Evaluating Part Load Performance

Sometimes a plant engineer may like to know what kind of steam temperature can be expected at part loads for which information may not have been provided by the boiler supplier. It will be helpful if the plant engineer can check the performance and compare it with the field data.

Example 3.8

See how the boiler discussed earlier performs at 50% load. The exit gas temperature was measured as 130°C. Ambient temperature is 21°C. Excess air is 15%, and the same

fuel is used, and hence, flue gas analysis is % volume CO 2 = 8.56, H 2 O = 17.3, N 2 = 71.65, O 2 = 2.48.

Solution

Let us use the simplified procedure for estimating the flue gas flow. Efficiency is obtained using the simplified formula as shown in Chapter 1. Boiler duty = 32.45 MM kcal/h. (50,000 kg/h steam at 399°C from 116°C feed water.) Natural gas fuel HHV = 12,879 kcal/kg, and LHV = 11,653 kcal/kg.

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Let us start from the furnace end and work toward the economizer so that plant engineers can appreciate both the methods—one working from the economizer end and the other from the furnace end.

Furnace Duty Since the fuel analysis, excess air, and boiler duty are known, let us obtain basic data

such as flue gas flow, furnace duty, and furnace exit gas temperature. From Chapter 1, Equation (1.16a).

Efficiency on HHV basis is η H = 89.4 – (0.002021 + 0.0351 × 1.15) × (130 – 21) – 0.5 =

84.28% and η L = 99 – (0.00202 + 0.0389 × 1.15) × (130 − 21) – 0.5 = 93.4% (note that an additional loss of 0.5% was used to account for higher heat losses at part load). Hence, heat input on HHV basis = 32.45/0.8428 = 38.5 MM kcal/h = 161.2 MM kJ/h. Air flow =

314 × 1.15 × 1.013 × 161.2 = 58,960 kg/h. Fuel flow = 38.5 × 10 6 /12,879 = 2,989 kg/h. Flue

gas flow = 58,960 + 2,989 = 61,949 kg/h. [A value for combustion on MM kJ/h basis was used for estimating air flow as discussed in Chapter 1. Correction for moisture in air was also used.]

Furnace effective projected area is 142 m 2 (as provided by boiler supplier; 9.76 m long, 3.96 m high, and 2.44 m wide completely water-cooled furnace except for burner open- ings). Net heat input NHI (it is based on LHV) = 2,989 × 11,653 × 0.985 = 34.3 MM kcal/h. (The value 0.985 is the casing heat loss of 1% plus unaccounted loss of 0.5%, neglecting

heat input by air as there is no air heater.) NHI/projected area = 34.3 × 10 6 /142 = 241,549 kcal/m 2 h = 280.9 kW/m 2 (effective projected area is referred to as EPRS (effective projected radiant surface).

From Figure 2.8 showing furnace exit gas temperature versus NHI/EPRS in Chapter 2, the furnace exit gas temperature is about 1180°C. Enthalpy of flue gas at

1180°C from Appendix F is 351 kcal/kg. Hence, furnace duty = 34.3 × 10 6 − 61,949 × 351 = 12.55 MM kcal/h.

Screen Section The U value may be calculated using Appendices C and D, or for quick estimates here,

let us use the 100% values and correct them. U = 117.8 × (0.5) 0.6 = 77.7 kcal/m 2 h °C (0.5

refers to the 50% load, and a power of 0.6 was used to correct for the load. One can do a detailed calculation to check this value, but for practical purposes, this should be good.)

The gas specific heat is assumed as 0.3252 kcal/kg °C. Screen surface area = 105 m 2 . Assume drum saturation temperature is 254°C due to the lower pressure drop. Ln[(t g1 –t s )/(t g2 –t s )] = UA/(W g C pg ) = 77.7 × 105/(61,949 × 0.3252) = 0.405 = ln[(1180 − 254)/ (t g2 − 254)]. Hence, t g2 = 871°C. Screen duty = 61,949 × 0.3252 × (1180 − 871) = 6.225 MM kcal/h.

