid fuels specifically butanol.

Butanol is a flammable liquid that is used as a fuel and as a solvent in industrial plants. It is a hydrocarbon with the chemical formula C4H9OH. It can be produced through petrochemical processes or through fermentation of sugars derived from agricultural crops such as corn. Most internal engines can burn butanol without experiencing problems. Due to this, studies were conducted on the use of butanol as a fuel additive and as an alternative fuel. According to Renewable

1

Energy World (2006), the interest in butanol as an alternative fuel emerged from the fact that it has more advantages over ethanol. Fuels must be vaporized before it can be burned in an engine. Butanol can be vaporized more easily at low temperatures than ethanol. It has also high energy content of 110000 Btu per gallon compared to the 84000 Btu per gallon for ethanol. On the other hand, gasoline has an energy content of 115000 Btu per gallon. Butanol is also less evaporative than ethanol and gasoline, making it safer to use as an oxygenate. These facts about butanol make it a good candidate as alternative fuel.

Since fuels are combustible substances capable of producing heat in an exothermic reaction, the amount of heat released from its combustion can be measured. The calorific value or heating value is the amount of heat obtained by the combustion of unit mass or volume of fuel with oxygen at a constant volume process. It is expressed as a unit of energy per unit mass or volume of the substance such as kcal/kg, kJ/kg, J/mol and Btu/m³ (Spectro Analytical Labs, 2015). There are two types of calorific value: the Gross Calorific Value and the Net Calorific Value.

The Gross Calorific Value or Higher Heating Value is the amount of heat produced by the complete combustion of a unit quantity of fuel. Hence, it is referred as the heat of combustion. The Gross Calorific Value is obtained when all products of combustion are cooled down to the pre-combustion temperature and when the water vapor produced from combustion process is entirely condensed (The Engineering Toolbox, n.d.).

However, most of the water vapor produced from combustion processes are entrained with other flue gases or remains as steam. This prevents the recovery of heat of the water vapor. Therefore, the Net Calorific Value or Lower Heating Value is measured. It is calculated by subtracting the latent heat of vaporization of water vapor from the Gross Calorific Value (The Engineering Toolbox, n.d.). The latent heat of vaporization is defined as the energy required to change the state of water from liquid to vapor at constant temperature (Perry and Green, 2008). Equation 1 shows the calculation of Net Calorific Value where W is the ratio of weight of water in the combustion products and weight of fuel burned. The latent heat of vaporization at the partial pressure of the vapor in the gas at standard temperature is represented by K (Perry and Green, 2008).

Net ∆ HC=Gross ∆H c−K·W (1)

Measurement of heating values of fuels is done using calorimeters. There are two types of calorimeters: the steady-flow calorimeter to measure the heating value of gaseous fuels and the constant-volume bomb calorimeter for liquid and solid fuels (Louisiana Tech University, 1999). Figure 1 shows a cross-section of a bomb calorimeter. The fuel sample is placed in the ignition cup and the bomb is pressurized by filling it with oxygen. The sample is combusted by passing current through the fuse wire which is in contact with the sample. When the combustion process is initiated, the temperature of the bomb and its surroundings increases and the change in temperature is measured.

The bomb calorimeter has an insulating jacket which makes it an adiabatic bomb calorimeter. Since the bomb calorimeter also operates at constant volume,

2

Figure 1. Cross-section of a bomb calorimeter

Figure 2. Example of graph of temperature as a function of time

the change in work and change in heat during combustion is both considered equal to zero. As shown in Equation 3, the change in internal energy will then be equal to zero. Equation 2 shows the mathematical expression of the Fist Law of Thermodynamics which states that the change in the internal energy of the system U is equal to the amount of heat Q transferred between the system and surroundings and work W done on or by the system (Smith et al., 2011).

∆U = ∆Q + ∆W (2)

∆U = 0 (3)

The insulating jacket isolates the bomb calorimeter from the surrounding environment. Therefore, the reactants inside the bomb (sample and oxygen) are considered as the system while the rest of the calorimeter are the surroundings (Roos et al.,2010). The change in the total internal energy ∆U total of the calorimeter is just the summation of change in internal energy of the system ∆U system and the change in internal energy of the surroundings ∆U surroundings as shown by Equation 4. It was previously stated that ∆U total is equal to zero hence Equation 4 can be simplified to Equation 5.

∆U total = ∆U system + ∆U surroundings

(4)∆U system = -∆U surroundings

(5)The insulating jacket surrounds the water bucket and the bomb cylinder. As

the temperature of the water bucket and the bomb cylinder increases due to combustion, the temperature of the insulating jacket is controlled such that it equals that of the bucket and the bomb cylinder. Moreover, the increase in temperature of the system due to combustion is related to the heat of combustion of the sample by a constant of proportionality (Lamp et al., 2013). The heat capacity of a bomb calorimeter is evaluated using Equation 6. For this experiment, the Equation 7 is used. If the heat capacity Cv of the bomb calorimeter is already determined by the previous experiments, the change in the internal energy of the calorimeter can be calculated using Equation 6.

