Combustion Example.docx

1.) A diesel plant consumes 650 liters of fuel at 26oC in 24 hours with 28oAPI. Find the fuel rate in kg/hr. SOLUTION: SGat 15.6oc = __ 141.5__ 131.5+oAPI o SGat 15.6 c = __ 141.5__ 131.5+28 SGat 15.6oc = 0.887 SGat 26oc = 0.887[1-0007(26-15.6)] SGat 26oc =0.88 Density of fuel = 0.88(1kg/li) Density of fuel =0.88kg/li W=m V V= 650/24 V=27.0833 li/hr 0.88= m/27.0833 m=23.83 kg/hr 2.) A boiler burns fuel oil with 15% excess air. The fuel oil may be represented by C 14 H30. Calculate the model air fuel ratio. SOLUTION: Fuel + Air  product combustion C14H30 +O2+3.76N2CO2+H2O+ 3.76 N2 C14H30+21.5O2+21.5(3.76)N214CO2+15H2O+21.5(3.76)N2 Theoretical A/F =21.5+21.5(3.76) 1 Theoretical A/F= 102.34 Actual A/F= 102.34(1.15) Actual A/F=117.69 mol air/mol fuel

3.) A diesel power plant uses fuel that has a density of 892.74 kg/m3 at 15.67oC. Find the heating value of fuel. SOLUTION:

SG = Density of fuel Density of Water SG =892.74/1000 SG =0.89274 o API = 141.5 -131.5 0.89274 o API =27 Q= 41,130+ 139.6 oAPI Q= 41,130+ 139.6(27) Q= 44,889.31KJ/kg

4.) A certain coal has the following ultimate analysis C= 69% H2= 2.5% O2=3.5% N2=5% S=7% moisture=8% Ash=5%.Determine the heating value of fuel used. SOLUTION: Qh = 33,820C+144,212(H-0) +9,304S 8 Qh = 33,820(0.69) = 144,212(0.025-0.035)+9,304(0.07) 8 Qh = 26,961.45 KJ/kg 5.) A diesel power plant uses fuel with heating value of 45,038.8 KJ/kg. What is the density of fuel at 30oC? SOLUTION: Qh = 41,130 =139.6 oAPI 45,038.8 = 41,130 +139.6 ( oAPI) API = 28 SG15.6oC= 141.5_ 131.5+ oAPI o SG15.6 C= 0.8872 SG30.oc= 0.8872[1-0007(30-15.6)] SG30oc=0.8782 Density of fuel = 0.8782(1 kg/li) Density of fuel = 0.8782 kg/li

6.) A diesel engine consumed 945 liters of fuel per day at 30oC. If the fuel was purchased at 15.5oC and 30oC API at P5.00/li, determine the cost of fuel to operate the engine per day. SOLUTION: SG15.6oC= 141.5_

131.5+30 = 0.87616 = 0.87616[1-0007(30-15.6)] =0.8673 = SG15.6oc SG30oc

o 15.6 C o 30. c o 30 c o 30 c o 15.6 c

SG SG SG V V

945_ = 0.87616 V15.6oc 0.8673 V15.6oc =935.44 li Cost = P5.00/li (935.44 li) Cost = P4,677.20

7.) A cylindrical tank 3m long and 2m diameter is used for oil storage. How many days can the tank supply the engine having 27oAPI with fuel consumption of 60 kg/hr. SOLUTION: V = π /4 D2h V= π /4(2)2(3) V=9.42478 m3 SG15.6oC= 141.5_ 131.5+27 o SG15.6 C= 0.8927 Density of fuel =0.8927(1000 kg/m3) Density of fuel =892.74 kg/m3 W= m/v 892.74 = 60/V V=0.0672 m3/hr Number of days =9.42478/0.0672 Number of days =140.23 hrs Number of days=5.843 days

8.) Determine the minimum volume of day tank in m3 of 28oAPI fuel having a fuel consumption of 200 kg/hr? SOLUTION: SG15.6oC= 141.5_ 131.5+28

