Energy Efficient Process Heating



Energy-Efficient Process HeatingIntroduction Process heating is the application of heat to products. This chapter begins by discussing guiding principles for reducing process heating energy use. The body of the chapter discusses common methods, organized according to the inside-out approach, for improving the process heating efficiency. For each method, the fundamental equations for estimating savings are presented, and the method is illustrated with an example.Principles of Energy-Efficient Process HeatingHeat Exchanger Effectiveness ApproachIn many process heating applications, the primary energy conversion process is either the conversion of chemical energy in fuel to sensible energy via combustion or the conversion of electrical power to heat through electrical resistance, induction or arching. In both cases, the temperature of the heat is determined by the conversion process and is typically very high. Heat not transferred to the end product (or lost in other ways) is typically carried away in exhaust gasses. Thus, increasing the quantity of heat transferred to the process typically decreases the quantity of heat lost in exhaust gasses and improves the efficiency of the process.Conceptually, this can be understood by considering heat transfer from a hot medium, h, to a process, p.The rate of heating, Q, is:Q = UA (Th – Tp) (1)where UA is the overall heat transfer coefficient, Th is the average temperature of the heating medium, Tp is the average temperature of the process.Increasing heat transfer effectiveness, represented by UA, increases the rate of heating and decreases the temperature of energy carried away in the heating medium, Th2. This increases the efficiency of the heating process. Thus, increasing heat transfer effectiveness is a key mechanism for improving process heating efficiency. Heat transfer effectiveness can be increased by increasing the heat transfer surface area, increasing the turbulence or density of the heat carrying medium, increasing radiation exchange between the heat source and the product or employing counter-flow heat exchange. Reducing the process temperature Tp increases the temperature difference (Th – Tp). As (Th – Tp) increases, more heat is transferred to the process. As in the case of increased UA, this decreases the temperature of energy carried away by the heating medium and increases the efficiency of the heating process. Energy Balance ApproachMinimizing energy losses also improves process heating system efficiency. For example, energy flows into and out of a typical combustion-fired furnace are shown below. Part of the energy in the gross fuel input is transferred as useful heat to the load. The rest is lost as heat in the flue gasses, through the walls, through openings, absorbed by cooling, absorbed by conveyance devices, or stored in the walls. Reducing these losses reduced the gross fuel input, and improves the energy efficiency of the process. Source: Thekdi, 2004Opportunities for Improving The Energy-Efficiency of Process Heating SystemsThese principles can be organized using the inside-out approach, which sequentially reduces end-use energy, distribution energy, and primary conversion energy. Combining the heat exchanger effectiveness, energy balance and inside-out approaches, common opportunities to improve the energy efficiency of process heating systems include:Reduce end use loadsInsulate hot surfaces to reduce wall lossesCover openings to reduce radiation lossesLower an opening to decrease temperature-driven infiltrationSeal openings or use draft control to reduce infiltrationReduce cooling losses by controlling cooling water temperatureReduce conveyance losses in continuous and batch processesReduce storage losses in batch processesImprove efficiency of distribution systemConvert to counter flow heat exchangeImprove efficiency of energy conversionReduce excess combustion airConvert from atmospheric to oxygen burnersPre-heat combustion air Pre-heat the load Cascade waste heat to lower temperature processThe sections that follow discuss how to quantify these saving bustion EfficiencyMany process heating applications employ combustion to covert fuel energy into high temperature thermal energy. In most cases, the fuel is a hydrocarbon. This section describes natural gas combustion and how to calculate combustion air flow, combustion temperature and the efficiency of the process. These results are used extensively throughout this chapter.The minimum amount of air required for complete combustion is called the “stoichiometric” air. Air consists of about 1 mole of oxygen to 3.76 moles of nitrogen. Assuming that natural gas is made up of 100% methane, the equation for the stoichiometric combustion of natural gas with air is: CH4 + 2 (O2 + 3.76 N2) CO2 + 2 H2O +7.52 N2 (17)The ratio of the mass of air required to completely combust a given mass of fuel is called the stoichiometric air to fuel ratio, AFs. AFs can be calculated using the molecular masses of the air and fuel at stoichiometric conditions. For combustion of natural gas in air, AFs is about: AFs = Mair,s / Mng,s = 2[ (2 x 16) + (3.76 x 2 x 14)] / [12 + (4 x 1)] = 17.2 The quantity of air supplied in excess of stoichiometric air is called excess combustion air, ECA. Excess combustion air can be written in terms of the stoichiometric air to fuel ratio, AFs, the combustion air mass flow rate, mca, and natural gas mass flow rate, mng.ECA = [(mca / mng) / AFs] – 1 (18)Large quantities of excess air dilute combustion gasses and lower the temperature of the gasses, which results in decreased efficiency. The energy input, Qin, to a combustion chamber is the product of the natural gas mass flow rate, mng, and the higher heating value of natural gas, HHV, which is about 23,900 Btu/lbm.Qin = mng HHV (19)The mass flow rate of the combustion gasses, mg, is the sum of the natural gas mass flow rate, mng, and combustion air mass flow rate, mca.mg = mng + mca (20)The temperature of combustion, Tc, can be calculated from an energy balance on the combustion chamber, where the chemical energy released during combustion is converted into sensible energy gain of the gasses. The energy balance reduces to the terms of inlet combustion air temperature, Tca, lower heating value of natural gas (21,500 Btu/lbm), excess combustion air, ECA, stoichiometric air fuel ratio, AFs, and specific heat of combustion gasses, Cpg (0.26 Btu/lbm-F). Combustion temperature, Tc, is calculated in terms of these easily measured values as:Tc = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg] (21)The combustion efficiency, ?? is the ratio of energy delivered to the system to the total fuel energy supplied. The energy delivered to the system is the energy loss of combustion gasses. The energy loss of the combustion gasses can be expressed as the product of the mass flow rate, specific heat and temperature drop of the gasses. The total energy fuel energy supplied is the higher heating value of the fuel. Using this approach, the combustion efficiency, ? is:? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV (22)The dew-point temperature of products of combustion is about 140 F. If the products of combustion leave the process at temperature of less than the dew-point temperature the water vapor will condense to a liquid and release energy. To include this effect, the efficiency equation can be written:If Tex > 140 F then hfg = 0 Else hfg = HHV – LHV ? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex) + hfg] / HHV (22b)The three required input values for computing combustion efficiency, entering combustion air temperature, Tca, exhaust gas temperature, Tex, and excess combustion air, ECA, can be measured using a combustion analyzer. The quantity of excess air in the combustion gasses is sometimes expressed as fraction oxygen. For methane (natural gas) the conversion between fraction oxygen, FO2, and excess combustion air, ECA, are:FO2 = 2 ECA / (10.52 + 9.52 ECA) ECA = 10.52 FO2 / (2 – 9.52 FO2)(23)ExampleAn atmospheric burner consumes 100,000 Btu/hr of natural gas. An analysis of the exhaust gasses finds that the fraction of excess air is 30% and the temperature of the exhaust gasses is 500 F. Calculate combustion air flow (lb/hr), exhaust gas flow (lb/hr), combustion temperature (F) and the efficiency of the process. In process heating, the combustion efficiency is sometimes called “available heat”. A chart showing percent available heat (combustion efficiency) as a function of exhaust gas temperature and excess air for combustion of natural gas is shown below.Source: DOE, 2004.Reducing Heat Loss through WallsHot surfaces lose heat to the surroundings via convection and radiation. Insulation reduces the rate of heat flow, and as a consequence, virtually always results in energy savings. Insulating hot surfaces also reduces the risk of burns and may make the workplace more comfortable. The equation for heat loss, Q, to the surroundings at Ta, from a hot surface at Ts, with area A is:Q = h A (Ts – Ta) + ? A ? (Ts4 – Ta4) (1)where h is the convection coefficient, ?? is the Stefan-Boltzman constant (0.1714 x 10-8 Btu/ft2-hr-R4), ? is the emissivity of the surface (about 0.9 for dark surfaces). Assuming steady state conditions, the heat loss to the surroundings must equal the heat loss through the furnace walls:Q = h A (Ts – Ta) + ? A ? (Ts4 – Ta4) = A (Tf – Ts) / R(2)where Tf is the interior temperature of the furnace and R is the thermal resistance of the furnace wall including the interior convection coefficient.Exterior convection over a hot surface is caused when air is warmed, becomes less dense than the surrounding air, and rises. Thus, natural convection is a function of the temperature difference between the hot surface and the exterior air. Dimensional approximations for the convection coefficient in natural convection, as a function of the orientation of the surface and the temperature difference between the surface and the surrounding air, are listed below (ASHRAE Fundamentals, 1989). In these relations, L is the characteristic (vertical) length (ft), ?T is temperature difference between the surface and the surrounding air (F), and h is the convection coefficient (Btu/hr-ft2-F).(3)Laminar: L3 ?T < 63Turbulent: L3 ?T > 63Horizontal Surfacesh = 0.27 x (?T / L) 0.25h = 0.22 x (?T) 0.33Vertical Surfacesh = 0.29 x (?T / L) 0.25h = 0.19 x [?T(sin B)] 0.33Typical thermal resistances and costs of common types of insulation are listed in the table below.Insulation TypeThermal ResistanceCost3.5-inch fiberglass batt11 ft2-hr-F/Btu$0.30 / ft2?-inch rigid blue Styrofoam board5.2 ft2-hr-F/Btu$0.32 /ft2?-inch rigid polyisocyanurate with foil facing3.5 ft2-hr-F/Btu$0.23 /ft21-inch spray-on cellulose (meets fire-code)5 ft2-hr-F/Btu$0.75 /ft2 *2-inch spray-on polyurethane9 ft2-hr-F/Btu$4 /ft2 *2-inch steam pipe and tank insulation5 ft2-hr-F/Btu$10 /ft2 **includes installationInsulating Fire BrickBNZ-20BNZ-23BNZ-26BNZ-28BNZ-30BNZ-32Temperture Use Limit (F)2,3002,3002,6002,8003,0003,200Max Mean Brick Temperture (F)1,8002,1002,2002,2002,4002,600Density (lbm/ft3)363748556575Conductivity (Btu-in/ft2-hr-F)@500 F0.91.01.62.32.83.9@1,000 F1.21.31.92.42.94.1@1,500 F1.51.62.22.63.14.2@2,000 F1.71.82.62.73.34.3Specific Heat (Btu/lbm-F)0.260.260.260.260.260.26Source: BNZ Materials, Inc.Ceramic BlanketsDensity8 lb/ft3 (128 kg/m3)Max Continuous use limit2,000 F (1,093 C)k (Btu-in./hr-ft2-°F)R (hr-ft2-F/Btu)/inchat 500°F (260°C)0.442.27at 1000°F (538°C)0.871.15at 1500°F (816°C)1.450.69at 1800°F (982°C)1.830.55at 2000°F (1093°C)2.090.48Source: Thermal Ceramics, Inc. In many cases, the interior and exterior temperatures of an oven/furnace and the temperature of the surroundings are known or can be measured. If so, Equations 1, 2 and 3 can be manipulated to calculate heat loss from hot surfaces and the savings from adding insulation to reduce heat loss.ExampleConsider a rectangular heat treat oven with dimensions of 10 ft x 10 ft x 10 ft, an inside air temperature of 1,600 F, an external surface temperature of 250 F, and an outer surface emissivity of 0.9. The combustion efficiency of the oven is 50%. The temperature of the surrounding air and surfaces is 70 F. Calculate heat lost from the oven’s sides and the savings from insulating the sides with R = 4 hr-ft2-F/Btu insulation. From Equation 3, air flow is turbulent and the current convection coefficient, h1, is:L3 ?T = L3 (Ts – Ta) = 103 (250 – 70) = 180,000 > 63 hence turbulenth1 = 0.19 (Ts – Ta)0.33 = 0.19 (250 – 70)0.33 = 1.054 Btu/hr-ft2-F From Equation 1, the current heat loss, Q1, is:Q1 = h1 A (Ts1 – Ta) + ? A ? (Ts14 – Ta4) Q1 = 1.054 · 400 · (250 – 70) + (0.1714 · 10-8) · 400 · 0.9 · [(250 + 460)4 – (70 + 460)4] Q1 = 184,028 Btu/hr From Equation 2, the current thermal resistance of the oven wall, R1, is:R1 = A (Tf – Ts) / Q1R1 = 400 · (1,600 – 250) / 184,028 = 2.93 hr-ft2-F/Btu With insulation of thermal resistance, R2, the new surface temperature of the wall will be Ts2 and the new convection coefficient will be h2. Equations 2 and 3 become:h2 A (Ts2 – Ta) + ? A ? (Ts24 – Ta4) = A (Tf – Ts2) / (R1 +R2)h2 = 0.19 (Ts2 – Ta) 0.33This system of two equations and two unknowns can be solved to give:Ts2 = 169.33 Fh2 = 0.867 Btu/hr-ft2-F Substituting Ts2 and h2 into Equation 1, the new heat loss, Q2, would be:Q2 = h2 A (Ts2 – Ta) + ? A ? (Ts24 – Ta4) Q2 = 0.867 · 400 · (180 – 70) + (0.1714 · 10-8) · 400 · 0.9 · [(169.33 + 460)4 – (70 + 460)4] Q2 = 82,527 Btu/hrThe heat loss savings, Qs, and fuel energy savings, Es, from adding insulation would be: Qs = Q1 – Q2 = 184,028 Btu/hr –82,527 Btu/hr = 101,501 Btu/hr Es = Qs / ? = 101,501 Btu/hr / 50% = 203,001 Btu/hrThus, this measure would reduce heat loss through the walls by 55.2% Reducing Radiant Heat Loss from WallsHot surfaces lose heat to