C v=−[ (∆U sample ) (msample )+(∆U fuse ) (mfuse ) ]

∆T

(6)

C v=−[ (∆U butanol) (mbutanol combusted )+ (∆U fuse) (mburnt fuse )+(∆U gelatin capsule) (mburnt gelatin capsule ) ]

∆T

(7) ∆U = -Cv ∆T

(8)However, the bomb calorimeter is not entirely adiabatic because in reality, a

small amount of heat escapes the insulating jacket. Therefore, the change in heat is not exactly equal to zero. The stirrer also does work on the calorimeter. Nonadiabaticity of the bomb calorimeter is corrected for with an empirical radiative correction RC.

Figure 2 shows a graph of temperature as a function of time. The time that the bomb is fired is the time that makes the shaded areas of Figure 2 to be equal. It is estimated to be 7

3

minutes for a Parr calorimeter. Therefore, the temperature at 6 minutes must be extrapolated forward 1 minute by the pre-firing slope, and the temperature at 12 minutes must be extrapolated backward 5 minutes by the post-firing slope (Parr Instrument Company, 2007).The following equations show the calculation of the change of temperature accounting for the radiative correction. The subscripted T in the equations corresponds to the temperature of the water at a particular time.

RC = (5 ) (T18−T 12 )+(T 6−T 0)

6

(9)∆T = T 12 - T 6 – RC

(10)It should be further noted that bomb calorimeters can only measure the

change in internal energy rather than the change in enthalpy because it operates at constant volume. Nonetheless, the relationship of internal energy and enthalpy can be determined as shown by Equations 11 and 12. When the pressure inside the bomb calorimeter remains constant, the change in enthalpy will be equal to change in internal energy. These two values differ by an amount equal to the difference between the number of moles of the combustion products and the number of moles of the reactants. Assuming that the gaseous components inside the bomb behave according to the ideal gas law, Equation 12 can be simplified to Equation 13 which is the same equation used for calculating the Gross Calorific Value of substances.

H = U + PV (11)

∆H = ∆U + ∆(PV) (12)

∆ H=∆U+RT ∆ngas (13)

The calculation for the percentage of excess air and volumes of cold water and ambient water to be used are usually encountered in experiments which determines the heating values of fuels using bomb calorimeter. Equation 14 shows the calculation of the percentage of excess air. Combustion chambers are usually supplied with excess air to ensure complete combustion. The presence of excess air in a combustion chamber increases the amount of oxygen and nitrogen entering the chamber and increases the probability that the oxygen will react with the fuel. It also increases turbulence which means increased mixing of air and fuel (Biarnes et al, n.d.). As more excess air enter a combustion chamber, more amount of the fuel is burned until it finally reaches complete combustion.

% Excess air = (Molesof air supplied )−(Molesof theoretical air)

(Molesof theoretical air) x 100

(14)On other hand, Equation 15 shows the energy balance for water used in

calculating the amount of cold water and ambient water to be used in the experiment. The volumes of cold water and ambient water are calculated and mixed in order to make sure that the temperature of the water in the bucket is at 25OC.

mcoldCpcold(25oC−T cold) = - mambientCpambient (25

oC−T ambient) (15)

4

III.MethodologyBomb Calorimeter Platinum fuse TowelAnalytical Balance Butanol GlovesGelatin capsule Water BasinThermometer Ice MaskDigital thermometer Graduated cylinder (1000mL) RulerOxygen tank Stirring rod Watch glass

A. Preparation of the sample1. The gelatin capsule was inspected first to ensure that it was not damaged. It

was also weighed and handled with the use of gloves. 2. An amount of 4mL of butanol was prepared in a small beaker and it was

covered at all times with an aluminium foil.3. The gelatin capsule was tested if it can accommodate 0.6mL of liquid

sample. The two halves of the capsule were closed tightly against each other. Using a syringe, the capsule was then tested by injecting 0.6mL of tap water. The volume of the fuel in the experiment was lessened if it was necessary.

4. The tested gelatin capsule was thrown away and another capsule was used for the experiment.

B. Filling of the calorimeter with water 1. Cold water was prepared using ice.2. The calorimeter used 3000mL of water at 25OC. Using energy balance; the

volumes of cold water and ambient water were calculated to make sure that the water temperature is 25OC.

3. The calculated volumes of cold water and ambient water were mixed to obtain a 3000mL of water at 25OC.

C. Preparing the platinum fuse and securing the sample in the bomb head1. A 10cm length of platinum fuse was weighed and cut. The bomb head was

set on the support stand and the gelatin capsule was placed on the steel capsule vertically with the larger half of the capsule at the bottom.

2. The vertical position of the capsule was secured using the fuse as shown in Figure 3. It was made sure that the two halves of the capsule were tightly closed against each other.

