Convection heat transfer

Convection Heat Transfer
The HRSG primarily absorbs heat from the hot flue gases by convection heat transfer. In the convection section, heat is transferred by both radiation and convection. The convection transfer coefficients for fin and stud tubes are explored here as well as bare tube transfer. The short beam radiation is treated separately from the convection transfer below.
This section of the HRSG Design is divided into five main areas, which can be selected from the jump links below to go to a section.

Convection Transfer, Bare Tubes

Convection Transfer, Fin Tubes

Convection Transfer, Stud Tubes

Short Beam, Reflective Radiation

Convection Section Design

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Convection Transfer, Bare Tubes
Overall Heat Transfer Coefficient, U_o:
U_o = 1/R_to Where,

Uo = Overall heat transfer coefficient, Btu/hr-ft2-F

Rto = Total outside thermal resistance, hr-ft2-F/Btu

And,

Rto = Ro + Rwo + Rio

Ro = Outside thermal resistance, hr-ft2-F/Btu

Rwo = Tube wall thermal resistance, hr-ft2-F/Btu

Rio = Inside thermal resistance, hr-ft2-F/Btu

And the resistances are computed as,

Ro = 1/he

Rwo = (tw/12*kw)(Ao/Aw)

Rio = ((1/hi)+Rfi)(Ao/Ai)

Where,

he = Effective outside heat transfer coefficient, Btu/hr-ft2-F

hi = Inside film heat transfer coefficient, Btu/hr-ft2-F

tw = Tubewall thickness, in

kw = Tube wall thermal conductivity, Btu/hr-ft-F

Ao = Outside tube surface area, ft2/ft

Aw = Mean area of tube wall, ft2/ft

Ai = Inside tube surface area, ft2/ft

Rfi = Inside fouling resistance, hr-ft2-F/Btu

Inside film heat transfer coefficient, hi:
The inside heat transfer coefficient calculation procedure is covered in detail, elsewhere in this course.

Effective outside heat transfer coefficient, he
h_e = 1/(1/(h_c+h_r)+R_fo) Where,

hc = Outside heat transfer coefficient, Btu/hr-ft2-F

hr = Outside radiation heat transfer coefficient, Btu/hr-ft2-F

Rfo = Outside fouling resistance, hr-ft2-F/Btu

Outside film heat transfer coefficient, hc:

The bare tube heat transfer film coefficient, h_c, can be described by the following equations.
For a staggered tube arrangement,
h_c = 0.33*k_b(12/d_o)((c_p*m_b)/k_b)^1/3((d_o/12)(G_n/m_b)))^0.6 And for an inline tube arrangement,
h_c = 0.26*k_b(12/d_o)((c_p*m_b)/k_b)^1/3((d_o/12)(G_n/m_b)))^0.6 Where,

hc = Convection heat transfer coefficient, Btu/hr-ft2-F

do = Tube outside diameter, in

kb = Gas thermal conductivity, Btu/hr-ft-F

cp = Gas heat capacity, Btu/lb-F

mb = Gas dynamic viscosity, lb/hr-ft

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

We can use this typical superheater coil as a sample bare tube bank. We will assume the first two rows are bare:

Process Conditions: Gas flow, lb/hr = 800,000 Gas temperature in, °F = 980 Gas temperature out, °F = 968.9 Compostion, moles N2, % = 72.55 O2, % = 12.34 CO2, % = 3.72 H2O, % = 10.52 Ar, % = 0.87 Mechanical Conditions: Tube Diameter, in = 2.000 Tube Spacing, in = 4 Number Tubes Wide = 28 Tube Effective Length, ft = 28.000 Number Of Tubes = 56 Tube Arrangement = Inline Pitch

Gas Properties
For the gas properties, we can use the JavaScript we previously used to get the properties of the gas at the average temperature.