the surroundings via convection and radiation. As noted before, the equation for heat loss, Q, to the surroundings at Ta, from a hot surface at Ts, with area A is:Q = h A (Ts – Ta) + ? A ? (Ts4 – Ta4) (1)where h is the convection coefficient, ? is the Stefan-Boltzman constant (0.1714 x 10-8 Btu/ft2-hr-R4), ? is the emissivity of the surface (about 0.9 for dark surfaces). Covering walls with low-emissivity coatings reduces radiation. Rustoleum “metallic finish” paint, has an emissivity of about 0.30. The BASF spray-on coating Radiance? has an emissivity of about 0.25. ExampleConsider a furnace wall at 250 F with emissivity = 0.90 in a facility with walls and ceilings at 70 F. The combustion efficiency of the furnace is 50%. Assuming the surface temperature remains constant, calculate the savings from applying a coating with emissivity = 0.25.Q1 / A = ? ?? (Ts4 – Ta4)Q1 / A = 0.90 · 0.1714 x 10-8 Btu/ft2-hr-R4 · [(250 + 460)4 – (70 + 460)4] Q1 / A = 270 Btu/hr-ft2 Substituting ?? into Equation 1, the new heat loss, Q2, would be:Q2 / A = ? ?? (Ts24 – Ta4) Q2 / A = 0.25 · 0.1714 x 10-8 Btu/ft2-hr-R4 · [(250 + 460)4 – (70 + 460)4] Q2 / A = 75 Btu/hr-ft2The heat loss savings, Qs, and fuel energy savings, Es, from adding insulation would be: Qs = Q1 – Q2 = 270 Btu/hr-ft2 – 75 Btu/hr-ft2 = 195 Btu/hr-ft2 Es = Qs / ? = 195 Btu/hr-ft2 / 50% = 390 Btu/hr-ft2Thus, this measure would reduce heat loss through the walls by 72% Reducing Radiation Loss through OpeningsOpenings in furnace walls allow heat to radiate outward. Radiation heat loss can be reduced by covering openings. The figures below show openings in a heat treat furnace. The first opening shows “room for improvement”. The second opening is covered with flaps to reduce radiation loss to the surroundings. Heat is radiated from the hot interior temperature of a furnace through openings to the surroundings. Due their geometries, both the interior and the surroundings can be approximated as black bodies with emissivities equal to 1.0. Thus, heat loss, Q, through an opening of area, A, from the interior of a furnace at temperature, Tf, to the surroundings at temperature, Ta, is:Q = ? A Ffa (Tf4 – Ta4) (4)where ?? is the Stefan-Boltzman constant (0.1714 x 10-8 Btu/ft2-hr-R4) and Ffa is the view factor between the inside of the furnace and the surroundings. If the opening is approximated as a circle with radius r through a wall of thickness L, the view factor can be calculated from the following two equations (Cengal, 1998).S = 2 + (L /r)2Ffa = 0.5 [ S – (S2 – 4)0.5 ] (5)If the opening were blocked by a radiation shield with emissivity ?shield, the heat transfer would be reduced to (Cengal, 1998):Q = ? A Ffa (Tf4 – Ta4) [?sheild / 2 ](6)ExampleConsider a furnace with an opening 1 ft in diameter through a 1.0 ft thick furnace wall. The temperature inside the furnace is 1,600 F, and combustion efficiency of the furnace is 50%. The temperature of the surrounding surfaces is 70 F. Calculate the radiation heat loss through the opening, and the savings if the opening is covered by a radiation shield with emissivity 0.9. The area of the opening is:A = ? r2 = ? (1/2)2 = 0.785 ft2From Equation 5, the view factor for radiation escaping through the opening is:S = 2 + (L /r)2 = 2 + (1 /0.5)2 = 6.0Ffa = 0.5 [ S – (S2 – 4)0.5 ] = 0.5 [ 6.0 – (6.02 – 4)0.5 ] = 0.172From Equation 4, the current heat loss, Q1, is:Q1 = ? A Ffa (Tf4 – Ta4) Q1 = (0.1714 · 10-8) · 0.785 · 0.172 · [(1600 + 460)4 – (70 + 460)4] = 4,141 Btu/hr From Equation 6, the heat loss with the radiation cover, Q2, would be:Q2 = ? A Ffa (Tf4 – Ta4) [?shield / 2] = Q1 [?shield / 2] = 4,141 Btu/hr [ 0.9 / 2] = 1,863 Btu/hr The heat loss savings, Qs, and fuel energy savings, Es, from adding a radiation shield would be: Qs = Q1 – Q2 = 4,141 Btu/hr – 1,863 Btu/hr = 2,278 Btu/hr Es = Qs / ? = 2,278 Btu/hr / 50% = 4,555 Btu/hr Thus, this measure would radiation heat loss through the opening by 55% Reducing Heat Loss Due to Infiltration Most high temperature furnaces and ovens operate at a negative air pressure relative to ambient air pressure. Thus, openings in the furnace wall allow air to infiltrate into the furnace. Infiltrating air is heated to the operating temperature of the furnace before being exhausted. The heat removed by infiltrating air, Q, at volume flow rate, V, with temperature rise from ambient temperature, Ta, to the furnace exhaust temperature, Tex, is:Q = V pcp (Tex – Ta)(24)where pcp is the product of the density and specific heat of air (0.018 Btu/ft3-F).The energy lost due to infiltration can also be quantified by recognizing that the quantity of excess air in the exhaust is the sum of the excess combustion air, ventilation air and infiltrating air. Thus, measures to reduce infiltration and ventilation air will reduce quantity of excess air in the exhaust increase overall combustion efficiency.Reducing Infiltration By Moving Opening to FloorIn an oven with vertical openings, warm air rises to the oven’s ceiling due to buoyancy forces and exfiltrates out of the top of vertical openings. An equal amount of cool ambient air infiltrates into the oven through the bottom half of the vertical openings. Figure A shows a typical velocity profile of infiltration and exfiltration air through a vertical oven opening. The velocities are greatest at the top and bottom of the openings. A balance point occurs near the center of the opening where air leaks neither into nor out of the oven. The velocity of infiltration and exfiltration can be measured with by performing a traverse from the top of the opening to the bottom of the opening with an anemometer. Buoyancy driven infiltration can be practically eliminated by moving the opening to the floor of the oven (Figure B).Figure A. Vertical entrance with infiltration and exfiltration air.Figure B. Horizontal entrance with negligible infiltration and exfiltration air.ExampleConsider a cure oven operating at 500 F located on the second floor of a plant. The total area of the entrance is 100 ft2. The average exfiltration velocity is measured to be 100 ft/min over the upper half of the entrance. If ambient air temperature is 70 F and the combustion efficiency of the oven is 70%, calculate heat loss and fuel energy savings if the entrance was oriented horizontally on the floor of the oven.The heat loss and fuel energy savings from reducing infiltration would be:Qs = Ve A pcp (Tex – Ta) = 100 ft/min x (100 ft2 / 2) x 60 min/hr x 0.018 Btu/ft3-F x (500 F -70 F)Qs = 2,322,000 Btu/hrEs = Qs / ? = 2,322,000 Btu/hr / 0.70 = 3,317,000 Btu/hrReducing Infiltration by Lowering OpeningsThe quantity of infiltration air through a vertical opening is a function of the height of the opening and the temperature difference between the oven air at the opening and the ambient plant air. In some ovens, vertical openings are near the top of ovens and have room to be moved lower. If the openings are moved lower, infiltration would be reduced because the temperature difference between ambient plant air and oven air at the opening would be smaller. Moving the opening usually requires moving the monorail or conveyor; however, the energy savings may be sufficient to justify the project. The figure below shows the position of a monorail opening located near the top of the oven and the new position after an energy-savings retrofit.