3. With the gelatin capsule secured, 0.6mL of butanol was drawn from the small beaker using the syringe and it was carefully injected at the top half of the capsule as shown in Figure 4. The capsule was then inspected for any leaks.

5

6

Figure 3. Securing the gelatin capsule with fuse

Figure 4. Injecting butanol at the top half of the capsule

Figure 5. Bomb head inserted to the bomb cylinder

Figure 6. Ignition wires connected to terminal sockets of the bomb head

D. Closing the bomb1. The bomb head was moved from the support stand to the bomb cylinder. It

was carefully inserted to the bomb cylinder as shown in Figure 5. 2. The screw cap was then firmly set on the cylinder and the bomb cylinder

was closed. The gas release valve was closed while the oxygen inlet valve was opened.

E. Filling the bomb with oxygen1. Assistance from the laboratory technician was needed in filling the bomb

with oxygen. The hose from the oxygen tank was securely attached to the bomb. The oxygen control valve of the tank was slowly opened to supply oxygen to the bomb and the bomb pressure was observed.

2. After the desired pressure was reached, the oxygen control valve was closed and the hose was removed.

F. Pre-firing1. The oxygen-filled bomb was then placed in the center of the bucket of water.

It was made sure that there was no continuous flow of gas bubbles after the bomb was submerged in the water.

2. The two ignition wires were then attached to the terminal sockets of the bomb head as shown in Figure 6 and the cover of the calorimeter was closed.

3. The digital thermometer was inserted in the calorimeter cover and made sure that it touched the water. It was also made sure that there was enough space for the stirrer.

4. The stirrer was then turned on for six minutes while the temperature for every 30 seconds was recorded.

G. Firing and recovery of the combustion products1. After 6 minutes, the firing button was pressed for 1-2 seconds on the ignition

unit and the temperature was recorded for every 30 seconds up to 12 minutes after firing.

2. The stirrer was turned off and the calorimeter cover was opened. The bomb was lifted and it was dried using a clean towel. It was then taken outside into an open area.

7

3. Before the screw cap of the bomb cylinder was removed, the gas release valve was slowly opened to get rid of residual gas pressure and observations were taken.

4. The cap was unscrewed and the bomb head was lifted. The interior of the bomb cylinder was inspected for any signs of incomplete combustion.

5. The mass of the capsule after combustion and the mass of the unburned fuse were measured. The interior of the bomb and the combustion capsule were washed and dried using the towel.

6. The same procedure was followed for Trial 2.

IV. Results and Discussion

Table 1. Calculated values for the desired parametersParameters Trial 1 Trial 2

Mass of combusted butanol (g) 0.4770 0.4798Mass of burned fuse (g) 0.0062 0.0113Mass of soot (g) 0.0054 0.0026Radiative correction, RC (OC) 0 0.0167Change in temperature accounting RC (OC)

1.31.2833

Heat of combustion of fuse (cal/mol) -12576.44 -12573.93Heat of combustion of butanol (cal/g) -8803.98 -8628.18Percent of excess air (%) 234.48% 231.86%

Table 2. Heating value of butanolHeating Value Trial 1 Trial 2

Gross ΔHC (cal/g) -8819.78 -8644.09Net ΔHc (cal/g) -8120.76 -7925.05Average Gross ΔHc (cal/g) -8731.94Average Net ΔHc (cal/g) -8022.90Theoretical Net ΔHc (cal/g) -7912.20Percent difference % 1.54%

Table 3. Temperature of water while combusting butanol

Time (min)Butanol

Trial 1 Trial 2Pre-firing T 0 25.0 25.2

8

Figure 7. Graph of temperature as a function of time for the combustion of butanol

T 1 24.9 25.3T 2 24.9 25.3T 3 24.9 25.3T 4 24.9 25.3T 5 24.9 25.3T 6 25.0 25.3

Post - firing

T 7 25.9 26.2T 8 26.1 26.5T 9 26.2 26.5T 10 26.3 26.6T 11 26.3 26.6T 12 26.3 26.6T 13 26.3 26.6T 14 26.3 26.6T 15 26.3 26.6T 16 26.3 26.6T 17 26.3 26.6T 18 26.3 26.6

T h e s a m p l e u s e d i n t h e e x p e r i m e n t i s t h e l i q u i d f u e l b u t a n o l . P r i o r t o t h e experiment, the Materials Safety Data Sheet (MSDS) of butanol was checked in order to avoid accidents while conducting the experiment. According to the MSDS, the Personal Protective Equipment (PPE) that should be used in the experiment are splash goggles, gloves, protective apron and vapor respirator. The group used gloves, mask, safety goggles and laboratory gown as PPE. Proper attire such as pants and closed shoes were also worn during the experiment. Only 4mL of butanol was prepared in a small beaker and it was made sure that it was covered with aluminium foil at all times because butanol is highly flammable.