Final Superheater Let us assume that we can obtain the 399°C final steam temperature. Assume the

inlet steam temperature as, say, 300°C. Assumed duty Q a = 50,000 × (766.1 − 705) =

3.055 MM kcal/h (766.1 and 705 kcal/kg refer to enthalpy of steam at exit and inlet conditions). Assume gas specific heat here as 0.313 kcal/kg °C. The exit gas tempera-

ture = 871 − 3.055 × 10 6 /(0.313 × 61,949) = 713°C. Let us use a different method here

just to show that there are other ways to evaluate the performance of superheaters and economizers other than the NTU method, and the plant engineer can select whatever he is comfortable with. (Note that all calculations are iterative in nature as one has to correct for the gas properties based on the final gas temperature profile.)

A = 103 m 2 . For the counter-flow superheater, LMTD = [(871 − 399) – (713 − 300)]/ln[(871 − 399)/ (713 − 300)] = 442°C. U = 109.2 × (0.5) 0.6 = 72 kcal/m 2 h °C. The transferred duty Q t = UAΔT = 72 × 103 × 442 = 3.28 MM kcal/h. Since the assumed and transferred duties do not match and that the transferred is higher, we can assume,

148 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers

say, a lower steam temperature at the inlet and repeat the exercise. Use 295°C as the

steam inlet temperature and try again. Q a = 50,000 × (766.1 − 701.6) = 3.22 MM kcal/h. Exit gas temperature = 871 − 3.22 × 10 6 /(0.313 × 61,949) = 705°C. LMTD = [(871 − 399) –

(705 − 295)]/ln[(871 − 399)/(705 − 295)] = 440°C. Q t = 72 × 103 × 440 = 3.26 MM kcal/h. This is close enough. Hence, duty = 3.26 MM kcal/h.

Primary Superheater Let us see if we need the desuperheater. U = 105 × (0.5) 0.6 = 69.3 kcal/m 2 h °C. A = 77 m 2 .

Assume C pg = 0.316 kcal/kg °C. Saturation temperature = 254°C, and enthalpy = 668.6 kcal/kg.

Q a = 50,000 × (701.6 − 668.6) = 1.65 MM kcal/h. Exit gas temperature = 705 − 1.65 × 10 6 /

(0.316 × 61,949) = 620°C. LMTD = [(705 − 295) – (620 − 254)]/ln[(705 − 295)/(620 − 254)] = 388°C. Q t = 77 × 69.3 × 388 = 2.07 MM kcal/h. Hence, we do need the desuperheater as the steam temperature at superheater exit will be much higher than 295°C.

Let us assume we can go to 305°C. The spray required to cool from 305°C to 295°C has to be estimated. Let W = flow in the superheater and so (50,000 – W) is the spray quantity. By heat balance around the desuperheater, W × 708.2 + (50,000 – W) × 117 = 50,000 × 701.6 or W = 49,441 kg/h. (117 kcal/kg is the enthalpy of spray water, and 708.2 and 701.6 kcal/ kg are the enthalpies at 305°C and 295°C, respectively.) Hence, spray flow = 559 kg/h.

Q a = 49,441 × (708.2 − 668.6) = 1.957 MM kcal/h. Exit gas temperature = 705 − 1.957 × 10 6 /(0.316 × 61,949) = 605°C. LMTD = [(705 − 305) – (605 − 254)]/ln[(705 − 305)/(605 − 254)] = 375°C. Q t = 375 × 77 × 69.3 = 2.00 MM kcal/h. The assumed and transferred duties are close, and hence, there is no need to repeat the exercise. With a computer program, one can improve the accuracy of these calculations.

Evaporator U = 105 × (0.5) 0.6 = 69.3 kcal/m 2 h °C, A = 322 m 2 . Let C pg = 0.2928, t s = 254°C.