From this program, we get the following properties,
k_b, Btu/hr-ft-F = 0.0324
c_p, Btu/lb-F = 0.2805
m_b, cp = 0.0359 = 0.0359*2.42 = 0.0869 lb/hr-ft
To calculate the mass velocity, G_n, we need to first calculate the net free area of the tube bank. For these calculations, we are going to assume the tube rows are corbelled, so the net free area, NFA:
NFA = N_wide*t_spc/12*t_lgth-N_wide*t_Od/12*t_lgth = 28*4/12*28-28*2/12*28 = 130.667 ft²
Therefore,
G_n = W_gas / NFA = 800000/130.667 = 6122.448
And using our formula for h_c,
h_c = 0.26*0.0324 (12/2)((0.2805*0.0869 )/0.0324)^1/3((2/12)(6122.448/0.0869 )))^0.6 = 12.7152 To get a feel for the values of the coeffcient, use the following script to run various designs.

Tube Diameter, in:	Tube Spacing, in:
Number Tubes Wide:	Tube Layout: Staggered Inline
Tube Eff. Length, ft:
Gas Flow, lb/hr:	Gas Visc, lb/hr-ft:
Therm.Cond., Btu/hr-ft-F:	Spec. Heat, Btu/lb-F:

Outside Film Coefficient, Btu/hr-ft2-F:

Net Free Area, ft2:

Mass Velocity, lb/hr-ft2:

The radiation transfer coefficient, h_r is described later in this section. Fouling resistances, R_fi and R_fo are allowances that depend upon the process or service of the heater and the fuels that are being burned.

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Convection Transfer, Fin Tubes

You will notice that the heat transfer equations for the fin tubes are basically the same as for the bare tubes untill you reach the h_e factor, where a new concept is introduced to account for the fin or extended surface. The procedure presented herein are taken from the Escoa manual which can be downloaded in full from the internet.

Overall Heat Transfer Coefficient, U_o:
U_o = 1/R_to Where,

Uo = Overall heat transfer coefficient, Btu/hr-ft2-F

Rto = Total outside thermal resistance, hr-ft2-F/Btu

And,

Rto = Ro + Rwo + Rio

Ro = Outside thermal resistance, hr-ft2-F/Btu

Rwo = Tube wall thermal resistance, hr-ft2-F/Btu

Rio = Inside thermal resistance, hr-ft2-F/Btu

And the resistances are computed as,

Ro = 1/he

Rwo = (tw/12*kw)(Ao/Aw)

Rio = ((1/hi)+Rfi)(Ao/Ai)

Where,

he = Effective outside heat transfer coefficient, Btu/hr-ft2-F

hi = Inside film heat transfer coefficient, Btu/hr-ft2-F

tw = Tubewall thickness, in

kw = Tube wall thermal conductivity, Btu/hr-ft-F

Ao = Total outside surface area, ft2/ft

Aw = Mean area of tube wall, ft2/ft

Ai = Inside tube surface area, ft2/ft

Rfi = Inside fouling resistance, hr-ft2-F/Btu

Inside film heat transfer coefficient, hi:
The inside heat transfer coefficient calculation procedure is covered in detail, elsewhere in this course.

Effective outside heat transfer coefficient, h_e:
h_e = h_o(E*A_fo+A_po)/A_o Where,

ho = Average outside heat transfer coefficient, Btu/hr-ft2-F

E = Fin efficiency

Ao = Total outside surface area, ft2/ft

Afo = Fin outside surface area, ft2/ft

Apo = Outside tube surface area, ft2/ft

And,
Average outside heat transfer coefficient, h_o:
h_o = 1/(1/(h_c+h_r)+R_fo) Where,

hc = Outside heat transfer coefficient, Btu/hr-ft2-F

hr = Outside radiation heat transfer coefficient, Btu/hr-ft2-F

Rfo = Outside fouling resistance, hr-ft2-F/Btu

Outside film heat transfer coefficient, hc:
h_c = j*G_n*c_p(k_b/(c_p*m_b))^0.67 Where,

j = Colburn heat transfer factor

Gn = Mass velocity based on net free area, lb/hr-ft2

cp = Heat capacity, Btu/lb-F

kb = Gas thermal conductivity, Btu/hr-ft-F

mb = Gas dynamic viscosity, lb/hr-ft

Colburn heat transfer factor, j:
j = C₁*C₃*C₅(d_f/d_o)^0.5((T_b+460)/(T_s+460))^0.25 Where,