9

0 2 4 6 8 10 12 14 16 18 2024

24.5

25

25.5

26

26.5

27

Trial 1Trial 2

Time(min)

Tem

pera

ture

(°C

)

This experiment aims to determine the heating value of butanol, namely Gross Calorific Value and Net Calorific Value, using the bomb calorimeter. It also aims to compare the experimental Net Calorific Value of butanol to its theoretical value. Two trials were conducted in determining the heating value of butanol. Table 1 shows the calculated values of the desired parameters in the experiment. On the other hand, the calculated values for Gross Calorific Value and Net Calorific Value are shown in Table 2.

The Gross Calorific Value (Gross ΔHC) of butanol is equal to the change in enthalpy of combustion of butanol Hbutanol because it corresponds to the total amount of heat released during the combustion of the sample. The calculation for the Gross ΔHC of butanol uses the values of the change in internal energy of butanol Ubutanol and the change of molar amount of gases during its combustion ngas as shown by Equation 13. The values of these parameters should be determined in order to calculate the Gross ΔHC of butanol.

For the calculation of the Ubutanol, Equation 7 was rearranged and used. From the equation, the only known values are the heat capacity of the bomb calorimeter Cv and change in internal energy of the gelatin capsule Ugelatin capsule

which are 3665.2136 cal/K and -4600 cal/g, respectively. The values for the mass of burned fuse, mass of burned gelatin capsule, mass of combusted butanol, change in temperature and change in internal energy of fuse were determined using the data gathered in the experiment. As shown in Table 1, the mass of the burned fuse and the mass of the burned gelatin capsule for Trial 1 are 0.0062g and 0.1228g, respectively while for Trial 2 are 0.0113g and 0.1224g, respectively. The mass of butanol used in the experiment was determined by multiplying the 0.6mL of butanol injected to the gelatin capsule and the density of butanol at 25OC which is 0.804036 g/mL. The mass of butanol used for each trial was determined to be 0.4824g. The mass of the combusted butanol was then calculated by subtracting the mass of the sample butanol and the mass of soot. As shown in Table 1, the mass of combusted butanol for Trials 1 and 2 are 0.4770g and 0.4798g, respectively.

The change in internal energy of the platinum fuse ∆ Ufuse was determined using the equation below where T2 is the melting point, T1 is the stabilized temperature and Cp is the heat capacity of platinum. This equation was integrated and calculated as shown in the sample calculations to determine the change in internal energy of fuse. The calculated values of ∆ Ufuse for Trials 1 and 2 are -12576.436 cal/mol and -12573.92961 cal/mol, respectively.

∆ Ufuse = ∫T 1

T 2

C pdT

The changes in the temperature of water during the combustion of butanol for Trials 1 and 2 are shown in Table 3. On the other hand, the graphs of temperature as a function of time for the two trials are shown in Figure 7. The graph for Trial 1 is represented by color blue while the graph for Trial 2 is represented by the color red. It can be observed from Figure 7 that the recorded temperatures for both trials during the pre-firing stage were constant. Trial 1 obtained a constant temperature of 24.9OC from 1 to 5 minutes of stirring and the temperature slightly increased to 25OC after 5 minutes of stirring. The slight increase of temperature at the last minute before firing must be due to the work done by the stirrer on the system. For Trial 2, a constant temperature of 25.3OC was obtained for the pre-firing period. The temperature for both trials escalated as

10

the combustion process was initiated 6 minutes of stirring. For Trial 1, the temperature gradually increases from 6 to 10 minutes and becomes constant at 26.3OC after 4 minutes from firing. On the other hand, the temperature in Trial 2 gradually increases between 6 and 9.30 minutes and becomes constant at 26.6OC after 3 minutes from firing. The increase in temperature after firing suggests that the reaction inside the bomb calorimeter is exothermic. Therefore, the temperature of water increases as the reaction releases heat to the surrounding water.

However, the change in temperature of water is not directly used for the calculation of Ubutanol. As previously stated, the bomb calorimeter is not entirely adiabatic because in reality, a small amount of heat escapes from the insulating jacket. This nonadiabaticity is corrected by the radiative correction. Radiative correction was determined using Equation 9 while Equation 10 was used to determine the temperature accounting radiative correction. The change in temperature accounting radiative correction was then determined to be 1.3OC for Trial 1 and 1.2833OC for Trial 2.

With the calculated values of mass of burned fuse, mass of burned gelatin capsule, mass of combusted butanol, change in temperature and change in internal energy of fuse; the Ubutanol can be determined. Substituting the known values to Equation 7, the Ubutanol for Trial 1 is -8803.98/g and for Trial 2 is –8628.18cal/g. Both values of Ubutanol are negative since the reaction inside the bomb is exothermic.

For the change of molar amount of gases ngas, the combustion reaction of butanol shown below was utilized. The ngas was determined by subtracting the moles of gases in the combustion products and the moles of gases in the reactants. The ngas was calculated to be -0.01287 moles for Trial 1 and -0.01295 moles for Trial 2.

