Wednesday, May 4, 2011
Heater Design: Heater Stack Draft Analysis
Fired heaters come in numerous configurations and designs. These configurations include systems that are forced draft, induced draft or a combination of both, but many are natural draft. The natural draft designs are what we will concentrate on in this section.
Regardless of the mechanics involved, the stacks purpose is the same, to safely disperse the products of combustion into the atmosphere. For environmental reasons, many stacks are required to discharge at a particular height. But in most cases they are designed only to meet the needs of the furnace or furnaces they are designed for. In the case of the natural draft furnace, the stack serves another purpose, that of assuring that the furnace stays below atmospheric pressure throughout the setting.
For most stack designs, a gas velocity at the exit of about 15 to 25 ft/sec is sufficient to discharge the gasses into the atmosphere at a rate that will assure they disperse properly. Additionally, most natural stack are designed for 125% of the design flue gas flow to assure that if the furnace is operated above the design point that it will still operate safely.
We can use the sketch of a typical horizontal cabin heater to look at the important features affecting the heater draft.
 | Natural Draft In this sketch, the area marked "A" is the height available, for the differences in the density of the ambient air and the flue gas, to create the draft required. Normally the draft required is that which will result in a slightly negative pressure at point "B". It should be noted that, for most heaters, the draft at point "C" required by the burners to induce combustion air is not considered in setting the stack height. The burners are normally sized to use only the draft in the firebox. |
Net loss across convection:
The calculation of this pressure loss was covered in detail on page 4 of this section. This loss is discounted by the draft gain across the convection section.
Pressure loss across stack entry:
This pressure loss can normally be considered as a sudden entry since the area of the outlet gas plenum in the heater is usually much greater than the area of the inlet to the transition. A sudden entry pressure loss can be approximated by the following equation.
Where,
| Dp = Pressure drop, inH2O |
| Vh = Velocity head at inlet area, inH2O |
Pressure loss across stack transition:
This pressure loss can normally be considered as a gradual contraction since the area of the inlet and the outlet are usually close in area. A gradual contraction pressure loss can be approximated by the following equation.
Where,
| Dp = Pressure drop, inH2O |
| Vh = Velocity head at outlet area, inH2O |
| Ca = Coefficient based on included angle |
And the coefficient can be described as,
| Included Angle | Ca |
| 30 | 0.02 |
| 45 | 0.04 |
| 60 | 0.07 |
Pressure loss across stack damper:
This pressure loss is normally accounted for by rule of thumb. This may be 0.5 or 0.25 velocity head. We will use 0.25.
Where,
| Dp = Pressure drop, inH2O |
| Vh = Average velocity head of stack, inH2O |
Stack friction loss:
For the stack friction loss, we can use the following equation.
Where,
| Dp = Pressure drop, inH2O |
| Vg = Average velocity of stack, ft/sec |
| rg = Density of flue gas, lb/ft3 |
| Ds = Stack diameter, ft |
| Ls = Stack length, ft |
Stack draft gain:
The draft gain will be taken based on the height, "A" on above sketch.
Where,
| Gd = Draft Gain, inH2O |
| rg = Density of flue gas, lb/ft3 |
| ra = Density of ambient air, lb/ft3 |
| A = Height of gas path, ft |
Pressure loss across stack exit:
This pressure loss, since it normally exits to atmosphere, can be considered as a sudden exit. A sudden exit pressure loss can be approximated by the following equation.
Where,
| Dp = Pressure drop, inH2O |
| Vh = Velocity head at inlet area, inH2O |
Velocity head of gas:
Using these formulas, we can now determine the height that A is needed to be to provide the draft required. It should be noted that many heaters are not this simple, in that they may have multiple stacks, collection ducts, etc.
Heater Design: Gas Side Pressure Drop Across Tubes, Online Calculation
The gas side pressure drop may be calculated by any number of methods available today, but the following procedures should give sufficient results for heater design.
| Bare Tube Pressure Loss | ||
| Fin Tube Pressure Loss | ||
| Stud Tube Pressure Loss |
Bare Tube Pressure Loss:
For bare tubes we can use the method presented by Winpress(Hydrocarbon Processing, 1963),
Where,
| Dp = Pressure drop, inH2O |
| Pv = Velocity head of gas, inH2O |
| Nr = Number of tube rows |
And the velocity head can be described as,
Where,
| Gn = Mass velocity of gas, lb/hr-ft2 |
| rg = Density of gas, lb/ft3 |
The Mass velocity is described as,
Where,
| Wg = Mas gas flow, lb/hr |
| An = Net free area, ft2 |
For staggered tubes without corbels,
For staggered tubes with corbels or inline tubes,
Where,
| Ad = Convection box area, ft2 |
| do = Outside tube diameter, in |
| Le = Tube length, ft |
| Pt = Transverse pitch of tubes, in |
| Nt = Number of tubes per row |
We can now use the following script to try some calculations,
Fin Tube Pressure Loss:
For the fin tube pressure drop, we will use the Escoa method.