C1 = Reynolds number correction

C3 = Geometry correction

C5 = Non-equilateral & row correction

df = Outside diameter of fin, in

do = Outside diameter of tube, in

Tb = Average gas temperature, F

Ts = Average fin temperature, F

Reynolds number correction, C₁:
C₁ = 0.25*R_e^-0.35 Where,

Re = Reynolds number

Geometry correction, C₃:
For segmented fin tubes arranged in,
a staggered pattern,

C₃ = 0.55+0.45*e^{(-0.35*lf/Sf)}
an inline pattern,

C₃ = 0.35+0.50*e^{(-0.35*lf/Sf)}
For solid fin tubes arranged in,
a staggered pattern,

C₃ = 0.35+0.65*e^{(-0.25*lf/Sf)}
an inline pattern,

C₃ = 0.20+0.65*e^{(-0.25*lf/Sf)} Where,

lf = Fin height, in

sf = Fin spacing, in

Non-equilateral & row correction, C₅:
For fin tubes arranged in,

a staggered pattern,

C₅ = 0.7+(0.70-0.8*e^{(-0.15*Nr^2)})*e^(-1.0*Pl/Pt)
an inline pattern,

C₅ = 1.1+(0.75-1.5*e^{(-0.70*Nr^2)})*e^(-2.0*Pl/Pt) Where,

Nr = Number of tube rows

Pl = Longitudinal tube pitch, in

Pt = Transverse tube pitch, in

Mass Velocity, G_n:
G_n = W_g/A_n Where,

Wg = Mass gas flow, lb/hr

An = Net free area, ft2

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,

Ad = Nt * Le * Pt / 12

Ac = (do + 2 * lf * tf * nf) / 12

tf = fin thickness, in

nf = number of fins, fins/in

Surface Area Calculations:
For the prime tube,
A_po = Pi * d_o (1- n_f * t_f) / 12 And for solid fins,
A_o = Pi*d_o(1-n_f* t_f)/12+Pi*n_f(2*l_f(d_o+l_f)+t_f(d_o+2*l_f))/12 And for segmented fins,
A_o = Pi*d_o(1-n_f* t_f)/12+0.4*Pi*n_f(d_o+0.2)/12+Pi*n_f (d_o+0.2)((2*l_f-0.4)(w_n+t_f)+w_s*t_f)/(12*w_s) And then,
A_fo = A_o - A_po Where,

ws = Width of fin segment, in

In the O-Frame Evaporator example we are using here, we should point out, that some of the tube rows can be used for downcomers instead of risers. Or , alternatively, the downcomers may be outside the gas pass. If they are part of the tube bank, they normally would not be finned, even if the riser tubes are finned. The sketch shown here would indicate the center two rows are downcomers since they are outside the collection baffle which directs the water/vapor mixter, coming from the riser tubes, through the primary separators. In this example, we are not going to consider any of the tubes as downcomers, but if we did, they would absorb heat, but would need to be rated separately since they would have a different inside rate and overall heat transfer coefficient. We can describe a sample fin tube bank as follows:

Process Conditions: Gas flow, lb/hr = 800,000 Gas temperature in, °F = 898 Gas temperature out, °F = 533 Average fin temperature, °F = 701 Compostion, moles N2, % = 72.55 O2, % = 12.34 CO2, % = 3.72 H2O, % = 10.52 Ar, % = 0.87
Mechanical Conditions: Tube Diameter, in = 2.00 Tube Spacing, in = 4 Number Tubes Wide = 28 Tube Effective Length, ft = 28.00 Number Of Tubes = 392	Tube Arrangement = Inline Pitch Fin Height, in = 0.75 Fin Thickness, in = 0.049 Fin Density, fins/in = 6 Fin Type = Segmented Fin Segment Width, in = 0.3125

Gas Properties
For the gas properties, we can use the JavaScript we used previously to get the properties of the gas at the average temperature.

