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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.

The resistance to the draft is the pressure loss across the tubes in the convection, the entry to the stack, the transition to the stack, the damper obstruction, the friction loss, and the stack exit.


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.



D_p = 0.34 * V_h
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.



D_p = C_a * V_h


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.


D_p = 0.25 * V_h


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.



D_p = (0.002989 * 0.018 * r_g * V_g²) / D_s * L_s


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.



G_d = (r_a - r_g)/5.2 * A


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.
D_p = 1.0 * V_h


Where,

Dp = Pressure drop, inH2O

Vh = Velocity head at inlet area, inH2O

Velocity head of gas:



V_h = V_g² * r_g / 2 / 32.2 / 144 * 27.67783


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.

Stack Data
Stack inside daimeter, ft:	Include stack damper: Yes No
Transition height, ft:	Convection plenum height, ft:
Transition included angle, °: 30 45 60	Transition inlet area, ft2:
Process Data
Flue gas flow, lb/hr:	Density of flue gas, lb/ft3:
Density of ambient air, lb/ft3:	Net Convection loss, inH2O:
Pressure under shield, inH2O:	Flue gas multiplier:

Stack Length above transition, ft:


Stack velocity, ft/sec:


Design, ft/sec:

Labels: heater, stack

Gas Side Pressure Drop Across Tubes


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),


D_p = P_v /2 * N_r


Where,

Dp = Pressure drop, inH2O

Pv = Velocity head of gas, inH2O

Nr = Number of tube rows

And the velocity head can be described as,


P_v = 0.0002307 * (G_n /1000)² / r_g


Where,

Gn = Mass velocity of gas, lb/hr-ft2

rg = Density of gas, lb/ft3

The Mass velocity is described as,



G_n = W_g / A_n
Where,

Wg = Mas gas flow, lb/hr

An = Net free area, ft2

And,



A_n = A_d - d_o/12 * L_e * N_t


For staggered tubes without corbels,



A_d = ((N_t +0.5) * P_t/12) * L_e


For staggered tubes with corbels or inline tubes,



A_d = (N_t * P_t/12) * L_e


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,

Coil Data
Tube outside dia., in:	Tube length, ft:
Number of tubes wide:	Number of rows:
Transverse pitch of tubes, in:	Corbels: Yes No
Process Data
Mass flow, lb/hr:	Density of gas, lb/ft3:

Pressure Drop, inH2O:

---

Fin Tube Pressure Loss:


For the fin tube pressure drop, we will use the Escoa method.





D_p = ((f+a)*G_n²*N_r)/(r_b*1.083E+10⁹)


And,
For staggered layouts,


f = C₂ * C₄ * C₆ * (d_f/d_o)^0.5
For inline layouts,


f = C₂ * C₄ * C₆ * (d_f/d_o)^1.0


And,



a = ((1+B²)/(4*N_r))*r_b*((1/r_out)-(1/r_in))


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,


B = A_n / A_d


For staggered tubes without corbels,


A_d = ((N_t +0.5) * P_t/12) * L_e


For staggered tubes with corbels or inlune tubes,


A_d = (N_t * P_t/12) * L_e


Net Free Area, A_n:


A_n = A_d - A_c * L_e * N_t


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:


C₂ = 0.07 + 8 * R_e^-0.45


And,


R_e = G_n * d_o/(12*m_b)


Where,

mb = Gas dynamic viscosity, lb/ft-hr

Geometry correction, C₄:


For segmented fin tubes arranged in,



a staggered pattern,
C₄ = 0.11*(0.0 5*P_t/d_o)^{(-0.7*(lf/sf)^0.23)}







an inline pattern,
C₄ = 0.08*(0. 15*P_t/d_o)^{(-1.1*(lf/sf)^0.20)}


For solid fin tubes arranged in,



a staggered pattern,
C₄ = 0.11*(0.0 5*P_t/d_o)^{(-0.7*(lf/sf)^0.20)}




an inline pattern,
C₄ = 0.08*(0. 15*P_t/d_o)^{(-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,



a staggered pattern,
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)