And,
For staggered layouts,
For inline layouts,
And,
Where,
| Dp = Pressure drop, inH2O |
| rb = Density of bulk gas, lb/ft3 |
| rout = Density of outlet gas, lb/ft3 |
| rin = Density of inlet gas, lb/ft3 |
| Gn = Mass gas flow, lb/hr-ft2 |
| Nr = Number of tube rows |
| do = Outside tube diameter, in |
| df = Outside fin diameter, in |
And,
For staggered tubes without corbels,
For staggered tubes with corbels or inlune tubes,
Net Free Area, An:
Where,
| Ad = Cross sectional area of box, ft2 |
| Ac = Fin tube cross sectional area/ft, ft2/ft |
| Le = Effective tube length, ft |
| Nt = Number tubes wide |
| And, |
| Ac = (do + 2 * lf * tf * nf) / 12 |
| tf = fin thickness, in |
| nf = number of fins, fins/in |
Reynolds correction factor, C2:
And,
Where,
| mb = Gas dynamic viscosity, lb/ft-hr |
Geometry correction, C4:
For segmented fin tubes arranged in,
C4 = 0.11*(0.0 5*Pt/do)(-0.7*(lf/sf)^0.23)
C4 = 0.08*(0. 15*Pt/do)(-1.1*(lf/sf)^0.20)
For solid fin tubes arranged in,
C4 = 0.11*(0.0 5*Pt/do)(-0.7*(lf/sf)^0.20)
C4 = 0.08*(0. 15*Pt/do)(-1.1*(lf/sf)^0.15)
Where,
| lf = Fin height, in |
| sf = Fin spacing, in |
Non-equilateral & row correction, C6:
For fin tubes arranged in,
C6 = 1.1+(1.8-2.1*e(-0.15*Nr^2))*e(-2.0*Pl/Pt) - (0.7*e(-0.15*Nr^2))*e(-0.6*Pl/Pt)
C6 = 1.6+(0.75-1.5*e(-0.70*Nr))*e(-2.0*(Pl/Pt)^2)
Where,
| Nr = Number of tube rows |
| Pl = Longitudinal tube pitch, in |
| Pt = Transverse tube pitch, in |
We can now use the following script to try some calculations,
Stud Tube Pressure Loss:
For the stud tube pressure loss we will use the Muhlenforth method,
The general equation for staggered or inline tubes,
Where,
| Dp = Pressure drop across tubes, inH2O |
| Nr = Number of tube rows |
| Cmin = Min. tube space, diagonal or transverse, in |
| do = Outside tube diameter, in |
| ls = Length of stud, in |
| G = Mass gass velocity, lb/sec-ft2 |
| Tg = Average gas Temperature, °F |
Correction for inline tubes,
And,
Where,
| Wg = Mass flow of gas, lb/hr |
| An = Net free area of tubes, ft2 |
| Le = Length of tubes, ft |
| Nt = Number of tubes wide |
| Pt = Transverse tube pitch, in |
| ls = Length of stud, in |
| ts = Diameter of stud, in |
| rs = Rows of studs per foot |
We can now use the following script to try some calculations,
Heat Exchanger Tube Pressure Drop Calculation, Online Calculator
The intube pressure drop may be calculated by any number of methods available today, but the following procedures should give sufficient results for heater design. The pressure loss in heater tubes and fittings is normally calculated by first converting the fittings to an equivalent length of pipe. Then the average properties for a segment of piping and fittings can be used to calculate a pressure drop per foot to apply to the overall equivalent length. This pressure drop per foot value can be improved by correcting it for inlet and outlet specific volumes.
Friction Loss:
Where,
| Dp = Pressure drop, psi |
| di = Inside diameter of tube, in |
| G = Mass velocity of fluid, lb/sec-ft2 |
| Vlm = Log mean specific volume correction |
| F = Fanning friction factor |
| Lequiv = Equivalent length of pipe run, ft |
And,
For single phase flow,
| V1 = Specific volume at start of run, ft3/lb |
| V2 = Specific volume at end of run, ft3/lb |
For mixed phase flow,
Where,
| Vi = Specific volume at point, ft3/lb |
| Tf = Fluid temperature, °R |
| Pv = Press. of fluid at point, psia |
| MWv = Molecular weight of vapor |
| Vfrac = Weight fraction of vapor %/100 |
| rl = Density of liquid, lb/ft3 |
Fanning Friction Factor:
The Moody friction factor, for a non-laminar flow, may be calculated by using the Colebrook equation relating the friction factor to the Reynolds number and relative roughness. And the Fanning friction factor is 1/4 the Moody factor. For a clean pipe or tube, the relative roughness value for an inside diameter given in inches is normally 0.0018 inch.
With this, we can calculate the factor,
Equivalent Length Of Return Bends:
The equivalent length of a return bend may be obtained from the following curves based on Maxwell table and can be corrected using the Reynolds number correction factor.