Hence, ln[(605 − 254)/(t g2 − 254)] = 69.3 × 322/(61,949 × 0.2928) = 1.23 or t g2 = 357°C. Duty of evaporator = 61,949 × (605 − 357) × 0.2928 = 4.5 MM kcal/h.

Economizer We can try the NTU method here as we know both the gas inlet and water inlet temper-

atures. Note that the water flow will be steam flow less spray water flow = 49,441 kg/h as zero blowdown assumed.

W g = 61,949 kg/h. C pg = 0.277 kcal/kg °C. t wi = 116°C. Let t w2 = 192°C. Duty = 49,441 × (h 192 –h 116 ) = 49,441 × (195.36 − 117) = 3.874 × 10 6 kcal/h (enthalpy of water is taken from steam tables). Average C pw = (195.36 − 117)/(192 − 116) = 1.031 kcal/kg °C. The exit gas temperature = 357 − 3.874 × 10 6 /(61,949 × 0.277) = 131°C. U = 40.61 × (0.5) 0.6 = 26.8 kcal/m 2 h °C. Using the NTU method, WC p for flue gas is 61,949 × 0.277 = 17,160 and for water is 49,441 × 1.031 = 50,974. WC min = 17,160 and C = WC min /WC max = 17,160/50,974 = 0.3366. (1 – C) = 0.6634.

NTU = UA/C min = 26.8 × 2,395/17,160 = 3.74.

∈ = [1 – exp{–NTU × (1 – C)}]/[1 – C exp{–NTU × (1 – C)}] = [1 – exp{–3.74 × 0.6634}]/[1 − 0.3366 × exp{–3.74 × 0.6634}] = 0.939 (counter-flow arrangement).

Duty Q = 0.939 × 17,160 × (357 − 116) = 3.88 × 10 6 kcal/h. Exit gas temperature may be

again checked = 3.88 × 106/(61,949 × 0.277) = 131°C. Exit water temperature = 116 + 3.88 ×

10 6 /49,441/1.031 = 192°C. These numbers agree with those assumed. Total boiler duty = 12.55 + 6.225 + 3.26 + 2.0 + 4.5 + 3.88 = 32.4 MM kcal/h.

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This also matches with the required duty for generating 50,000 kg/h at 399°C from 116°C feed water. Hence, even if performance at a particular load is not provided, plant engineers using the aforementioned approach can estimate the boiler performance. However, it is better to demand and get all the pertinent data from the boiler supplier and then cross-check the results and question the boiler supplier if large differences are seen.

Performance without Economizer

Example 3.9

What happens if the economizer is removed? Can we generate the same amount of steam?

Solution

Applying the same procedure as Example 3.7, one arrives at the results shown in Table 3.14. The following points are noted:

a. The exit gas temperature is much higher, about 512°C versus 154°C at full load with economizer. This is due to the lower efficiency of the boiler, which results in larger fuel flow and hence flue gas flow.

b. The flue gas flow is also much larger than what we had at 100% load with economizer. c. The energy absorbed by the superheaters is also much higher resulting in a large amount of spray water for steam temperature control, 8500 kg/h versus 3260 kg/h. Also the temperature after spray is getting closer to saturation temperature, which is not a good situation. The superheater can have wet steam in pockets if the mixing of water and steam at the desuperheater is not good and solids can deposit inside the tubes of final superheater, increasing its tube wall temperature leading to its failure.

d. The gas temperature is higher along the entire gas flow path compared to the case with the economizer in operation. Hence, the superheater tube wall temperature will be higher.

e. The flue gas pressure drop will be higher due to the higher mass flow of flue gas and higher flue gas temperatures, 480 mm wc versus 370 mm wc. f. The boiler efficiency is also much lower due to the higher exit gas temperature of 508°C versus 154°C when economizer is used. A drop of 16% is seen from Figure 3.23.

In conclusion, one should use the economizer or air heater at full load operation; else, the boiler will be operating with larger flue gas flow and higher gas temperature through- out the unit. The other option is to reduce the boiler load to about 70%.

Evaluating Part Load Performance