Water Dew Point of Flue Gases

Significant amount of energy can be recovered from flue gas containing water vapor if it is cooled below the water dew point. Water dew point is a function of the partial pressure of water vapor in the flue gas as discussed in Chapter 1. For every kg of natural gas com- busted, about 2.15 kg of water vapor is produced. This represents not only a significant amount of latent heat energy, but also a significant amount of water that can also be recovered. From combustion calculations of fuels, the volume fraction of various components

such as CO 2 ,H 2 O, N 2 ,O 2 , and SO 2 is obtained. From the partial pressure of H 2 O, one may

Miscellaneous Boiler Calculations 311

estimate the water dew point, which is an important variable. For example, flue gas from the combustion of typical natural gas has about 17%–18% volume of water vapor while fuel oil has about 11%–12%. If flue gas pressure is atmospheric, the partial pressure of water vapor in natural gas products of combustion is 17.2 kPa or 0.17 × 14.7 = 2.5 psia. The water dew point corresponding to this pressure from steam tables is 134°F or 57°C. Oil products of combustion contain about 12% volume of water vapor, and the water dew point is about 50°C (122°F). If the flue gas is cooled below this temperature, then water in the flue gas starts condensing, and if a condensing economizer is used, the latent plus sensible heat is transferred to the water, which can be a significant amount. A large low-temperature heat sink is required for this purpose such as makeup water or condensate. The following example shows how much energy can be recovered in condensation process.

Energy Recoverable through Condensation Example 6.1

Determine the amount of energy recovered below the water dew point temperature of 100,000 kg/h of flue gases from natural gas combustion that is cooled to 20°C. Flue gas

analysis is % volume CO 2 = 8, H 2 O = 17, N 2 = 72, O 2 = 3.

Solution

One can easily calculate the amount of energy that can be recovered while sensible heat

is recovered from the flue gas by using the following equation: Q = [W g h 1 − (W g − C)h 2 ], where W g is the gas flow, C the amount of water condensed, and h 1 ,h 2 are the enthalpies of the flue gas corresponding to temperatures t 1 and t 2, the initial and final tempera-

tures of the flue gas. Note that due to condensation, the flue gas quantity and analysis will change. The % volume of water vapor after condensation will be lesser than 17%. Estimating the duty below the dew point is quite involved as shown later, as C, the amount of water condensed must be first estimated.

At the inlet temperature, the partial pressure of water vapor = 0.17 × 14.7 = 2.5 psia. This corresponds to 57°C or 134°F water dew point from steam tables. We have to estimate the energy recovered when 100,000 kg/h flue gases are cooled from 57°C to 20°C. Assume that a suitable heat sink is available.

The flue gas molecular weight = 0.08 × 44 + 0.17 × 18 + 0.72 × 28 + 0.03 × 32 = 27.7 The weight fraction or mass flow of each constituent is then

CO 2 = . 0 08 44 27 7 0 127 × / . = . or , 12 700 kg/h . HO 2 = . 0 17 18 27 7 0 11 × / . = . or 111 000 , kg/h

N 2 = . 0 72 28 27 7 0 728 × / . = . or , 72 800 kg/h

O 2 = 100 000 12 , − , 70 11 000 72 800 3 500 − , − , = , kg/h Let C kg/h of water condense when flue gas is cooled to 20°C.

Moles of flue gas after condensation at 20°C = (12,700/44) + {(11,000 − C)/18} + (72,800/28) + (3,500/32) = 288.63 + 611.1 − (C/18) + 2,600 + 109.4 = 3,609.1 − (C/18) Volume fraction of H 2 O after condensation = (11,000 − C)/18/(3,609.1 − 0.0556C)

Corresponding to 20°C from steam tables, the saturation pressure = 0.34 psia or frac-

tion volume of H 2 O = 0.34/14.7 = 0.023. Solving for C,

. 0 023 = ( , 11 000 − C//, ) 18 3 609 1 0 0556 ( . − . ) C or C = , 9 730 kg/h

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

TABLE 6.1

Results of Condensation Calculations

Gas flow, kg/h

90,270 % volume CO 2 8 9.4

17 2.3 N 2 72 84.8 O 2 3 3.5

Gas temperature, °C

Energy recovered, kW

The total moles of the flue gases leaving the economizer: 3,609.1 − (9,730/18) = 3,068.5, and the gas flow after condensation = 100,000 − 9,730 = 90,270 kg/h. Let us obtain the flue gas analysis after condensation:

% volume CO 2 = 100 × 12,700/44/3,068.5 = 9.4% % volume H 2 O = 100 × (11,000 − 9,730)/18/3,068.5 = 2.3%

% volume N 2 = 100 × 72,800/28/3,068.5 = 84.73% % volume O 2 = 100 × 3500/32/3068.5 = 3.56% The sensible energy recovered from 57°C to 20°C = 100,000 × 10.8 − 90,270 × 1.07 =

983,411 kcal/h = 1,144 kW (3.9 MM Btu/h). (The enthalpy of flue gas corresponding to the flue gas analysis at 57°C is 10.8 kcal/kg, and 1.07 kcal/kg is the gas enthalpy at 20°C; see Appendix F on gas properties for the calculation of enthalpy of flue gases.) The latent energy recoverable = 9,730 × 569 = 5,536,370 kcal/h = 6,437 kW (21.9 MM Btu/h). An average latent heat of 569 kcal/kg (1024 Btu/lb) was used from steam tables. Total energy recovered = 14 + 6437 = 7581 kW.

Table 6.1 summarizes the results. Figure 6.2 shows the amount of sensible and latent energy recovery from the flue gases cooled from water dew point temperature of 57°C to 20°C. The latent energy contribution is significant. However, there should be a large low temperature heat sink such as cooling water at about 5°C–10°C for condensation to occur.

One may estimate the total energy recovered and compute the boiler efficiency by using the following expression: efficiency = total energy recovered/fuel input in consistent units. Depending on final flue gas temperature, lower heating value efficiency can range from 95% to even 100%.

Water Dew Point of Flue Gases