Analysis of an Evaporator

In an evaporator, the large mass of metal and water inventory results in a longer startup time, but the residual energy in the metal also helps to respond to load changes faster when the energy from flue gas or heat input to the boiler is cut off. Drum-level fluctuations are also smoothed out by a large water inventory.

The basic equation for energy transfer to the evaporator is

(6.45) where

Q = Wh s fg + ( h l − hW f ) f + W C dt/dp W dh/dp dp/dz m p + w

W m is the mass of metal, kg W s ,W f are mass flow of steam, feed water, kg/h W w is the amount of water inventory in the boiler system including drums, tubes dh/dp is the change of enthalpy to change of steam pressure, kcal/kg/kg/cm 2 dt/dp is the change of saturation temperature change to change in pressure, °C/kg/cm 2 Q is the energy transferred to evaporator, kcal/h

dp/dz is the rate of pressure change, kg/cm 2 /h

Let the steam space between the drum level and the first valve = V m 3 . The change in pressure may be written as pv = C = pV/m where C is a constant m is the mass of steam, kg in volume V or,

= pV/m C

dp/dz = pv/V W − W

)( s l ) (6.46)

where p is the steam pressure W s ,W l are the steam generated and the steam withdrawn by process

When steam withdrawn is equal to steam generated, the steam pressure is unchanged. Pressure fluctuations occur when the steam generated and steam demand differ.

Miscellaneous Boiler Calculations 349

Example 6.22

A waste heat boiler evaporator has the following data: Gas flow = 158,800 kg/h. Gas inlet temperature = 537°C. Gas exit temperature = 266°C. Steam pressure = 42 kg/cm 2 a. Feed water entering = 105°C. Tubes: 50.8 × 2.7 mm, 30 tubes/row, 20 deep, 3.66 m with 177 fins/m, 19 mm high ×

1.27 mm thick serrated fins. Steam drum has an ID of 1371 mm, mud drum ID = 914 mm, 3.96 m long. Boiler gen- erates 20,400 kg/h steam. Weight of steel tubes, drums = 38,000 kg. Weight of water in

evaporator = 8600 kg. Volume of steam space = 3.26 m 3 . Feed water temperature = 105°C.

Energy transferred to evaporator in normal operation = 11.67 mm kcal/h. What happens to the steam pressure and steam generation when the heat input and feed water supply are turned off?

Solution

For steam in the pressure range 42–44 kg/cm 2 a, from steam tables we have enthalpy

of saturated liquid as 261.7 and 265 kcal/kg; saturation temperature: 252°C and 255°C;

average latent heat h fg = 405.3 kcal/kg; average specific volume of steam = 0.047 m 3 /kg Hence,

dh/dp 2 = ( 265 261 7 2 1 65 − . ) / = . kcal/kg/kg/cm dt/dp =

2 ( 255 252 2 1 − ) / = .55 ° h C/kg/cm

From Equation 6.45, when Q = 0, W f = 0,

W s × 405 3 . +

, 38 000 0 12 1 5 8 600 1 65 × . × . + , × . × dp/dz = 0 or 405 3 . W s + 21 , 0030 dp/dz = 0 ( )

dp/dz from (6.46) = 43 × 0.047/3.26 (W s −W l ) = 0.62 (W s −W l ). Combining this with earlier equation,

s × 405 3 21 030 0 62 . + × . × ( W s − W l ) = 0 . If W l = , 20 400 kg/h W , s = , 19 784 kg / /h

From (6.46), pressure decay = dp/dz = 0.62 × (19,784 − 20,400) = −382 kg/cm 2 /h or −0.106 kg/cm 2 /s. This situation has often been experienced by plant engineers. If for

some reason the gas flow to the HRSG is diverted and the feed water also cut off, there is decay in steam pressure till the heat input is restored. In the case of fresh air–fired HRSGs, there is a time lag between the bypassing of exhaust gases and the start of the fresh air fan. During this period, there is a pressure decay that may be correctly esti- mated by this method.