From this program, we get the following properties,
k_b, Btu/hr-ft-F = 0.0278
c_p, Btu/lb-F = 0.2719
m_b, cp = 0.0314 = 0.0314*2.42 = 0.0760 lb/hr-ft
To calculate the mass velocity, G_n, we need to first calculate the net free area of the tube bank. For these calculations, we are going to assume the tube rows are corbelled, so the net free area, A_n:
A_d = 28*28*4/12 = 261.333
A_c = (2+2*0.75*0.049*6)/12 = 0.2034 So,
A_n = 261.333 - 0.2034 * 28 * 28 = 101.8674 And,
G_n = 800000 / 101.8674 = 7853.3466 Now we can calculate the reynolds number, R_e,
R_e = 7853.3466*2/(12*0.0760) = 17222.251 And,
C₁ = 0.25*17222.251^-0.35 = 0.00823 For,
s_f = 1/6-.049=0.1177
C₃ = 0.35+0.50*e^{(-0.35*0.75/0.1177)} = 0.4037 And,
P_l = 4.0
C₅ = 1.1-(0.75-1.50*e^{(-0.70*14^2)})*e^(-2.0*4/4) = .9985 Now we can calculate the Colburn factor,
j = 0.00823*0.4037*.9985(3.5/2)^0.5((715.5+460)/(701+460))^0.25 = 0.0044 And finally,
h_c = 0.0044*7853.3466*0.2719(0.0278/(0.2719*0.0760))^0.67 = 11.475
To get a feel for the values of the coeffcient, use the following script to run various designs.

Tube Diameter, in:	Tube Spacing, in:
Number Tubes Wide:	Tube Layout: Staggered Inline
Number Tube Rows:	Tube Eff. Length, ft:
Fin Height, in:	Fin Density, fins/in:
Fin Thickness, in:	Fin Segment Width, in:
Average Fin Temperature, F:	Fin Type: Segmented Solid
Gas Flow, lb/hr:	Average Gas Temperature,F:
Therm.Cond., Btu/hr-ft-F:	Spec. Heat, Btu/lb-F:
Gas Visc, lb/hr-ft:

Outside Film Coefficient, Btu/hr-ft2-F:

Net Free Area, ft2:

Mass Velocity, lb/hr-ft2:

The radiation transfer coefficient, h_r is described later in this section. Fouling resistances, R_fi and R_fo are allowances that depend upon the process or service of the HRSG and the fuels that are being burned.

Fin Efficiency, E:

For segmented fins,
E = x * (0.9 + 0.1 * x) And for solid fins,
E = y * (0.45 * ln(d_f / d_o) * (y - 1) + 1) Where,
y = x * (0.7 + 0.3 * x) And,
x = tanh(m * B) / (m * B) Where,
B = l_f + (t_f /2) For segmented fins,
m = (h_o (t_f + w_s) / (6 * k_f * t_f * w_s))^0.5 And for solid fins,
m = (h_o / (6 * k_f * t_f))^0.5 Fin Tip Temperature, T_s:

The average fin tip temperature is calculated as follows,
T_s = T_g + (T_w - T_g) * 1/((e^1.4142mB+e^-1.4142mB)/2) Maximum Fin Tip Temperature, T_fm:

The maximum fin tip temperature is calculated as follows,
T_sm = T_wm + q(T_gm - T_wm) Where,

Tsm = Maximum Fin Tip Temperature, F

Tgm = Maximum Gas Temperature, F

Twm = Maximum Tube Wall Temperature, F

And,
The value for theta, q, can be described by the following curve.

