How Is Life of Superheater Affected by High Tube Wall Temperatures?

Many plant engineers and consultants ignore the fact that though the material selected may be acceptable for the temperature in consideration, the life of the tubes can be lower with higher tube wall temperature. The life of the superheater tubes is often estimated using Larsen Miller chart (Figure E.3).

Appendix E: Calculations with Finned Tubes 427

25 i 20 15 ss 1000 ps 10 5 Stre

LMP parameter (T + 460) (20 + log t)/1000

FIGURE E.3

Larsen Miller parameters for T11 and T22 materials.

Let the maximum operating pressure be 39 kg/cm 2 g (555 psig).The stress inside the tubes due to the steam pressure = Pr/t = 555 × 1/0.12 = 4625 psig. The LMP factor from Figure E.3 is 37,300. Using the equation, LMP = (460 + T w ) × (20 + log t), where t is the num- ber of hours to failure and T w is the tube wall temperature, °F.

With 78 fins/m, the operating tube wall temperature = 510°C. Use 15°C margin. T w = 977°F = 1437°R. Hence, (20 + log t) = 37,300/1437 = 25.95 or t = 912,000 h = 114 years! When the fin density is 216 fins/m, the maximum tube wall temperature is, say, 555°C (1031°F). Then, (20 + log t) = 37,300/(1,031 + 460) = 25.0 or t = 100,000 h or less than 12.5 years, using 8,000 h of operation in a year. One can see the concern with using higher than required fin density in a superheater. These may be estimates but are given to show the effect of tube wall temperature on the life of tubes. Even a 10°C change can bring about a large decrease in the operating life of the tubes. If the temperature of tubes had been 545°C maximum, then (20 + log t) = 37,300/1,473 = 25.32 or t = 210,100 h = 26 years. That is, a 10°C decrease has doubled the life of the tubes! Hence, boiler designers should not flippantly select the fin geometry in superheaters, and plant engineers should understand the impli- cations of the high fin density and challenge the boiler supplier, who may not even furnish the tube wall temperature calculations at the time of purchase unless asked!

There is another important point to be made here. If the tube wall temperature data and the overall heat transfer coefficient U are not shown in the table, unsuspecting customers or plant engineers who are purchasing the boiler would assume that they are getting a lot

of surface area with the high fin density design, 908 m 2 versus 582 m 2 , and may even reject the option with the 78 fins/m design assuming it will not be adequate for the duty! It will

be shown later that U decreases as the fin density and the external surface area increases. Plant engineers should look at the product of U × A (overall heat transfer coefficient and surface area) rather than A in isolation for a given duty!

The heat flux inside tubes is an important parameter. For example, for the 216 fins/m case, the heat flux inside the tubes = 32.7 × (700 − 424) × 983/(π × 0.0446 × 24 × 3.1 × 8) = 106,430 kcal/m 2 h. Table E.5 shows the summary of results with varying fin densities.

Example E.3

Plant engineers are noticing high tube wall temperatures in a superheater. The design and performance details of the superheater supplied are shown in Table E.6 in column 1.

Exit steam pressure is 38 kg/cm 2 g. The present design is counter-flow. Plant engineers

want to know what will happen to the performance with same gas and steam inlet

428 Appendix E: Calculations with Finned Tubes

TABLE E.6

Performance of Counter-Flow and Parallel-Flow Superheaters

Gas flow, kg/h

80,000 Gas inlet temperature, °C

800 Gas exit temperature, °C

501 Duty, MM kcal/h

8.31 7.21 Gas pressure drop, mm wc

47 48 Steam flow, kg/h

60,000 Steam temp. in, °C

249 Steam exit temp., °C

218,000 Estimated max tube wall temp., °C

Max heat flux, kcal/m 2 h 129,168

U, kcal/m 2 h °C

908 No rows deep

Surface area, m 2 908

8 8 Steam pressure drop, kg/cm 2 1.3 1.2

Steam-side coefficient, kcal/m 2 h °C

conditions but with parallel-flow configuration. Flue gas is from combustion of natural

gas.% volume CO 2 = 8, H 2 O = 18, N 2 = 71.5, O 2 = 2.5. Tube size: 50.8 × 44.6,24 tubes/row, 8 rows deep, 3.1m long, 101.8 mm spacing, staggered.

Solution

These are typical problems in an operating plant. One has access to lot of valuable field data and should know how to use the information to predict the performance at other operating cases. Here, since we have the data on gas and steam temperatures and hence

the duty, we can estimate the U value. Surface area is 908 m 2 . Ratio of external to inter- nal area = 908/(3.14 × 0.0446 × 24 × 8 × 3.1) = 10.9

Q = 60,000 × (808.7 − 669) = 8.38 × 10 6 kcal/h. LMTD = [(800 − 473) – (453 − 249)]/ ln[800 − 473)/(453 − 249)] = 261°C. Hence, U = 8.38 × 10 6 /(261 × 908) = 35.3 kcal/m 2 h °C. We can use these data for starters.

