Simplified Approach to Evaluating Performance of Fire Tube Boilers
Simplified Approach to Evaluating Performance of Fire Tube Boilers
Plant engineers may use the following procedure discussed to evaluate quickly the design or performance of fire tube boilers. Plants, for example, re-tube their old boilers or modify the existing ones and would like to know how the performance will be affected when tube geometry data or steam parameters change. It is a fact that a computer program gives more accurate results. However, plant engineers may not have access to such programs, and
246 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers
hence, shortcuts such as these shown later help them in evaluating new designs or perfor- mance of existing ones. Based on these shortcuts, they may also develop some computer code or Excel worksheet. Note that this method neglects the contribution of nonluminous heat transfer coefficient, which is anyway not significant in fire tube boilers due to the small inner diameter of tubes, which is the beam length. The overall heat transfer coef- ficient U i may be written as
w = flow per tube in kg/s = W/N, where N is the number of tubes and W the total flow in kg/s. Q/ΔT = U i A i = 0.0278A i Cw 0.8 /d i 1.8 = 0.0873C d i N LW 0.8 /(N 0.8 d i 1.8 ) (substituting A i = π d i LN). Rewriting,
Q = duty in W, L is the length in m, d i the tube ID in m, and C = (C p /µ) 0.4 k 0.6 , where C p is in J/kg K, µ in kg/m s, and k in W/m K. For the evaporator, ΔT = (T 1 −T 2 )/ ln[(T 1 −T s )/(T 2 −T s )] and Q = WC p (T 1 −T 2 ) neglecting heat loss. Substituting these in (4.17) and simplifying, we have
( 0.2 T 1 − T s ) 0.0873CN L
ln
Cd p i W )
Example 4.15
Let us check the results of Example 4.7 using this shortcut. W = 45,370 kg/h =
12.6 kg/s. N = 600, L = 6.1 m, C p = 1,204 J/kg K, C = 182, d i = 0.045 m, T 1 = 815°C, T s = 185°C. Solve for T 2 .
or T 2 = (815 − 185)/8.03 or T 2 = 263°C. Close to 260°C using the detailed calculations. If the gas pressure drop equation is considered, we have
∆P = 0 08262 . fLvw e 2
where w is the flow/tube in kg/s L e is the equivalent tube length in m d i is the tube inner diameter in m f is the friction factor
v is the specific volume in m 3 /kg
ΔP in mm wc as discussed in Appendix B
Waste Heat Boilers 247
If inlet and exit losses are included, L e = L + 60d i. Also substituting for w = W/N, . 0 08262 f (L + 60 i ) d vW ∆P 2 =
2 5 † ( (4.21) Nd i )
Substituting for N from this in (4.19) and simplifying, we have
) vL 01 ln .
01 ( . T
01 . L
1 T ) . 0 068 Cf ( + 60 di
( ∆ P di 01 . . C 11 3 p )
This formula relates the duty with gas pressure drop. Equations 4.19 and 4.22 help one arrive at quick solutions to the major boiler configuration or performance.
In the conventional method, the following are the steps (though the use of a computer program makes things easier).
1. Given the gas flow, and inlet and exit gas temperatures, one first assumes w, flow per tube, or N from which w is obtained. w, the flow per tube, is based on experience.
2. h c and h n and then U, ΔT are computed. 3. The required surface area is obtained, and the tube length L is computed. 4. The gas pressure drop is then evaluated. If one does not like these values, steps
1–4 are repeated. The simplified approach gives a reasonable geometry for a desired performance quickly
or the performance for a given tube geometry in one step.
Example 4.16
In a fire tube boiler with 600 tubes with tube size 38 × 33 mm and length 5 m, 30,000 kg/h of flue gas with the same analysis shown in Example 4.7 has to be cooled from 700°C.
Steam pressure is 12 kg/cm 2 g. Determine the exit gas temperature and gas pressure drop. T 1 = 700°C. T s = 191°C. L = 5 m, d i = 0.033 m, W = 8.33 kg/s. Let C p = 1195 J/kg K, C = 181 as before. (The gas properties may be corrected after obtaining the average gas tempera- ture.) Average gas specific volume is 2.19 m 3 /kg.
Using (4.19),
ln ( T 1 − T s )
= . 0 0873 CN L
02 . ( T 2 − T s )
( Cd p i W )
ln
( T 2 − 191 ) 1195 0 033 × . 08 . × . 8 33 . 00 2 ( )
Or T 2 = 238°C. The duty and steam generation may be obtained from this as shown earlier. Using (4.22),
( T 1 − T s ) . 0 068 Cf 01 . +
( L 60 d )
P di 01 . ( . ∆ 1 13 C p )
( ∆P ∆ 01 . × . 0 033 13 . × 1195 )
or
ΔP = 136 mm wc.
These values may be checked out using detailed calculations.
248 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers
Example 4.17
If L = 4 m and gas pressure drop = 100 mm wc due to plant space limitations, what is T 2
and N in the aforementioned situation? Other data such as gas flow and analysis may
be assumed to be the same. These types of what if studies may be quickly performed with these equations. Using (4.22),
ln
( T 2 − 191 )
Hence, T = °
= . 0 0873 181 × × N 02 . × 2 4 265 . C Using ( . 4 19 1 935 ) ,. or N = 648
Thus, the entire geometry is arrived at. One may obtain the duty and steam generation as shown earlier. These methods are suggested for quick evaluation of performance or geometry or to study the impact of varying any tube geometry data.
Estimating Tube Sheet Thickness
The shell diameter for the boiler in Example 4.7 is obtained as follows: N = 600 tubes, 50.8 ×
45 mm. Tube pitch p = 70 mm triangular. D = minimum diameter of shell = 1.05 × 70 × √600 = 1800 mm. Use, say, 150 mm between the outer most tubes and the shell inner diameter. Hence, shell ID = 2100 mm. Let the design pressure including static head = 14 kg/cm 2 g = 199 psig. ASME code provides a simple formula for the tube sheet thickness; t = p√(P/2.2/S), where P is the design pressure, psig, and S the allowable stress. Up to 650°F, for carbon steel plates, S = 17,500 psi may be used. p = 70 mm.
Hence, t = 70 × (199/2.2/17,500) = 5.03 mm. Considering corrosion allowances on gas and steam side and rigidity, let us use 25 mm plate.