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Convection Transfer, Stud Tubes
Overall Heat Transfer Coefficient, U_o:
U_o = 1/R_to Where,

Uo = Overall heat transfer coefficient, Btu/hr-ft2-F

Rto = Total outside thermal resistance, hr-ft2-F/Btu

And,

Rto = Ro + Rwo + Rio

Ro = Outside thermal resistance, hr-ft2-F/Btu

Rwo = Tube wall thermal resistance, hr-ft2-F/Btu

Rio = Inside thermal resistance, hr-ft2-F/Btu

And the resistances are computed as,

Ro = 1/he

Rwo = (tw/(12*kw))(Ao/Aw)

Rio = ((1/hi)+Rfi)(Ao/Ai)

Where,

he = Effective outside heat transfer coefficient, Btu/hr-ft2-F

hi = Inside film heat transfer coefficient, Btu/hr-ft2-F

tw = Tubewall thickness, in

kw = Tube wall thermal conductivity, Btu/hr-ft-F

Ao = Outside surface area, ft2/ft

Aw = Mean area of tube wall, ft2/ft

Ai = Inside tube surface area, ft2/ft

Rfi = Inside fouling resistance, hr-ft2-F/Btu

Effective outside heat transfer coefficient, h_e:
For staggered and inline pitch,
h_e = (h_so*E*A_fo+h_t*A_po)/A_o Where,

ht = Base tube outside heat transfer coefficient, Btu/hr-ft2-F

hso = Stud outside heat transfer coefficient, Btu/hr-ft2-F

Ao = Total outside surface area, ft2/ft

Afo = Stud outside surface area, ft2/ft

Apo = Tube outside surface area, ft2/ft

Inline pitch correction,
h_e = h_e*(d_o/P_l)^0.333 Where,

do = Outside tube diameter, in

Pl = Longitudinal pitch of tubes, in

Base tube outside heat transfer coefficient, h_t:
h_t = (0.717/d_o^0.333)(G_n/1000)^0.67(T_b+460)^0.3 And the stud coefficient,
h_s = 0.936*(G_n/1000)^0.67(T_b+460)^0.3 With fouling,
h_so = 1/(1/h_s+R_fo) Where,

hs = Stud outside heat transfer coefficient, Btu/hr-ft2-F

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

Tb = Average gas temperature, F

Stud efficiency, E:
E = 1/((e^x+e^-x)/1.950) Where,
X = L_s/12((2*h_so)/(k_s*D_s/12))^0.5 And,

Ls = Length of stud, in

Ds = Diameter of stud, in

ks = Conductivity of stud, Btu/hr-ft-F

The following script will allow us calculate the coeffcient for stud tubes.

Tube Diameter, in:	Number Tubes Wide:
Tube Longitudinal Pitch, in:	Tube Transverse Pitch, in:
Tube Layout: Staggered Inline
Number Tube Rows:	Tube Eff. Length, ft:
Stud Length, in:	Studs Per Plane:
Stud Diameter, in:	Planes Per Foot:
Conductivity Of Stud, Btu/hr-ft-F:	Outside Fouling Factor, hr-ft2-F/Btu:
Gas Flow, lb/hr:	Average Gas Temperature,F:

Outside Film Coefficient, Btu/hr-ft2-F:

Net Free Area, ft2:

Mass Velocity, lb/hr-ft2:

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Short Beam, Reflective Radiation

The gas radiation factor, h_r, can be calculated from the following correlations. This factor is used in calculating the overall heat transfer coefficient for bare tubes and fin tubes. The formulas for the stud tubes has this factor built into the equations.

For bare tubes,
h_r = 2.2*g_r*(pp*mbl)^0.50 And for fin tubes,
h_r = 2.2*g_r*(pp*mbl)^0.50(A_po/A_o)^0.75 Where,

hr = Average outside radiation heat transfer coefficient, Btu/hr-ft2-F

gr = Outside radiation factor, Btu/hr-ft2-F

pp = Partial pressure of CO2 & H2O, , atm

mbl = Mean beam length, ft

Apo = Bare tube exposed surface area, ft2/ft

Ao = Total outside surface area, ft2

Outside radiation factor, g _r:

The outside radiation factor can be described by the following curves:

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Convection Section Design
The following calculator will allow you to calculate the overall heat transfer coefficient for fin tubes, stud tubes, or bare tubes. This calculator uses the methods described above.