SG15.6oC= 0.877 Density of fuel =0.887(1000) Density of fuel =887 kg/m3 W= m/v 887= 200/v V=0.22548 m3/hr x 24 hrs/day V= 5.41 m3 9.) Given the following ultimate analysis; C=70% N2=5% H2=3% O2=4% S2=6% ASH=5% moisture=8%. Using 25% excess air, determine the actual air fuel ratio. SOLUTION: A/F = 11.5C + 34.5(H-O/8) + 9.3S A/F =11.5(0.7) +34.5(0.03-0.04/8) +4.3(0.06) A/F = 9.1705(1.25) A/F = 11.46 kg air/kg fuel 10.) A 650Bhp diesel engine uses fuel oil of 28oAPI gravity, fuel consumption is 0.65 lb/Bhp-hr. Cost of fuel is P7.95 per liter. For continuous operation, determine the minimum volume of cubical day tank in cm3, ambient temperature is 45oC. SOLUTION: SG15.6oC= 141.5_ 131.5+28 o SG15.6 C= 0.877 SG45.oc= SG15.6oC [1-0007(t-15.6)] SG45oc=0.877[1-0.0007(45-15.6)] SG45oC=0.869 Density of fuel=0.869(1kg/li) Density of fuel=0.869kg/li Solving for fuel consumption: mf= 0.65(650) mf= 422.5 lb/hr mf= 191.61kg/hr vf= 191.61/0869 vf= 220.495 li/hr Volume of day tank = 220.495 x 24 hrs Volume of day tank = 5,291.88 li x 1m3/1000 li Volume of day tank = 5.291.88m3 x (100)3 cm3/m3 Volume of day tank = 5,291,880 cm3

11.) A 650Bhp diesel engine uses fuel oil of 28oAPI gravity, fuel consumption is 0.65 lb/Bhp-hr. Cost of fuel is P7.95 per liter. For continuous operation, determine the cost of fuel per day at 45oC. SOLUTION: SG15.6oC= 141.5_ 131.5+28 SG15.6oC= 0.877 SG45.oc= SG15.6oC [1-0007(t-15.6)] SG45oc=0.877[1-0.0007(45-15.6)] SG45oC=0.869 Density of fuel=0.869(1kg/li) Density of fuel=0.869kg/li Solving for fuel consumption: mf= 0.65(650) mf= 422.5 lb/hr mf= 191.61kg/hr vf= 191.61/0869 vf= 220.495 li/hr Volume of day tank = 220.495 x 24 hrs Volume of day tank = 5,291.88 li Cost of fuel per day = 5,291.88 li x P7.95/li Cost of fuel per day = P42,070.45 12.) A logging firm in Isabela operates a Diesel Electric Plant to supply its electric energy requirements. During a 24 hour period a 24 hour period, the plant consumed 250 gallons of fuel at 80oF and produced 2700 KW-hrs.Industrial fuel used is 30oAPI and was purchased at P3.00/li at 60oF. Determine cost of fuel to produce one Kwh. SOLUTION: 60oF =15.6oC 80oF =26.6 oC SG15.6oC= 141.5_ 131.5+30 o SG15.6 C= 0.876 SG26.6oc= 0.876[1-0007(26.67-15.56)] SG26.6oc=0.869 At 26.6oC, mf=250gal/24hrsx3.785 li/gal At 26.6oC, mf=39.431 li/hr Load=2700/24 Load =112.5 KW Cost per KW-hr= 39.116xP3.00/li 112.5 Cost per KW-hr=P1.043/KW-h

13.) A logging firm in isabela operates of Diesel Electric Plant to supply its electric energy requirements. During a 24 hour period, the plant consumed 250 gallons of fuel at 80oF and produced 2700 KW-hrs.Industrial fuel used is 30oAPI and was purchased at P3.00/li at 60oF.Determine the overall thermal efficiency of the plant. SOLUTION: Qh=41,130+139.6xoAPI Qh=41,130+139(30) Qh=45,318 KJ/kg 60oF =15.6oC 60oF =26.6oC SG15.6oC= 141.5_ 131.5+30 SG15.6oC= 0.876 SG26.6oc= 0.876[1-0007(26.67-15.56)] SG26.6oc=0.869 At 26.6oC, mf=250gal/24hrsx3.785 li/gal At 26.6oC, mf=39.431 li/hr Load=2700/24 Load =112.5 KW mf=39.431 li/hr x 0.869kg/li x 1hr/3600sec mf=0.00952 kg/sec Overall Efficiency= POWER OUTPUT mfQh Overall Efficiency= 112.5___ 0.00952(45318) Overall Efficiency=26.08 14.) A diesel electric plant in one of the remote provinces in the South utilizes diesel fuel with an oAPI of 28 at 15.6oC. The plant consumes 680 liters of diesel fuel at 26.6oC. in 24 hrs, while the power generated for the same period amounts to 1,980 KW-hrs. Determine overall thermal efficiency of the plant. SOLUTION: SG15.6 oC = 141.5 131.5+28 o SG15.6 C = 0.887 SG26.6 oC = 0.887(1-0.0007(26.6-15.6) SG26.6 oC =0.88 Density=0.88x1kg/li Density=0.88kg/li Mass of fuel=680(0.88)/24