C4H 9OH (l )+6O2(g)→4CO2(g )+5H 2O(l )

With the calculated values of the Ubutanol and ngas during combustion, the enthalpy of combustion of butanol Hbutanol for both trials can now be calculated. Substituting the known values of Ubutanol and ngas to Equation 11, the Hbutanol for Trials 1 and 2 were determined to be -8819.78cal/g and -8644.093917cal/g, respectively. These values are also the Gross ΔHC of butanol for Trials 1 and 2. The average Gross ΔHC from both trials is -8731.94cal/g. This is the amount of heat produced by the complete combustion of 0.6mL of butanol.

The Net Calorific Value (Net ΔHC) of butanol was calculated by subtracting the latent heat of vaporization of water ∆ H v from the Gross ΔHC as shown by Equation 1. The Gross ΔHC for Trials 1 and 2 were previously calculated to be -8819.78cal/g and -8644.093917cal/g, respectively. The latent heat of vaporization of water was calculated using the equation below where the constants in the equation were obtained from Table 2-150 of the Perry’s Chemical Engineering Handbook 8th edition. The calculated ∆ H v for Trials 1 and 2 are 581.65cal/g and 581.46cal/g, respectively. Substituting the known values of Gross ΔHC and ∆ H v for each trial to Equation 1, the Net ΔHC for Trials 1 and 2 are -8120.69cal/g and -7941.14cal.g, respectively. These values were then averaged to obtain -8030.920cal/g.

∆ H v=C1(1−Tr)C 2+C 3T r+C 4Tr

2+C 5T r3

11

The average Net ΔHC was then compared to the theoretical Net ΔHC by determining the percent difference of these two values. The theoretical Net ΔHC of butanol was obtained from Table 2-179 of Perry’s Chemical Engineering Handbook 8th edition and was determined to be -7909.50cal/g. The calculated percent difference between the experimental and theoretical Net ΔHC of butanol is 1.54%. This small percent difference suggests that that there is no significant difference between the experimental and theoretical Net ΔHC of butanol. This also suggests that the Net ΔHC calculated using the data gathered in the experiment is accurate and acceptable.

Excess air is usually present in a combustion process to ensure that the reaction inside the bomb calorimeter will proceed to complete combustion. For this experiment, the percentage of excess air was calculated. Table 1 shows a 234.48% excess air for Trial 1 and 231.86% excess air for Trial 2. The high percentage of excess air suggests that a large amount of oxygen and nitrogen were supplied in order to achieve a complete combustion. This means that a large amount of oxygen reacted with naphthalene until it finally reaches a complete combustion. However the presence of soot suggests that the combustion was incomplete. There was also evolution of gas when the bomb cylinder was opened after combustion. The common gases that can evolve from a combustion process are carbon monoxide, carbon dioxide and hydrogen gas. It can be confirmed that carbon monoxide is one of the residual gases since the combustion process inside the bomb was incomplete. It was also observed that moisture was present in the bomb cylinder. This is because water is one of the products of combustion.

Recommendations

It should be made sure that the gelatin capsule is securely attached by the fuse to the electrodes. It should be noted that in securing the gelatin capsule with the fuse, the longer half of the capsule should be on bottom while the shorter half is on top. If the positions of the two parts of the capsule are interchange, there will be leakage of the butanol. This should be avoided.

V. Conclusion

The heating value of butanol was determined using the bomb calorimeter. The Gross Calorific Value of butanol for Trials 1 and 2 were determined to be -8819.7cal/g and -8644.09cal/g, respectively. The average Gross Calorific value for the two trials was calculated to be -8731.94cal/g. For the Net Calorific Value of butanol, -8120.70 was determined for Trial 1 and -7941.14cal/g for Trial 2. The average Net Calorific Value of butanol was then determined to be -803.92cal/g. It was compared to the theoretical value of – 7909.45cal/g by obtaining the percent difference. The 1.45% difference between the experimental and theoretical Net Calorific Values of butanol suggests that the results of the experiment are accurate and acceptable.

References

Biarnes, M., Freed, B. & Esteves, J. (n.d.). Combustion. E Instruments International LLC. Retrieved from http://www.e-inst.com/docs/Combustion-Booklet-2013.pdf

12

Bright Hub Engineering. (2012). Burning Coal in Power Plants – Calorific Value and Moisture. Retrieved from http://www.brighthubengineering.com/power-plants/22202-burning-coal-in-power-plants-calorific-value-and-moisture/

Cross-section of a bomb calorimeter[Image] (n.d.) Retrieved from October 4, 2015, from http://www.dartmouth.edu/~pchem/75/Bomb.html

Encylopedia Britannica. (2015). Combustion. Retrieved from http://www.britannica.com/science/combustion