an inline pattern,
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,

Coil Data
Tube outside dia., in:	Tube length, ft:
Number of tubes wide:	Number of rows:
Trans. pitch of tubes, in:	Long. pitch of tubes, in:
Fin height, in:	Fin thickness, in:
Fins density, fins/in:	Fin type: Segmented Solid
Tube layout: Staggered Inline	Corbels: Yes No
Process Data
Mass flow, lb/hr:	Bulk density of gas, lb/ft3:
Inlet density of gas, lb/ft3:	Outlet density of gas, lb/ft3:
Bulk viscosity of gas, lb/hr-ft:

Pressure Drop, inH₂O:

---

Stud Tube Pressure Loss:
For the stud tube pressure loss we will use the Muhlenforth method,
The general equation for staggered or inline tubes,





D_p = N_r*0.0514*n_s((C_min-d₀-0.8*l_s)/((n_s*(C_min-d_o-1.2*l_s)²)^0.555))^1.8*G²*((T_g+460)/1460)


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,



D_p = D_p*((d_o/C_min)^0.333)²


And,



G = W_g/(A_n*3600)





A_n = L_e*N_t*(P_t-d_o-(l_s*t_s*r_s)/12)/12
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,

Coil Data
Tube outside dia., in:	Tube length, ft:
Number of tubes wide:	Number of rows:
Trans. pitch of tubes, in:	Long. pitch of tubes, in:
Stud height, in:	Stud diameter, in:
Stud rows per foot:	Tube layout: Staggered Inline
Process Data
Mass flow, lb/hr:	Average gas temperature, °F:

Pressure Drop, inH₂O:

Labels: heater, online calculator

Intube Pressure Drop




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:







D_p = 0.00517/d_i*G²*V_lm*F*L_equiv


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,









V_lm = (V₂-V₁)/ln(V₂/V₁)


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,









V_i = 10.73*(T_f/(P_v*MW_v)*V_frac+(1-V_frac)/r_l


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,

Reynolds number =
Inside Diameter, inches =

Friction factor, F:

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.







L_equiv = Fact_Nre*L_rb


Where,

FactNre = Reynolds number correction

Lrb = Equivalent length of return bend, ft

Return Bend Equivalent Length:


Reynolds Correction:
Where,

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.

Coil Data
Tube inside dia., in:	Pipe straight length, ft:
Bend radius, in:	Number of returns:
Process Data
Mass vel., lb/sec-ft2:	Viscosity, cp:
Spec. vol. at start, ft3/lb:	Spec. vol. at end, ft3/lb:

 Pressure Drop, psi:

Labels: exchanger, heater, online calculator, pressure drop

Heat Transfer Coefficients


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 R_e =>10,000,




h_l = 0.023(k/d_i)R_e^0.8*P_r^0.33(m_b/m_w)^0.14

And for vapor flow with R_e =>15,000,




h_v = 0.021(k/d_i)R_e^0.8*P_r^0.4(T_b/T_w)^0.5

Where the Reynolds number is,




R_e = d_i*G/m_b

And the Prandtl number is,




P_r = C_p*m_b/k

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,



h_tp = h_lW_l + h_vW_v

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.

Tube diameter, in:		Mass flow, lb/hr-ft2:
Percent vapor, %:		Bulk Temp., °F:
Liquid Properties		Vapor Properties
Thermal cond., Btu/hr-ft-°F:		Thermal cond., Btu/hr-ft-°F:
Visc. bulk, lb/ft-hr:		Visc. bulk, lb/ft-hr:
Spec. Heat, Btu/lb-°F:		Spec. Heat, Btu/lb-°F:
Visc. @ wall, lb/ft-hr:		Temp. @ wall, °F:

h_l Coefficient, Btu/hr-ft²-°F:


h_v Coefficient, Btu/hr-ft²-°F:


h_tp Coefficient, Btu/hr-ft²-°F:


Reynolds number:LiquidVapor

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.

Labels: exchanger, heat transfer, heater, online calculator

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

Labels: heat loss, heat losses, heat transfer