(a) (b)Figure 6. Oven face with high vertical opening (a) and more energy-efficient low vertical opening (b).The net pressure difference, Pnet, between the bottom and the top of the column is a function of column height, h, temperature inside the column, Ti, and temperature outside the column, To. The constants in the equation are the acceleration due to gravity, g (32.2 ft/s2), atmospheric pressure, Patm (14.7 psi), and the gas constant for air, R. Pnet = h g Patm [(1 / To) – (1 / Ti)] / R (25)Assuming that friction is negligible, Pnet can be used in Bernoulli’s equation to calculate the velocity, Ve, through the opening as:Ve = (26)To obtain the most conservative result, the density, ρ, in Equation 26 can be assumed to be the density of air at the temperature when it exits the stack. If the internal temperature profile over the oven’s height is known, the internal oven temperature near the top of the opening before the retrofit, Toven,1, and after the retrofit, Toven,2, are known. The temperature profile could be found by taking temperature measurements along the oven’s height. If velocity and temperature of infiltration air, Ve1 and Ta, are known, velocity of infiltration air after lowering the opening, Ve2, can be found by combining Equation 25 and 26 and creating a velocity ratio. The resultant equation for the change in infiltration is:Ve2 = Ve1 (27)ExampleConsider a cure oven with a high vertical entrance. The temperature of air is measured to be 500 F near the ceiling of the oven 400 F near the mid-height of the oven. Oven air at a temperature of 500 F is measured to be exfiltrating the oven at an average velocity of 300 ft/min over an area of 8 ft2. If ambient air temperature is 70 F and the combustion efficiency of the oven is 70%, calculate heat loss and fuel energy savings if the entrance were lowered to mid-height. The current heat loss, Q1, is:Q1 = V pcp (Tex – Ta) = 300 ft/min x 8 ft2 x 60 min/hr x 0.018 Btu/ft3-F x (500 F -70 F)Q1 = 1,114,560 Btu/hrThe velocity of infiltrating air after the oven is lowered would be:Ve2 = Ve1 Ve2 = 300 ft/min Ve2 = 300 ft/min · 0.926 = 278 ft/minThe heat loss, Q2, and fuel energy savings after the oven is lowered would be:Q2 = V2 pcp (Tex – Ta) = 278 ft/min · 8 ft2 · 60 min/hr · 0.018 Btu/ft3-F x (400 F -70 F)Q2 = 791,695 Btu/hrQs = Q1 – Q2 = 1,114,560 Btu/hr - 791,695 Btu/hr = 322,865 Btu/hrEs = Qs / ? = 322,865 Btu/hr / 0.70 = 461,235 Btu/hrThus, this measure would reduce infiltration heat loss by 29.0%.Reducing Infiltration by Sealing Leaks or Installing a Back Pressure DamperAir infiltrates into ovens and furnaces that are not tightly sealed. To minimize infiltration, leaks should be sealed and doors and ports should be well maintained. To further reduce infiltration, backpressure dampers can be installed on exhaust stacks to control the furnaces pressure so that it is neutral or slightly positive. Back pressure dampers can be sophisticated mechanisms with active pressure control, or as simple as blocking a small part of the exhaust stack with a ceramic brick. All air infiltrating a furnace leaves in the exhaust gasses and increases the excess air measured by a combustion analyzer. If excess air is higher than would be expected from combustion air alone, infiltration most likely takes place in the system. If no ventilation air is required in the system, the target would be to eliminate infiltration and reduce total excess air in the flue gasses to 10% or less as required for combustion. ExampleA heat treat furnace burns 2 mmBtu/hr of natural gas. A combustion analyzer measures 90% excess air in the flue gasses and the temperature of the flue gasses to be 1,000 F. The temperature of the ambient air is 70 F. Inspection of the furnace shell and the high excess air content suggested that air is infiltrating into the furnace. Sealing leaks in the furnace and installing a backpressure damper will reduce infiltration and bring excess air down to 10%. Calculate the heat energy and fuel savings from sealing openings and installing a backpressure damper. From Equations 22 and 23 the current combustion efficiency of the annealing furnace is:Tc1 = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg]Tc1 = 70 F + 21,500 Btu/lbng / [{1 + 1.90 · 17.2 lba/lbng} 0.26 Btu/lba-F] = 2,525 F?? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV ?? = [{1 + 1.90 · 17.2 lba/lbng} 0.26 Btu/lba-F (2,525 – 1,000 F)] / 23,900 Btu/lbm?? = 55.9%The heat delivered to the system, Qout, would be:Qout = Qf1 ?? = 2 mmBtu/hr · 55.9% = 1.12 mmBtu/hr If excess air were reduced to 10%, the combustion efficiency would increase to:Tc2 = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg]Tc2 = 70 F + 21,500 Btu/lbng / [{1 + (1.1) 17.2 lba/lbng} 0.26 Btu/lba-F] = 4,221 F?? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV ?? = [{1 + 1.1 · 17.2 lba/lbng} 0.26 Btu/lba-F (4,221 F – 1,000 F)] / 23,900 Btu/lbm?? = 69.8%The fuel energy input, Qf2, to delivered the same energy output to the system would would be:Qf2 = Qout / ?? = 1.12 mmBtu/hr / 69.8% = 1.60 mmBtu/hrThe fuel energy savings, Es, from reducing excess combustion air would be: Es = Qf1 – Qf2 = 2.00 mmBtu/hr – 1.60 mmBtu/hr = 0.40 mmBtu/hrThus, this measure would reduce fuel use by 19.9% Reducing Heat Loss to CoolingDoors, conveyors, and other equipment in high temperature furnaces are sometimes cooled to prevent warping or failure. The heat removed by cooling water, Q, at volume flow rate, V, with temperature rise from Tw1 to Tw2 is:Q = V p cp (Tw2 – Tw1)(7)where p is the density of water (8.32 lb/gal) and cp is the specific heat of water (1.0 Btu/lb-F). The heat removed by the cooling water is also equal to the heat loss from the furnace. The heat loss from the furnace can be approximated as: Q = UA [Tf – (Tw2 + Tw1) / 2 ](8)where UA is the overall heat transfer coefficient between the furnace and the water, Tf is the temperature of the furnace. Heat removed by the cooling water must be made up by fuel energy; thus reducing heat removed by the cooling water saves energy. ExampleConsider a furnace with inside air temperature of 2,000 F and combustion efficiency of 50%. The furnace is cooled by 100 gpm of water from a cooling tower. During winter, the cooling tower supplies water to the furnace at 67 F, and the water exits the furnace at 100 F. During summer, the cooling tower delivers 90 F water to the furnace and internal parts are not overheated. Calculate the rate of heat removed by the cooling water in the summer and