Convection Section Design

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.
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Overall Heat Transfer Coefficient, U_o:

U_o = 1/R_to

Where,

And,

And the resistances are computed as,

Where,

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,

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,

We can describe a sample bare tube bank as follows:

Gas Properties
For the gas properties, we can use the script we used in the radiant section design 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 = 0.0315
c_p, Btu/lb-F = 0.2909
m_b, cp = 0.0340 = 0.0340*2.42 = 0.0823 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 = 8*8/12*13-8*4.5/12*13 = 30.333 ft²
Therefore,
G_n = W_gas / NFA = 100000/30.333 = 3296.739
And using our formula for h_c,

h_c = 0.33*0.0315 (12/4.5)((0.2909 *0.0823 )/0.0315 )^1/3((4.5/12)(3296.739/0.0823 )))^0.6 = 8.1115