There are two ways to solve this problem without doing elaborate calculations. The U value in both cases will remain nearly the same, though slightly higher in the parallel-flow case due to the higher average gas temperature.

Method 1

This is the traditional method. We assume a certain steam temperature and compute the duty and the exit gas temperature and LMTD and then check the transferred duty. Assume that in the parallel-flow case, the steam temperature is 445°C. The enthalpy of

steam at 38 kg/cm 2 g and at 450°C is 793.4 kcal/kg from steam tables. Using the same steam-side pressure drop of 1.3 kg/cm 2 , the saturation temperature at 39.5 kg/cm 2 g is 249°C and enthalpy is 669 kcal/kg. Q a = energy absorbed by steam is 60,000 × (793.4 − 669) = 7.46 × 10 6 kcal/h. Average gas

specific in previous case is 0.3057 kcal/kg from Appendix F. Gas temperature drop = 7.46 × 10 6 /(80,000 × 0.99 × 0.3057) = 308°C. Hence, exit gas temperature is 800°C − 308°C = 492°C.

LMTD = [(800 − 249) – (492 − 445)]/ln[(800 − 249)/(492 − 445)] = 205°C. Q t = energy transferred = UAΔT = 35.3 × 908 × 205 = 6.57 MM kcal/h. Hence, the assump-

tion we made is incorrect as we cannot transfer the assumed duty with this surface area, and actual duty will be lesser than 7.46 MM kcal/h.

Appendix E: Calculations with Finned Tubes 429

Let us try 435°C as steam temperature. Q a = (787.5 − 669) × 60,000 = 7.1 × 10 6 kcal/h.

Gas temperature drop = 234°C. Hence, exit gas temperature = 800°C − 294°C = 506°C. LMTD = 234°C. Q t = 35.3 × 908 × 234 = 7.5 MM kcal/h. Hence, the assumed steam temperature is lower than what it should be. More iterations such as these are required, but

assuming 439°C, we get Q a = 60,000 × (789.7 − 669) = 7.24 MM kcal/h. Gas temperature

drop = 299°C. Exit gas temperature is 501°C. LMTD = 224°C. Q t = 908 × 35.3.2 × 224 =

7.18 MM kcal/h. Now Q a and Q t agree closely. One should now perform detailed calcu-

lations as shown in Example E.1 to arrive at the duty more accurately. The U value will

be slightly higher due to the higher average gas temperature and can be shown to be 35.8 kcal/m 2 h °C. Also the steam-side coefficient will be slightly higher due to the lower steam temperature (see Appendix B).

Method 2

The NTU method discussed in Appendix A is another way of handling this problem. The specific heat of steam may be estimated as C ps = (789.7 − 669)/(439 − 249) = 0.635 kcal/kg °C using the steam temperatures obtained earlier. Let us use 0.3061 as specific heat of gas from Appendix F at 650°C for natural gas products of combustion.

Q =∈ C t t min ( g 1 − t s 1 )

∈ = [1 − exp {−NTU (1 + C)}]/(1 + C) for parallel-flow exchanger.

C min = 80,000 × .99 × 0.3061 = 24,243. C max = 60,000 × 0.635 = 38,100 C = 24,243/38,580

= 0.636. (1 − C) = 0.364. NTU = UA/C min = 35.8 × 908/24,243 = 1.34 Hence, ∈ = [1 − exp(−1.34 × 1.636)]/1.6284 = 0.543.

Hence, Q t = 0.543 × 24,243 × (800 − 249) = 7.25 MM kcal/h Exit gas temperature = 800 − 7.25 × 10 6 /(80,000 × 0.3061 × 0.99) = 501°C

Exit steam temperature = 249 + 7.25 × 10 6 /(60,000 × 0.636) = 439°C. The predicted performance is shown in column 2. The other values may be calculated as shown in Example E.1.

In counter-flow arrangement, the steam exit end will always see the higher tube wall temperature. However, in parallel-flow arrangement, one has to calculate the tube wall temperatures at both the steam inlet end and exit end.

Heat flux inside the tube at the exit in parallel-flow mode = 35.8 × (501 − 439) × 10.9 =

24193 kcal/m 2 h. Inner wall temperature = 439 + 24,193 (0.0002 + 1/1,363) = 462°C. At the gas inlet end, heat flux = 35.8 × (800 − 249) × 10.9 = 215,011 kcal/m 2 h. Inner wall temperature = 249 + 215,011/(0.0002 + 1/1,363) = 450°C.

With counter-flow arrangement, the steam exit end will see the hottest temperature.

Heat flux = 35.3 × (800 − 472) × 10.9 = 125,820 kcal/m 2 h. Tube wall temperature = 473

+ 125,820/(0.0002 + 1/1,333) = 593°C. If we consider the actual heat transfer coefficients at the inlet and exit conditions, the heat flux will be higher and hence the tube wall temperatures. Also, some margins over calculated values are used in practice. These are shown in the Table E.6. One can see that the wall temperatures are much lower with parallel-flow arrangement.

How Is Life of Superheater Affected by High Tube Wall Temperatures?