Encylopedia Britannica. (2015). Photosynthesis. Retrieved from http://www.britannica.com/science/photosynthesis

Lamp, B. D., Humphry, T., Pultz, V. M., & McCornick, J. M. (2013). Enthalpies of Solution. Retrieved from http://chemlab2.truman.edu/files/2015/07/Enthalpies-of-Solution.pdf

Litwinienko, G. (n.d.). Determination of Heat of Combustion of Biofuels. Retrieved from http://www.chem.uw.edu.pl/people/AMyslinski/nowy/zarzadzanie/31ME_th_inst.pdf

Louisiana Tech University. (2015). Heating Value of a Fuel. Retrieved from

http://www2.latech.edu/~hhegab/pages/me354/Lab2/Lab2_bombcal_Sp97.htm

Martinez, I. (2015). Fuels. Retrieved from http://webserver.dmt.upm.es/~isidoro/bk3/c15/Fuels.pdf

Preserve Articles. (2012). What are the advantages and disadvantages of Liquid and Gaseous Fuels over Solid Fuels? Retrieved from http://www.preservearticles.com/201012302061/avantage-and-disadvantage-of-liquid-and-gaseous-fuels-over-solid-fuels.html

Renewable Energy World. (2006). Why is Butanol More Prevalent? Retrieved from http://www.renewableenergyworld.com/articles/2006/09/why-isnt-butanol-more-prevalent-45946.html

Roos, M., Kucerova, G., & Brimaud, S. (2010). Experiement 9 – Calorimetry. Ulm University. Retrieved from http://www.uni-ulm.de/~hhoster/pc_lecture/Calorimetry_1.pdf

Smith, J. M., Van Ness, H., & Abbott, M. (2011). Introduction to Chemical Engineering Thermodynamics (8th ed). New York: McGraw-Hill Book Company

Spectro Analytical Labs. 2015. Coal Testing Definition. Retrieved from http://www.spectro.in/Coal-definition.html

The Engineering Toolbox. (n.d.). Heating Value. Retrieved from http://www.engineeringtoolbox.com/gross-net-heating-value-d_824.html

13

AppendicesRaw Data

Table 4. Data Gathered for ButanolParameters Trial 1 Trial 2

Mass of gelatin capsule (g) 0.1228 0.1224Volume of butanol injected (mL)

0.6 0.6

Mass of fuse (g) 0.0161 0.0164Mass of steel capsule (g) 12.6696 12.8046Mass of capsule and soot (g) 12.6750 12.8072Mass of unburned fuse (g) 0.0099 0.0051Pressure (psig) 120 120Volume of the bomb (mL) 0.345 0.345

Table 5. Temperature of Water for Combusting ButanolTime (min)

Temperature (OC)

Trial 1 Trial 2

Pre-Firing

T0 0.00 25.0 25.20.30 24.9 25.3

T1 1.00 24.9 25.31.30 24.9 25.3

T2 2.00 24.9 25.32.30 24.9 25.3

T3 3.00 24.9 25.33.30 24.9 25.3

T4 4.00 24.9 25.3

14

4.30 24.9 25.3T5 5.00 24.9 25.3

5.30 25.0 25.3T6 6.00 25.0 25.3

Post-Firing

6.30 25.5 25.8T7 7.00 25.9 26.2

7.30 26.0 26.4T8 8.00 26.1 26.5

8.30 26.2 26.5T9 9.00 26.2 26.5

9.30 26.2 26.6T10 10.00 26.3 26.6

10.30 26.3 26.6T11 11.00 26.3 26.6

11.30 26.3 26.6T12 12.00 26.3 26.6

12.30 26.3 26.6T13 13.00 26.3 26.6

13.30 26.3 26.6T14 14.00 26.3 26.6

14.30 26.3 26.6T15 15.00 26.3 26.6

15.30 26.3 26.6T16 16.00 26.3 26.6

16.30 26.3 26.6T17 17.00 26.3 26.6

17.30 26.3 26.6T18 18.00 26.3 26.6

15

Sample Calculations Calculations for the water used in the experiment

For Trial 1Using energy balance

(mCp∆T )cold + (mCp∆T )ambient = 0(mCp∆T )cold = - (mCp∆T )ambientmcoldCpcold(25

oC−T cold) = - mambientCpambient (25oC−T ambient)

Since mass = (densityρ) (volume V), therefore ρcoldCpcoldV cold(25

oC−T cold) = −ρambient CpambientV ambient (25oC−T ambient)

Assumptions: a. heat capacities of cold and ambient water have a small difference b. densities of cold and ambient water have a small difference

V cold (25oC−T cold ) = −V ambient (25oC−T ambient )

The measured temperatures are Tcold = 9OC, Tambient=29OC V cold (25oC−T cold ) = −V ambient (25oC−T ambient )

Vcold = −3000mL(25oC−29oC )

(9oC−29oC )