winter, and the savings from supplying 90 F water to the furnace instead of 67 F water during winter.From Equation 7, the winter heat loss, Q1, is:Q1 = V p cp (Tw2– Tw1)Q1 = 100 gpm · 8.32 lb/gal · 1.0 Btu/lb-F (100 F – 67 F) · 60 min/hr = 1,647,000 Btu/hr From Equation 8, the overall heat transfer coefficient between the furnace and the average temperature of the cooling water, UA, is:UA = Q / [Tf – (Tw2 + Tw1) / 2] UA = 1,647,000 Btu/hr / [2,000 F – (100 F + 67 F) / 2 ] = 859.6 Btu/hr-FCombining Equations 7 and 8 gives:Q = V p cp (Tw2 – Tw1) = UA [Tf – (Tw2 + Tw1) / 2 ]During summer, when Tw1 = 90 F, this equation can be solved to give:Tw2 = 122.6 FHence, From Equation 7, the summer heat loss, Q2, is:Q2 = V p cp (Tw2– Tw1)Q2 = 100 gpm · 8.32 lb/gal · 1.0 Btu/lb-F (122.6 F – 90 F) · 60 min/hr = 1,628,000 Btu/hr The heat loss savings, Qs, and fuel energy savings, Es, from increasing cooling water temperature would be: Qs = Q1 – Q2 = 1,647,000 Btu/hr – 1,628,000 Btu/hr = 19,600 Btu/hr Es = 19,600 Btu/hr / 50% = 39,200 Btu/hr Thus, this measure would reduce cooling heat loss by 1.2% Reducing Heat Loss to Conveyance EquipmentContinuous heating processes often use conveyors to move products into and out of the oven. Batch processes often use racking to move products into and out of the oven. In both cases, the heat absorbed and removed by the conveyance equipment must be made up by additional thermal energy.Continuous ProcessThe energy lost to a conveyor in a continuous heating system, Q, is the product of conveyor velocity, V, mass per linear length, m, specific heat of conveyor material, cp, and the temperature difference of the conveyor leaving, Tc2, and entering, Tc1 the system.Q = V m cp (Tc2 – Tc1)(9)When the conveyor moves slowly, the temperature of the conveyor leaving the system, Tc2, frequently approaches the interior temperature of the oven/furnace, and the temperature of the conveyor entering the system, Tc1, approaches room temperature. If conveyor velocity, V, or mass, m, were reduced, the energy savings would be the difference between energy absorbed by the conveyor before and after the change.ExampleA brazing oven with an operating temperature of 1,900 F and combustion efficiency of 50% has a stainless steel conveyor belt weighing 5 lbs/ft traveling 42 ft per hour when loaded with parts. The specific heat of stainless steel is 0.12 Btu/lb-F. To prevent overheating, the conveyor must always be moving. Calculate the savings from slowing the conveyor speed to 18 ft per hour when no parts are being brazed.Assuming the conveyor belt approaches the interior temperature of the oven when leaving the oven, and approaches room temperature before reentering the oven, the conveyor heat losses at 42 ft per hour, Q1, and 18 ft per hour, Q2, are:Q1 = V1 m cp (Tc2 – Tc1) = 42 ft/hr · 5 lb/ft · 0.12 Btu/lb-F · (1,900 – 70) F = 46,116 Btu/hr Q2 = V2 m cp (Tc2 – Tc1) = 18 ft/hr · 5 lb/ft · 0.12 Btu/lb-F · (1,900 – 70) F = 19,764 Btu/hr The heat loss savings, Qs, and fuel energy savings, Es, from slowing the conveyor during non-production hours would be: Qs = Q1 – Q2 = 46,116 Btu/hr – 19,764 Btu/hr = 26,352 Btu/hr Es = Qs / ? = 26,352 Btu/hr / 50% = 52,704 Btu/hrThus, this measure would reduce conveyor heat loss by 57.1% Batch ProcessCarts and racking that hold or transport the product in batch processes absorb energy. In this case, energy absorbed, Q, can be calculated from the mass, m, specific heat, cp, and temperature rise (T1-T2) as:Q = m cp (T2 – T1)(10)If the material of a component is highly conductive, such as metal, the temperature is approximately uniform throughout the component. In these cases, the final temperature of the component can be calculated using the lumped capacitance method. In the lumped capacitance method, the final temperature of an object, T2, initially at temperature, T1, immersed in an atmosphere at temperature, Ta, after time, t, is:k = h A / (m cp)T2 = Tf + (T1-Tf) exp( -k t ) (11)where h is the convection coefficient, A is the area, m is the mass and cp is the specific heat. Heat loss to conveyance devices in batch processes can be reduced by reducing the weight of the conveyance devices, or by reducing the number to batch cycles.ExampleSteel racking with mass 500 lb and surface area 200 ft2 is used to carry products into a batch drying oven at 400 F with combustion efficiency of 80%. The specific heat of steel is 0.12 Btu/lb-F, and the convection coefficient is 2 Btu/hr-ft2-F. If the racking enters the oven at room temperature of 70 F and the drying process takes 0.25 hours, calculate the exit temperature of the racking, and the heat and fuel energy loss to the racking with each drying cycle. The exit temperature would be:k = h A / (m cp) = 2 Btu/hr-ft2-F · 200 ft2 / ( 500 lb 0.12 Btu/lb-F) = 7 /hrT2 = Tf + (T1-Tf) exp[ -(h A / (m cp)) t ] = 400 F + (70 F – 400 F) exp[ -7 · 0.25 ] = 338 FThe heat absorbed by the racking,?Q, and the fuel energy loss, E, are:Q = m cp (T2 – T1) = 500 lb/cycle 0.12 Btu/lb-F (338 F – 70 F) = 16,060 Btu/cycleE = Q / ? = 16,060 Btu/cycle / 80% = 20,075 Btu/cycleA plot of racking temperature over time as it heats up in the oven is shown below.Reducing Energy Storage Losses in Batch ProcessesThe interior structure of batch ovens and furnaces absorb and release energy during heating cycles. If the heat absorbing material is highly conductive, such as metal, the temperature is approximately uniform throughout the component. In this case, energy absorbed can be calculated using the lumped capacitance method shown above.However, if the heat absorbing material is less conductive, such as refractory, the temperature of the refractory will not be uniform, and the lumped capacitance method would not accurately predict energy storage. In this case, a finite-difference model can determine the temperature profile throughout the material at the end of a cycle. The energy absorbed by the mass can then be determined from this profile. Energy savings can be realized if the quantity of thermal mass can be reduced. Finite difference models are constructed by dividing a material into a number of finite nodes, calculating temperature of each node at the end of a short time period from energy balances on each node, then repeating the process over the entire time frame. For example, consider a layer of fire brick that is exposed to interior of an oven on one side and insulated on the other. The thermal mass is divided into eight layers called nodes of thickness dx. Each node is assumed to have a uniform temperature.Firebrick wall of oven divided into 8 nodesThe thermal resistance between the oven atmosphere and node 1, R1, is the sum of the convection and conduction thermal resistances. The convection thermal resistance is the inverse of the product of convection coefficient, h, and the surface area, A. The conduction thermal resistance is the distance between the center of node 1 and the surface, divided by thermal conductivity, k, and cross-sectional area, A.R1 = 1 / (h A) + dx / (2 k A) (12)The thermal resistance between each consecutive node is the distance between each node, dx, divided by thermal conductivity, k, and the cross-sectional area, A. If material properties of the thermal mass is constant throughout, the thermal resistances between all interior nodes are equal.R2 = R3 = R4 = R5 = R6 = R7 = R8 = dx / (k A) (13)From the First Law of Thermodynamics, the energy into each node minus the energy out of each node equals the change of energy stored in each node.