Where,
| FactNre = Reynolds number correction |
| Lrb = Equivalent length of return bend, ft |
Return Bend Equivalent Length:
Reynolds Correction:
| G = Mass velocity, lb/sec-ft2 |
| Di = Inside tube diameter, in |
| Visc = Viscosity, cp |
Now that we have all the details described, we can calculate the pressure drop for some typical heater coils.
Tuesday, May 3, 2011
Online Heat Transfer Coefficient Calculator
The inside film coefficient needed for the thermal calculations may be estimated by several different methods. The API RP530, Appendix C provides the following methods,
For liquid flow with Re =>10,000,
And for vapor flow with Re =>15,000,
Where the Reynolds number is,
And the Prandtl number is,
Where,
| hl = Heat transfer coefficient, liquid phase, Btu/hr-ft2-°F |
| k = Thermal conductivity, Btu/hr-ft-°F |
| di = Inside diameter of tube, ft |
| mb = Absolute viscosity at bulk temperature, lb/ft-hr |
| mw = Absolute viscosity at wall temperature, lb/ft-hr |
| hv = Heat transfer coefficient, vapor phase, Btu/hr-ft2-°F |
| Tb = Bulk temperature of vapor, °R |
| Tw = Wall Temperature of vapor, °R |
| G = Mass flow of fluid, lb/hr-ft2 |
| Cp = Heat capacity of fluid at bulk temperature, Btu/lb-°F |
For two-phase flow,
Where,
| htp = Heat transfer coefficient, two-phase, Btu/hr-ft2-°F |
| Wl = Weight fraction of liquid |
| Wv = Weight fraction of vapor |
The following script will allow us to try these formulas out using our browser.
It should be stressed at this time, that there are many ways to calculate the inside heat transfer coefficient, and a lot of care should be taken in the procedure selected for use in heater design. Other methods, such as HTRI, Maxwell, Dittus-Boelzer, or others may be more appropriate for a particular heater design.
Monday, May 2, 2011
Heat Loss Calculations and System Design
Heat Loss Calculations and System Design
Note: Urecon offers a complete computer assisted engineering service providing such information as: heat loss, time to freeze, fluid outlet temperature, minimum flow rates, tracing wattage required, heat gain, etc., all based on the specific requirements of each project. Basic information needed for typical heat trace design includes: project name/location; minimum ambient temperature; above and/or below ground; depth of bury if applicable; core pipe material and diameter; pipe length per circuit; insulation thickness; required maintain temperature; flow direction for sensor positioning; power point location (one or both ends, middle etc.); voltage available.
Contact Urecon for more info and help with custom design/calculations.
It is recommended that a safety factor of 10 to 25% be added to allow for such field conditions as voltage drop, under voltage condition, etc.
Heat Flow Chart Watts/ft/hr/100°F 
T(2)
T(2)| Dia.(1) | Urethane Insulation Thickness | ||||
| 25 mm (1 in) | 40 mm (1½ in) | 50 mm (2 in) | 63 mm (2½ in) | 75 mm (3 in) | |
| 1 | 1.8 | 1.4 | 1.2 | 1.1 | 1.0 |
| 2 | 2.9 | 2.2 | 1.8 | 1.6 | 1.4 |
| 4 | 4.9 | 3.6 | 2.9 | 2.5 | 2.2 |
| 6 | 6.9 | 4.9 | 3.9 | 3.3 | 2.9 |
| 8 | 8.9 | 6.3 | 4.9 | 4.1 | 3.6 |
| 10 | n/a | 7.6 | 5.9 | 4.9 | 4.2 |
| 12 | n/a | 9.0 | 6.9 | 5.8 | 4.9 |
| 14 | n/a | 10.4 | 7.9 | 6.5 | 5.6 |
| 16 | n/a | n/a | 8.9 | 7.4 | 6.2 |
| 18 | n/a | n/a | 9.9 | 8.3 | 6.9 |
| 20 | n/a | n/a | 10.8 | 9.0 | 7.6 |
| 22 | n/a | n/a | 11.8 | 9.9 | 8.3 |
| 24 | n/a | n/a | 12.7 | 10.8 | 9.0 |
Heat flow in watts per lineal foot
The table is based upon the application of the following formula:
Where:
W = Watts/ft/hr (W x 3.414 = Btu/hr)
K = Btu/ft²/hr/1°F/ft = 0.0108 for Urethane
T = temperature differential°F
D = outside diameter of insulation
d = outside diameter of pipe
Diameters for (D/d) taken as 3/1 for 1" pipe +1" insulation and is typical for all other combinations.
For other than 100°F
T, divide by 100 and multiply by required T. Formula
The heat loss for an externally traced pipe may be calculated by the following formula:
Where:
W = Watts per foot of pipe
Tm = maintained temperature°F
Ta = ambient temperature°F
Ln = natural log
Di = outside diameter of insulation (in)
Dp = outside diameter of pipe (in)
Ki = K value of insulation (BTU • in / hr • ft² •°F)
Dj = outside diameter of jacket (in)
Kj = K value of jacket (BTU • in / hr • ft² •°F)
Sf = Safety Factor
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