Energy-Efficient Process Heating
Introduction
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 Heating
Heat Exchanger EffectivenessApproach
In 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 transferfrom 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 Tpincreases 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 Approach
Minimizing 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, 2004
Opportunities for Improving The Energy-Efficiency of Process Heating Systems
These 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 loads
- Insulate hot surfaces to reduce wall losses
- Cover openings to reduce radiation losses
- Lower an opening to decrease temperature-driven infiltration
- Seal openings or use draft control to reduce infiltration
- Reduce cooling losses by controlling cooling water temperature
- Reduce conveyance losses in continuous and batch processes
- Reduce storage losses in batch processes
- Improve efficiency of distribution system
- Convert to counter flow heat exchange
- Improve efficiency of energy conversion
- Reduce excess combustion air
- Convert from atmospheric to oxygen burners
- Pre-heat combustion air
- Pre-heat the load
- Cascade waste heat to lower temperature process
The sections that follow discuss how to quantify these saving opportunities.
Combustion Efficiency
Many 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)
Example
An 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 Walls
Hot 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)
whereTf 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 < 63 / Turbulent: L3 T > 63Horizontal Surfaces / h = 0.27 x (T / L) 0.25 / h = 0.22 x (T) 0.33
Vertical Surfaces / h = 0.29 x (T / L)0.25 / h = 0.19 x [T(sin B)]0.33
Typical thermal resistances and costs of common types of insulation are listed in the table below.
Insulation Type / Thermal Resistance / Cost3.5-inch fiberglass batt / 11 ft2-hr-F/Btu / $0.30 / ft2
¾-inch rigid blue Styrofoam board / 5.2 ft2-hr-F/Btu / $0.32 /ft2
½-inch rigid polyisocyanurate with foil facing / 3.5 ft2-hr-F/Btu / $0.23 /ft2
1-inch spray-on cellulose (meets fire-code) / 5 ft2-hr-F/Btu / $0.75 /ft2 *
2-inch spray-on polyurethane / 9 ft2-hr-F/Btu / $4 /ft2 *
2-inch steam pipe and tank insulation / 5 ft2-hr-F/Btu / $10 /ft2 *
*includes installation
Insulating Fire Brick / BNZ-20 / BNZ-23 / BNZ-26 / BNZ-28 / BNZ-30 / BNZ-32Temperture Use Limit (F) / 2,300 / 2,300 / 2,600 / 2,800 / 3,000 / 3,200
Max Mean Brick Temperture (F) / 1,800 / 2,100 / 2,200 / 2,200 / 2,400 / 2,600
Density (lbm/ft3) / 36 / 37 / 48 / 55 / 65 / 75
Conductivity (Btu-in/ft2-hr-F)
@500 F / 0.9 / 1.0 / 1.6 / 2.3 / 2.8 / 3.9
@1,000 F / 1.2 / 1.3 / 1.9 / 2.4 / 2.9 / 4.1
@1,500 F / 1.5 / 1.6 / 2.2 / 2.6 / 3.1 / 4.2
@2,000 F / 1.7 / 1.8 / 2.6 / 2.7 / 3.3 / 4.3
Specific Heat (Btu/lbm-F) / 0.26 / 0.26 / 0.26 / 0.26 / 0.26 / 0.26
Source: BNZ Materials, Inc.
Ceramic BlanketsDensity / 8 lb/ft3 (128 kg/m3)
Max Continuous use limit / 2,000 F (1,093 C)
k (Btu-in./hr-ft2-°F) / R (hr-ft2-F/Btu)/inch
at 500°F (260°C) / 0.44 / 2.27
at 1000°F (538°C) / 0.87 / 1.15
at 1500°F (816°C) / 1.45 / 0.69
at 1800°F (982°C) / 1.83 / 0.55
at 2000°F (1093°C) / 2.09 / 0.48
Source: 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.
Example
Consider a rectangularheat 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 turbulent
h1= 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) / Q1
R1 = 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) 0.33
This system of two equations and two unknowns can be solved to give:
Ts2 = 169.33 F
h2 = 0.867 Btu/hr-ft2-F
Substituting Ts2 and h2into 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/hr
The heat loss savings, Qs, and fuel energy savings, Es, from adding insulation would be:
Qs = Q1 – Q2 =184,028 Btu/hr –82,527Btu/hr = 101,501 Btu/hr
Es= Qs / = 101,501 Btu/hr / 50% = 203,001 Btu/hr
Thus, this measure would reduce heat loss through the walls by 55.2%
Reducing Radiant Heat Loss from Walls
Hot 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.
Example
Consider 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-ft2
The 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 / = 195Btu/hr-ft2 / 50% = 390Btu/hr-ft2
Thus, this measure would reduce heat loss through the walls by 72%
Reducing Radiation Loss through Openings
Openings 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 = AFfa (Tf4 – Ta4) (4)
whereis 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)2
Ffa = 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)
Example
Consider a furnace with an opening 1ft indiameter 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. Calculatethe 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 ft2
From Equation 5, the view factor for radiation escaping through the opening is:
S = 2 + (L /r)2 = 2 + (1 /0.5)2 = 6.0
Ffa = 0.5 [ S – (S2 – 4)0.5 ] = 0.5 [ 6.0 – (6.02 – 4)0.5 ] = 0.172
From 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)
wherepcp 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 Floor
In 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.
Example