(TN – T) / RN – (T – TS) / RS = [dx A Cp (T’ – T)] / dt(14)The energy storage term is the product of the distance between the nodes, dx, cross-sectional area, A, material density, , material specific heat, Cp, and the difference between node temperature after the iterative step, T’, and current node temperature, T, all divided by the time period of iteration, dt. Rearranging Equation 14, the temperature after each iterative step, T’, can be calculated as:T’ = dt (TN – T) / (RN dx A Cp) – dt (T – TS) / (RS dx A Cp) + T (15)After T’ is calculated for a given time step dt, it can be substituted into T in Equation 15 to calculate the next T’ value in the following time iteration. This step is repeated until enough time iterations have been performed to equal the time duration of a cycle. The temperatures of each node at the end of the cycle are calculated by performing all time iterations.The total energy absorbed by the thermal mass during the cycle, QTM, can be calculated by summing the energy at all nodes relative to their energy before the cycle. This calculation involves each node temperature at the end of the cycle, Tend, and at the beginning of the cycle, Tbeg. Equation 16 calculates the total energy absorbed by the thermal mass during the cycle, Qc.Qc = ? dx A p cp (Tend – Tbeg)(16)ExampleConsider a heat treat oven whose floor is mounted on rails so that it can be slid into and out of the oven. The floor is 100 ft2 in area, and consists of fire bricks on top of a thick layer of insulation. The bricks are laid vertically such that the firebrick floor is eight inches thick. The density of firebrick is 36 lb/ft3, the specific heat is 0.25 Btu/lb-F, and the thermal conductivity is 1.2 Btu-in/ft2-hr-F. The heat treat cycle consists of raising the temperature to 1,700 F for four hours, then decreasing the oven temperature by about 50 F per hour until the oven reaches 400 F. As an energy-savings measure, the 8-inch thick firebrick is replaced with 4-inch firebrick. Calculate the energy savings pwer cycle from replacing the 8-inch floor with a 4-inch floor.Using the methodology shown above, the temperature profile of 8-inch firebrick and 4-inch firebrick at the conclusion of the heating cycle is shown below. Temperature profile of 8-inch and 4-inch thick firebrick after heat treat cycleUsing Equation 16, the energy absorbed by the 8-inch firebrick after the cycle is 389,600 Btu and the energy absorbed by the 4-inch firebrick is 154,400 Btu. The savings would be 235,200 Btu per cycle, or 60% of the energy absorbed by the original floor.Reduce Heating Energy Use By Improving Heat TransferIn process heating applications where combustion gasses transfer heat to a product, improved heat transfer effectiveness increases heat delivered to the product and reduces the heat lost in exhaust. This improves system efficiency. Consider the following graphs of heat transfer effectiveness for counter flow, cross flow and parallel flow heat exchanger configuration. The effectiveness of counter flow heat exchange is always greater than the effectiveness of cross flow or parallel flow heat exchange. Figure 11. Heat transfer effectiveness for counter flow, cross flow and parallel flow heat exchange (Incropera and DeWitt, 1996). The equations for heat exchanger effectiveness are:Ch = mh * cph Cc = mc * cpc Cmin = min(Ch, Cc)Cmax = max(Ch, Cc)Cr = Cmin / CmaxNTU = UA/CminParallel flow: e = [1-exp( -NTU (1+Cr) )] / (1+Cr)Cross flow: e = 1-exp[(1/Cr)*(NTU0.22)*{exp((-Cr)*(NTU0.78))-1}]Counter flow: e = [1 - exp(-NTU (1-Cr))] / [1 – Cr*exp(-NTU (1 - Cr))] (use Cr = 0.999 when Cr = 1.0)Q = e*Cmin*(Th1-Tc1)ExampleA curing oven has burners at each end and an exhaust fan in the middle. The valve at the exit of the exhaust fan is adjusted so that no air leaks into or out of the openings at the ends of the oven. 400 lb/min of carbon steel (cp = 0.12 Btu/lb-F) enters the oven at 70 F. The surface area of the steel in the oven is 1,000 ft2 and the convection coefficient in the oven is 10 Btu/hr-ft2-F. In this configuration, each burner consumes 1 mmBtu of fuel. The temperature of the combustion gasses leaving the burners is 3,382 F and the specific heat of the combustion gasses is 0.30 Btu/lb-F. Calculate the fuel savings from reconfiguring the oven so that all of the combustion gasses travel in a direction counter-flow to the product as shown below.The mass flow rate of combustion air traveling through the first half of the oven is:In the first half of the oven, the combustion gasses primarily travel in the same direction as the product creating a parallel-flow heat exchanger. Using the heat exchanger equations shown above, the properties of the combustion gasses and steel can be calculated as shown below. By the midpoint of the oven, the temperature of the carbon steel increased to 355 F, the temperature of the combustion gasses decreased to 355 F, the quantity of heat transferred to the carbon steel is 822,049 Btu/hr. The effectiveness of the heat exchange process is 91.4%. To determine the fuel use if the oven was reconfigured as a counter-flow heat exchanger, the equation for parallel flow effectiveness is replaced with the equation for counter-flow effectiveness. Then the quantity of fuel use is decreased until the temperature of the carbon steel at the mid-point is 355 F. The results are shown below.In a counter-flow configuration, fuel use was decreased to 910,000 Btu/hr and the mass flow rate of combustion air through the first half of the oven is:By the midpoint of the counter-flow oven, the temperature of the carbon steel reaches 354 F, the temperature of the combustion gasses has decreased to 70 F, the quantity of heat transferred to the carbon steel is 818,619 Btu/hr. The effectiveness of the heat exchange process was 100%. The fuel savings are:Qf,sav = Qf1 – Qf2 = 1,000,000 Btu/hr – 910,000 Btu/hr = 90,000 Btu/hrThus, in this case, reconfiguring the oven from parallel to counter-flow decreased fuel use a by about 9%.Reducing Heat Loss Due to Excess Combustion AirThe minimum amount of air required for complete combustion is called the “stoichiometric” air. Combustion efficiency is maximized when stoichiometric air is supplied to the fuel. Supplying less than stoichiometric air causes some fuel to be exhausted without combusting, which increases fuel use, air