Mass of fuel=24.933 kg/hr Qh=41,130+139.6(oAPI) Qh=41,130+139.6(28) Qh=45,039KJ/kg Overall Efficiency=POWER OUTPUT mfQh Overall Efficiency= 82.5__________ (24.933/3600)(45.039) Overall Efficiency= 26.47% 15.) A steam generator burns fuel oil with 20% excess air. The fuel oil may be represented by C14 H30. The fuel gases leave the preheater at 0.31 Mpa. Determine the actual air-fuel ratio in kg air per kg fuel. SOLUTION: Fuel + Air  product combustion C14H30 +O2+3.76N2CO2+H2O+ 3.76 N2 C14H30+21.5O2+21.5(3.76)N214CO2+15H2O+21.5(3.76)N2 Theoretical A/F =21.5(32)+21.5(3.76)(28) 14(12)+30(1) Theoretical A/F=14.91 kg air/kg fuel Actual A/F=14.91(1.2) Actual A/F=17.89 kg air/kg fuel 16.) A steam generator burns fuel oil with 20% excess air. The fuel oil may be represented by C14 H30. The fuel gas leaves the preheater at 0.31 Mpa. Determine partial pressure of H2O. SOLUTION: Fuel + Air  product combustion C14H30 +O2+3.76N2CO2+H2O+ 3.76 N2 C14H30+21.5O2+21.5(3.76)N214CO2+15H2O+21.5(3.76)N2 Combustion reaction with 20% excess air: C14H30 +1.20(21.5)O2+1.2(21.5)(3.76)N214CO2+ 15 H2O+1.2(21.5)( 3.76) N2+0.20(21.5)O2 Total mols of product =14+15+1.2(21.5)(3.76)+0.2(21.5) Total mols of product =130.308 mols Partial pressure of H2O= (15/130.308)(0.31) Partial pressure of H2O=0.0357Mpa Partial pressure of H2O=35.7 kpa

17.) In the boiler design, it is desired to have the flue gas exit temperature above the dew point. Estimate the dew point temperature of the flue gas produced by combustion having the gravimetric analysis of: N2=71.84% CO2= 20.35% O2=3.61% H2O=4.20 SOLUTION: Converting the given analysis of volumetric: N2=0.7184/28=0.02565714 CO2=0.2035/44=0.004625 O2=0.0361/32=0.00112812 H2O=0.042/18=0.00233333 Total mols of product =0.025657+0.00462+0.001128+0.00233 Total mols of product =0.03374359 Partial Pressure of H2O= (0.0023333/0.337436)(101.325) Partial Pressure of H2O= 7.006 Kpa From steam table: At 0.007006 Mpa, tsat=39oC Dew point temperature= 39oC

18.) The dry exhaust gas from oil engine has the ff. gravimetric analysis: CO2=21.6% O2=4.2% N2= 74.2%. Specify heat at constant pressure for each component of the exhaust gas in Kcal/kg oC are CO2=0.203 O2=0.219 N2= 0.248. Calculate the gas constant in J/kgo K. SOLUTION: Converting the gravimetric analysis to volumetric: CO2 0.216 0.216/44 =0.004909 O2 0.042 0.042/32 =0.001312 N2 0.742 0.742/28 =0.026500 0.032721 mols/kg-mol Molecular Weight=1/0.032721 Molecular Weight=30.56 kg/kg-mol SG=30.56/28.97 SG=1.055

19.) The dry exhaust gas from oil engine has the ff. gravimetric analysis: CO2=21.6% O2=4.2% N2= 74.2%. Specific heats at constant pressure for each component of the exhaust gas in Kcal/kg oC are: CO2=0.203 O2=0.219 N2= 0.248. Calculate the specific heat of the gas in KJ/kg-oK.

SOLUTION: Cp= 0.216(0.203) + 0.042(0.219) + 0.742(0.248) Cp=0.237 Kcal/kg-oC x 4.187 Cp=0.992 KJ/kg-oC

20.) A bituminous coal has the ff. composition: C= 71.5% S= 3.6% H =5.0% ASH= 8.2% O= 7.O% W= 3.4 % N= 1.3%. Determine the theoretical weight of oxygen in lb/lb of coal. SOLUTION: Theo. A/F =11.5C+34.5(H-0/8) +4.3S Theo. A/F= 11.5(0.715) +34.5(0.05-0.07/8) +4.3(0.036) Theo. A/F= 9.8 lb air/lb coal O2 in air by weight= 23.2% Therefore: Theoretical weight of O2=0.232(9.8) Theoretical weight of O2=2.274lb/lb coal