Vcold = 600 mLV ambient = 3000mL-600 mLV ambient = 2400mL

Change in Internal Energy of fuse ∆U fuse

Using the following equation for Trial 1

∆ Ufuse = ∫T 1

T 2

C pdT

where: T2 = Melting Point temperature = 17550C +273.15 = 2028.15 K

T1 = Stabilized Temperature = 24.9 0C + 273.15 = 298.05 K Heat Capacity of Platinum = 5.92 + 00.116T Source: Perry’s Chemical EngineersHandbook, 8th edition (Table 2-

151)

∆ Ufuse = -∫T 1

T 2

(5.92+0.00116T )dT

∆ Ufuse = -[5.92T + 0.00116T2

2 ] ⃒T1T2

∆ Ufuse = -5.92 (T 2−T1) - 0.00116(T22−T 12)2

∆ Ufuse = -5.92 (2028.15K−298.05K) - 0.00116 (2028.15K2−298.05K2)

2

16

∆ Ufuse =-12576.436 calmol

Solving for radiative correction, R CFor the combustion of butanol in Trial 1

RC = (5 ) (T18−T 12 )+(T 6−T 0)

6

= (5 ) (26.3OC−26.3OC )+(25.0OC−25.0OC)

6 = 0

Solving for ∆T For the combustion of butanol in Trial 1

∆T = T 12 - T 6 – RC = 26.3OC – 25.0OC – 0 = 1.3OC

Density of butanol , ρFrom Table 2-32, Perry’s Chemical Engineering Handbook 8th ed

C1 = 0.98279 C3 = 563.1C2 = 0.26830 C4 = 0.25488

Solving for density of butanol

ρ = C1C2¿¿¿

= 0.982790.26830¿¿¿

= 10.8474779 mol

dm3 x ( 10dm

1m)3

x 74.122gmol

x 1kg1000g

= 804.0367571 kg/m3

= 804.036 kg/m3

Change in Internal Energy of Butanol For Trial 1Since heat capacity of bomb calorimter is Cv = 3665.213579 cal/K

C v=−[ (∆U butanol) (mbutanol combusted )+ (∆U fuse) (mburnt fuse )+(∆U gelatin capsule) (mburnt gelatin capsule ) ]

∆T

Rearranging the equation above to solve for ∆U naphthalene

∆U butanol = −C v∆T−(∆U fuse ) (mburnt fuse )−(∆U gelatin capsule ) (mburnt gelatin capsule )

mbutanolcombusted

=

−(3665.2136 calK ) (1.3 )−(−12576.436 calmol )( 1mol Pt195.08g Pt )(0.0062 g )−(−4600cal /g)(0.1228g)

0.477 g= -8803.978989 cal/g= -8803.9790 cal/g

Solving for n gas

17

From the combustion of butanol in Trial 1C4H 9OH (l )+6O2(g)→4CO2(g )+5H 2O(l )

ngas = nproduct – nreactants

= (moles of butanol)(4) – (moles of butanol)(6)

= (0.477g C10H8)( 1molC4H 9OH

74.122 g )(4) – (0.477g C10H8)( 1molC4H 9OH

74.122 g )(6)

= -0.01287067267 moles = -0.01287moles

Solving for Gross Calorific Value , Gross ΔH C For Trial 1

Hbutanol = Gross ΔHC Hbutanol = Ubutanol + RTngas

= (-8803.978989calg

) + (1.987 cal

mol K)(298.05K )(−0.01287067267moles)

0.4824 g = -8819.779857 cal/g = -8819.78 cal/g

For Trial 2Hbutanol = Gross ΔHC Hbutanol = Ubutanol + RTngas

= (-8628.178967calg

) + (1.987 cal

mol K)(298.45K )(−0.01294622379moles )

0.4824 g = -8644.093917cal/g = -8644.09cal/g

Solving for average Gross ΔHC from the two trials

Gross ΔHC ave = Gross ΔH c ¿Trial1+Gross ΔH c¿Trial 2 ¿2

= (−8819.779857cal/ g )+(−8644.093917cal /g)

2= - 8731.936887 cal/g= - 8731.94 cal/g

Latent heat of vaporization of water, ∆ H v

From Table 2-150, Perry’s Chemical engineering Handbook 8th edC1 = 5.2053x107 C2 = 0.3199C3 = -0.212 C4 = 0.25795

From Table 2-141, Perry’s Chemical Engineering Handbook 8th ed For Trial 1

TC = 647.096 K

Tr = TTc

= 298.05K647.096K

= 0.4605962639 Solving for latent heat of vaporization ∆ H v=C1(1−Tr)

C 2+C 3T r+C 4Tr2+C 5T r

3

= (5.2053x107)(1−0.4605962639)(0.3199 )+(−0.212) (0.4605962639 )+(0.25795)(0.4605962639)2