pollution and the danger of uncontrolled combustion in the exhaust stack. Supplying more than stoichiometric air dilutes the combustion gasses, lowers combustion temperature and reduces combustion efficiency. Thus, best practice is to supply only slightly more oxygen (combustion air) than the stoichiometric minimum, so that unburned fuel is avoided and the efficiency penalty from excess combustion air is minimized.The common target of excess combustion air to guarantee complete combustion is about 10% (EPA, 2001). This produces combustion gasses with about 1.7% O2 content when combusting natural gas. However, in well-controlled burners, it is possible to reduce excess combustion air even lower. This can be achieved by reducing the air fuel ratio until just before CO levels, which are indicators of unburned hydrocarbons, grow exponentially as shown in the graph below.Source: Thekdi, 2004ExampleA well-sealed melting furnace burns 10 mmBtu/hr of natural gas. Combustion air enters the burner at 70 F. A combustion analysis of exhaust gasses shows that the flue temperature is 1,600 F and the quantity of excess air is 30%. Calculate the current combustion efficiency, the combustion efficiency if the excess air were reduced to 10%, and the resulting fuel savings.From Equations 22 and 23, the current combustion efficiency of the melt furnace is:Tc1 = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg]Tc1 = 70 F + 21,500 Btu/lbng / [{1 + (1.30) 17.2 lba/lbng} 0.26 Btu/lba-F] = 3,610 F?? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV ?? = [{1 + (1.30) 17.2 lba/lbng} 0.26 Btu/lba-F (3,610 – 1,600 F)] / 23,900 Btu/lbm?? = 51.1%The heat delivered to the system, Qout, would be:Qout = Qf1 ?? = 10 mmBtu/hr x 51.1% = 5.11 mmBtu/hr If excess air were reduced to 10%, the combustion efficiency would increase to:Tc2 = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg]Tc2 = 70 F + 21,500 Btu/lbng / [{1 + (1.1) 17.2 lba/lbng} 0.26 Btu/lba-F] = 4,221 F?? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV ?? = [{1 + (1.1) 17.2 lba/lbng} 0.26 Btu/lba-F (4,221 – 1,600 F)] / 23,900 Btu/lbm?? = 56.8%The fuel energy input, Qf2, to delivered the same energy output to the system would would be:Qf2 = Qout / ?? = 5.11 mmBtu/hr / 56.8% = 8.99 mmBtu/hrThe fuel energy savings, Es, from reducing excess combustion air would be: Es = Qf1 – Qf2 = 10 mmBtu/hr – 8.99 mmBtu/hr = 1.01 mmBtu/hrThus, this measure would reduce fuel use by 10.1% Reducing Heat Loss with Oxygen Enhancement When atmospheric air is used in combustion, the nitrogen in the air is largely inert. However, this does not mean that it has no effect on combustion. The energy released during the combustion reaction is absorbed by the products of combustion. The nitrogen in the products of combustion absorbs a portion of the heat of combustion and lowers the adiabatic combustion temperature. The reduced combustion temperature reduces the efficiency of combustion.Burning pure oxygen instead of atmospheric air eliminates the dilutive effects of nitrogen. This increases the combustion temperature and the efficiency of combustion. Using the method developed earlier, the air to fuel ratio for stochiometric combustion of methane with oxygen combustion is:CH4 + 2 O2 > CO2 + 2 H2OMair / Mfuel = (2 x 2 x 16) / (12 + 4) = 4.0 This value can be used in conjunction with the previous simplified equations for combustion temperature and efficiency of combustion with oxygen. Combustion efficiencies calculated using the simplified method compare well with values from the graph below. The primary reason for the small discrepancies between the simplified method and the graph shown below is that the simplified method is based on combustion of pure methane while the graph below is for natural gas, which includes a small percentage of other hydrocarbons in addition to methane. The lower fuel costs from oxygen enhancement must be compared to the cost of oxygen to determine the cost effectiveness of this option. In general, oxygen enhancement is most cost effective in high temperature applications that use large quantities of gas.ExampleConsider a glass melting furnace using 10 mmBtu/hr of natural gas. The temperature of O2 entering the burner is 70 F, and the temperature of exhaust leaving the furnace is 2,000 F with 20% excess air. Calculate the fuel energy savings (mmBtu/hr) from converting from air-fired to oxy-fired burners assuming 0% excess oxygen in the exhaust and the temperature of exhaust leaving the furnace remains 2,500 F.From Equations 22 and 23, the current combustion efficiency of the glass melting furnace is:Tc1 = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg]Tc1 = 70 F + 21,500 Btu/lbng / [{1 + (1.20) 17.2 lba/lbng} 0.26 Btu/lba-F] = 3,891 F?? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV ?? = [{1 + (1.2) 17.2 lba/lbng} 0.26 Btu/lba-F (3,891 F – 2,000 F)] / 23,900 Btu/lbm?? = 44.5%The heat delivered to the system, Qout, would be:Qout = Qf1 ?? = 10 mmBtu/hr x 44.5% = 4.45 mmBtu/hr If the burner were converted to burn oxygen with 0% excess air and the same exhaust temperature, the combustion efficiency would increase to:Tc2 = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg]Tc2 = 70 F + 21,500 Btu/lbng / [{1 + (1.0) 4.0 lba/lbng} 0.26 Btu/lba-F] = 16,608 F?? = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV ?? = [{1 + (1.0) 4.0 lba/lbng} 0.26 Btu/lba-F (16,608 F – 2,000 F)] / 23,900 Btu/lbm?? = 79.5%The fuel energy input, Qf2, to delivered the same energy output to the system would would be:Qf2 = Qout / ?? = 4.45 mmBtu/hr / 79.5% = 5.60 mmBtu/hrThe fuel energy savings, Es, from reducing excess combustion air would be: Es = Qf1 – Qf2 = 10.00 mmBtu/hr – 5.60 mmBtu/hr = 4.40 mmBtu/hrThus, this measure would reduce fuel use by 44.0% Pre-heat Combustion Air Heat exchangers can be employed to reclaim heat escaping in exhaust gasses to pre-heat incoming combustion or ventilation air. Pre-heating combustion/ventilation air is frequently cost-effective because the close proximity of the exhaust and combustion air streams reduces the cost of the required heat reclaim equipment. Following the heat-exchanger effectiveness method derived in the Energy Efficient Steam Systems” chapter, actual heat transfer, Qact, is the product of heat exchanger effectiveness, e, the minimum mass capacitance of the two streams, mcp,min, and the difference between the entering temperatures of the hot and cold streams, Th1 and Tc1:Qact = e mcp,min (Th1 – Tc1)It follows that the exit temperatures of each stream are given by:Tc2 = Tc1 + e mcp,min (Th1 – Tc1) / mcpcTh2 = Th1 - e mcp,min (Th1 – Tc1) / mcphPreheating Combustion Air with Heat Exchanger with Known EffectivenessHeat exchangers are typically designed with sufficient heat transfer area such that the effectiveness of the heat exchanger is between about 0.6 and 0.9. At higher levels of heat exchanger effectiveness, the cost of the required