Instructions: Use black or blue ink. You must show your work. Due by the start o

Question

Instructions: Use black or blue ink. You must show your work. Due by the start of the MONDAY lecture. Exercise 1. The heat of combustion of a fuel is commonly referred to as the heating value or the amount of heat released when a unit amount of the fuel burns completely. The heatin unique characteristic of each specific fuel. g value isa The heating value may be given as either the higher heating value (HHV) or the lower heating value (LHV). The HHV is the amount of heat released when a unit amount of fuel at a given temperature (usually 25°C) is completely oxidized, at and constant pressure, and the products cooled to their initial temperature. As a result HHV includes the energy released (AHe or just H,) when the water vapor, from all s produced during combustion condenses into liquid. stoichiometric conditions The LHV has a similar definition, except that all the water remains as vapor, does not include H so LHV a. Coal-water-slurry (CWS) is a liquid consisting of water and finely pulverized coal. It is often used instead of solid coal as a fuel because liquids are much easier to work with than solids, and because it is fire-proof and explosion-proof (CWS ignition temperatures are usually about 800 C). Estimate the mass percent of water in a CWS which has a HHV of 25.58 MJ/kg and a LIHV of 24.429 MJ/kg. Assume coal is pure carbon. What is the percent composition of a hydrocarbon fuel if it has a HHV of 35.62 MJ/kg and a LHV of 31.33MJ/kg? What is the empirical formula of the hydrocarbon? c. What is the HHV and LHV for the combustion of one kilogram of ethanol (CH,CH OH) and one kilogram of gasoline? Assume gasoline is 100% octane (C8H48) which has a standard enthalpy of formation of-249.95 kJ/mol . Rank the fuels (CWS, hydrocarbon, ethanol, and gasoline) from the worst to the best motor vehicle fuel. Briefly explain your reasoning

Explanation / Answer

a)

known statistics,

HHV = 25.58MJ / Kg

LHV = 24.429MJ / Kg

Hv = 2.257MJ / Kg

the disparity in force of HHV and LHV = 25.58-24.429 = 1.151MJ

So the sum of hose = heat evolved / disparity in heat of low and high rich price

quantity of hose in Kg = 1.151 / 2.257 = 0.51 Kg

b)

the disparity in HHV as well as LHV = 35.62-31.33 MJ /Kg = 4.29MJ / Kg

the sum of water fashioned = 4.29 / 2.257 = 1.9 Kg

9grams of hose down is bent from 1 gram of hydrogen

so 3.72Kg have got to have obtain on or after = 1.9 / 9 Kg = 0.21Kg of hydrogen

mole of hydrogen = 210 / 1 = 210 moles

the gathering of carbon here = 1-0.21 = 0.79Kg

mole of carbon = Mass / Molecular burden =790/12 = 65.83

Let us separate the moles of hydrogen with mole of carbon = 658.3 / 210 = 3.13

So the empricial procedure = CH3

c)

warmth of combustion of ethanol = HHV

C2H5OH + 3O2 - --->> 2CO2 + 3H2O

Enthalpy of burning of ethanol = 2X enthalpy of configuration of CO2 + 3X ethanlpy of arrangement of hose - [ethanlpy of arrangement of ethanol]

Enthalpy of combusiton = [2 X (-393.5) + 3X(-285.3)] - [-277.7] = -1365.2 KJ / mole

for 46gram = -1365.2 KJ

so for 1Kg = 29.678MJ / Kg = HHV

likewise intended for octanol

C8H8(l) + 12.5O2 --> 8CO2(g) + 4H4O(l)

DeltaH of reaction = -5471 kJ x mol-1

for 114grams = 5471KJ

so for 1Kg = 47.991MJ / Kg

d)

the HHV for octane is uppermost so this is top oil

The organize willpower be

octane> hydrocarbon > Ethanol>CWS

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