18

= 43872459.93J

kmolx1kmol1000mol

x1cal4.1858J

x1mol

18.02 gH 2O

= 581.6459731cal/g For Trial 2

TC = 647.096 K

Tr = TTc

= 298.45K647.096K

= 0.4612144102

Solving for latent heat of vaporization ∆ H v=C1(1−Tr)C 2+C 3T r+C 4T r

2+C 5T r3

= (5.2053x107)(1−0.4612144102)(0.3199 )+(−0.212) (0.4612144102 )+(0.25795)(0.4612144102)2

= 43858096.31J

kmolx1kmol1000mol

x1cal4.1858J

x1mol

18.02g H 2O

= 581.4555452cal/g Solving for Net Calorific Value, Net ΔHC

For Trial 1Net ΔHC = Gross ΔHC - K·W

Where: Gross ΔHC = Hbutanol = -8819.779857cal/g K = ∆ H v=¿ 581.6459731cal/g

W = weight of water∈the combustion products

weight of butanolburned

= (0.477 gC4H 9OH )(1molC4H 9OH

74.122g )( 5mol H 2O

1molC4H 9OH )( 18.02 g H 2O

1mol H 2O )(0.6mL )(0.804036 g

mL)

= 1.201902659 = 1.20

Solving for Net ΔHC

Net ΔHC = Gross ΔHC - K·W = (-8819.779857cal/g) – (1.201902659)(- 581.6459731cal/g)

= - 8120.698015 cal/g = - 8120.6980cal/g

For Trial 2Net ΔHC = Gross ΔHC - K·W

Where: Gross ΔHC = Hbutanol = -8628.178967 cal/g K =∆ H v=¿ 581.4555452cal/g

W = weight of water∈the combustion products

weight of butanolburned

19

=(0.4798 gC4H 9OH )( 1molC4H 9OH

74.122g )( 5mol H 2O

1molC4H 9OH )( 18.02g H 2O

1mol H 2O )(0.6mL )(0.804036 g

mL)

= 1.208957853 = 1.21

Solving for Net ΔHC

Net ΔHC = Gross ΔHC - K·W = (-8644.093917 cal/g) – (1.208957853)(- 581.4555452cal/g) = - 7941.13867 cal/g = - 7941.1387 cal/g

Solving for average Net ΔHC from the two trials

Net ΔHC ave = Net ΔH c¿Trial1+Net ΔH c¿Trial2 ¿2

= (−8120.698015 calg )+(−7941.13867cal /g)

2= - 8030.918343 cal/g= - 8030.9183 cal/g

Percent difference of experimental Net ΔH C and theoretical Net ΔHC of naphthaleneFrom Table 2-179, Perry’s Chemical Engineering Handbook 8th ed

Theoretical Net Δ HC = -2.454 x109 Jkmol

x1cal4.1858J

x1kmol1000mol

x1mol74.122g

= -7909.45cal/g

% difference = Net ΔH cave−Net ΔH ctheo

Net ΔH c theo x 100

= (−8030.918343 calg )−(−7909.498637 cal

g)

−7909.498637 calg

x 100

= 1.535112547% = 1.54%

Moles of O 2 suppliedFor Trial 1Using the Ideal Gas Law

R = 82.057atm•cm3

mol •K (Perry’s Chemical Engineering Handbook, 8th edition)

Measured volume of bomb cylinder = 345mL PV = nRT

n = PVRT

20

moles O2 =

(120+14.7 psia)( 1atm14.7 psi

)(345mL)( 1L1000mL

)

(82.057 atm• cm3

mol •K)( 1m100cm )

3

( 1000 L1m3

)(298.05K )

moles O2 supplied = 0.1292601319 molesmoles O2 supplied = 0.1293 moles

Theoretical number of moles of oxygen for the combustion of butanol For Trial 1

C4H 9OH (l )+6O2(g)→4CO2(g )+5H 2O(l )

moles O2 = 0.477 gC4H 9OH×1molC4H 9OH

74.12 gC4H 9OH×

6mol O21molC4H 9OH

= 0.0386130599 moles = 0.03861 moles

For the combustion of fuse Pt (s) + O2 (g) → PtO2

moles O2 = 0.0062g Pt ×1mol Pt195.08g Pt

×1mol oxygen1mol Pt

= 0.00003178183309moles = 3.1782 x 10-5 moles

Theoretical moles O2 = moles O2 combusted with benzoic acid + moles O2

combusted with fuse = 0.0386130599moles + 0.0386130599moles = 0.03864484173 moles = 0.03864 moles

Percent of excess air For Trial 1

% excess air =supplied moles of O2−theoreticalmolesof O2

theoreticalmolesof O2 x 100

= 0.1292601319moles−0.03864484173moles

0.03864484173moles x 100

= 234.4822393% = 234.48%

21

Hazel Jan B. Fuertes ChE124 Sec 2Groupmates: Rodgin Grace D. Regencia

Viena Mae E. VillaruzDate Performed: October 2, 2015Date Submitted: October 9, 2015

22