Example 6.23

Let us assume that the boiler is operating at 20,400 kg/h and suddenly the steam demand goes up to 22,686 kg/h.

Case 1: What happens to steam pressure if we maintain the heat input and feed water supply? Case 2: What happens if feed water is cut off but heat input remains?

Case 1: Feed water enthalpy at 105°C = 105.8 kcal/kg, and feed water flow is the same as present steam flow assuming zero blowdown = 20,400 kg/h.

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

From (6.45), 20,400×(261.7 − 105.8) + W s × 405.3 + 21,030dp/dz = 11.67 × 10 6 also, dp/dz = 0.62 × (W s − 22,686)

. 3 18 10 × 6 + 405 3 . W s + , 21 030 0 62 × . × ( W s − , 22 686 ) = . 11 67 10 × 6

× 6 + × 6 = , 13 443 9 . W W s 295 79 10 . . 8 49 10 or W s , 22 630 kg/h Thus, dp/dz = 0.62 × (22,630 − 22,686) = −35 kg/cm 2 /h = −0.01 kg/cm 2 /s (pressure falls

slightly)

Case 2:

, 13 443 9 . W s = 295 79 11 67 + . ) × 10 ( 6 . or W s = , 22 869 kg/h

dp/dz = . 0 62 ×

2 2 ( 22 , 8869 22 686 − , ) = 113 kg/cm /h = . 0 0315 kg/cm /s

The pressure actually increases because the cooling effect of the feed water is not sensed. In practice, controls respond fast and restore the balance among heat input, feed water flow, and steam flow. The preceding equations represent the worst-case scenario when controls do not act. If we do not adjust the heat input, the steam pressure will slide if we withdraw more steam than generated. The preceding equations may be used to show the trend and are simplistic models. Programs can be developed for more accurate modeling of transients in any steam generator.

Drum-Level Fluctuations

In cogeneration plants, the process conditions dictate the steam demand and could vary

a lot from normal operating conditions. When the steam demand suddenly increases in a boiler, it may take a few milliseconds to a few seconds depending on the control system to adjust the flow and energy input to match the demand. When the evaporation rate in a boiler increases suddenly, water from the boiler tubes is displaced into the drum, which raises the water level momentarily or causes the swell effect. Momentarily, the pressure is also reduced. This swell occurs whether the increased evaporation is due to heat transfer in boiler surfaces or due to self-boiling as a result of falling steam pressure due to higher than ongoing steam generation. Similarly, a shrink effect occurs when the steam demand decreases and water level decreases momentarily. The change in water level is propor- tional to the change in evaporation. The effect of feed water flow will also have to be con- sidered. The mass of water in the drum increases at the following rate:

[(W i −W s ) + T (W s −W l )/dt] where T is the mass of water displaced into drum by unit increase in evaporation = 3600 ×

water weight × (h l −h fw )/[W s × (h v −h fw )], where W i ,W s ,W l are the water flow in, normal steam generation, and steam withdrawn in kg/h, respectively.

If water inventory = 40,000 kg in a boiler and from steam tables h l = 198, h fw = 105,

h v = 665 kcal/kg, and W s = 18,000 kg/h, then T = 3,600 × 40,000 × (198 − 105)/[18,000 × (665 − 105)] = 1,328 s. In a steady-state situation, W s =W l =W i (neglecting blowdown). Then,

dh/dt = level change/s = v[(W i −W s + T(W s −W l )/t]/A

where

A is the drum cross section, m 2

v is the specific volume of saturated liquid, m 3 /kg

Miscellaneous Boiler Calculations 351

Example 6.24

If in the boiler mentioned earlier, steam demand suddenly changes from 18,000 to

21,000 kg/h in 7 s, and if A = 10 m 2 , v = 0.001125 m 3 /kg, then momentary change in

level = 0.001125 × [1,328 × 3,000/3,600/7]/10 = 0.0177 m/s = 17.7 mm/s, which is signifi- cant assuming controls do not react immediately. In practice, the swell will be much smaller and will last for a much shorter period, say, for a few milliseconds with good controls such as three-element-level control system.

Analysis of an Evaporator