surface area frequently outweighs the additional performance benefits. In addition, heat exchanger designers must also ensure that the pressure drop through each side of the heat exchanger is acceptably small, and that the materials can withstand the temperatures, fouling and corrosiveness of the fluids involved. Finally, the effectiveness of the heat exchanger must be chosen so that the exit temperatures are within acceptable ranges. When pre-heating combustion air, the temperature of the pre-heated combustion air must be not exceed the maximum temperature allowed by the burner or else a new burner is needed. Many burners can accept combustion air up to about 500 F without modification. In addition, the effectiveness of the heat exchanger cannot be so large that water in the exhaust gasses condenses. The dew-point temperature of water vapor in combustion exhaust is about 180 F. ExampleConsider reclaiming heat from furnace exhaust at 1,000 F to preheat combustion air initially at 80 F. The furnace consumes 10,000,000 Btu/hr of natural gas. The fraction excess air in the exhaust is measured to be 20%. Calculate the rate of energy savings (Btu/hr) and the temperatures of the two streams leaving the heat exchanger if the heat exchanger is 50% effective.Thus, this would decrease fuel use by about 10.3%. Also note that the combustion air would be heated to 540 F, which is close to the maximum allowable combustion air temperature for many burners. The exit temperature of the combustion gasses is 561 F, which ensures that the water in the combustion gasses will not condense.Pre-heating Combustion Air Using a Tube-in-Tube Heat ExchangerAn inexpensive heat exchanger for reclaiming heat from exhaust gasses to pre-heat incoming combustion air can be fabricated by enclosing an un-insulated exhaust duct with a second duct and directing combustion air in a counter-flow direction through the annular space. In this case, the effectiveness of the fabricated heat exchanger must be calculated by calculating the convection coefficients and surface area.The reference diameter, Dr, to calculate the convection coefficient for a fluid travelling through a tube is the diameter of the tube, D1. The reference diameter Dr to calculate the convection coefficient for a fluid travelling through the annular space between two tubes is (Dr = D2 – D1) where D1 is the diameter of the inside tube and D2 is the diameter of the outside tube. The convection coefficient, h, for a fluid travelling through a tube or the annular space between two tubes is given by the following equations:Re = 4 M / (?*DV*Dr)Nu = .023*(Re^.8)*(Pr^.4)h = Nu*k/Dhwhere Dr is the reference diameter, Re is the Reynolds number, M is the mass flow rate of the fluid, DV is the dynamic viscosity of the fluid, Pr is the Prandtl number of the fluid and k is the conductivity of the fluid. The reference diameter, Dr, for a fluid travelling through a tube is the diameter of the tube, D1. The reference diameter Dr for a fluid travelling through the annular space between two tubes is (Dr = D2 – D1) where D1 is the diameter of the inside tube and D2 is the diameter of the outside tube. Assuming that the thermal resistance of the inner tube is negligible compared to the thermal resistance caused of the convection coefficients, the thermal resistance, R, and conductance, U, of the tube-in-tube heat exchanger is:R = (1/h1) + (1/h2)U = 1 / [ (1/h1) + (1/h2) ] Heat exchanger effectiveness, e, for a counter-flow heat exchanger between cold, c, and hot, h, streams can be calculated using the following equations:Ch = mh * cph Cc = mc * cpc Cmin = min(Ch, Cc)Cmax = max(Ch, Cc)Cr = Cmin / CmaxN = UA/Cmine = [1-exp(N(1-Cr))]/[1-Cr*exp(-N(1-Cr))]where m is the mass flow rate, cp is the specific heat, U is the conductance of the heat exchanger, and A is the area of the heat exchanger. ExampleConsider reclaiming heat from furnace exhaust at 1,100 F to preheat combustion air initially at 80 F. The furnace consumes 1,200,000 Btu/hr of natural gas. The fraction excess air in the exhaust is measured to be 44%. The heat will be reclaimed by fabricating a 25 ft long, 2 ft diameter duct around the outside of the existing 1 ft diameter exhaust duct. Calculate the rate of energy savings (Btu/hr) and the temperatures of the two streams leaving the heat exchanger.This would decrease fuel use by about 6.2%. Pre-heat Load In some furnaces, the load moves via conveyor into a heat treating or melting furnace. In these cases, an inexpensive heat exchanger for reclaiming heat from exhaust gasses to pre-heat the load can be fabricated by enclosing the load carrying conveyor and directing exhaust gasses in a counter-flow direction over the load.In this case, the heat transfer area is the exposed area of the load, and the conductance, U, is the convection coefficient, h. The heat exchanger effectiveness method can be used to calculate heat transfer to the load, Qsav. Fuel energy savings, Esav, should take into account the combustion efficiency, ?, since some fuel energy is lost with the exhaust gasses.Esav = Qsav / ?ExampleA conveyor carries 20,000 lb/hr of lead ingots at 80 F into an open melt furnace with a hood enclosure. Currently, 6,000 cfm of 500 F air is collected by the hood and exhausted to atmosphere. The overall efficiency of the melt furnace is 60%. Consider reclaiming heat by enclosing the lead-carrying conveyor and directing exhaust gasses in a counter-flow direction over the lead. The enclosure could be approximated as having a diameter of 4 ft, and the lead ingots on the conveyor can be approximated as a tube having a diameter of 2 ft. If so, the exposed area of the lead ingots on the conveyor is 160 ft2. Calculate the rate of fuel energy savings (Btu/hr) and the temperature of the lead ingots when they leave the heat exchanger.The temperature of the lead ingots would be 383 F, which is below the melting point of lead. Cascade HeatWhen reclaiming heat, it is generally preferable to use the heat in the process that created it in order to minimize transportation costs and losses and reduce control complexity. However, when exhaust heat from one process cannot be used internally, heat may be cascaded to another process operating at a lower temperature.ExampleConsider an oven operating at 500 F and an adjacent oven operating at 300 F. Fuel use in both ovens is 200,000 Btu/hr. Both exhaust air steams have 40% excess air and the combustion air and natural gas enter the burner at 70 F. Calculate the fuel savings (Btu/hr) from cascading the exhaust from the 500 F oven to the 300 F oven by installing a duct from the exhaust of the 500 F oven into the 300 F oven.The quantity of air exhausted by the 500 F furnace is:The efficiency of the 300 F furnace is:From an energy balance on the exhaust gasses entering and leaving the 300 F furnace, the savings are: ................
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