Calculation procedure for horizontal tube bank heat exchangers

In the calculation model, horizontal tube bank heat exchangers are modeled so that they consist of tube banks and small cavities. This is illustrated in figure 56.

Figure 56. Heat exchangers in back-pass.

In figure 56 starting from the top, there can be seen the last tube bank of the superheater, a small cavity between the superheater and the economizer, a first tube bank of the econo-mizer, a small cavity of the econoecono-mizer, a second tube bank of the econoecono-mizer, a small cavity between the economizer and the air heater and a first tube bank of the air heater. Small cavities between heat exchangers are distributed for adjacent heat exchangers in some pro-portions, usually so that the proportion for one of the heat exchangers is zero. For example, in figure 56 the small cavity between the superheater and the economizer belongs in practice entirely for the economizer. However, the superheater also has a cavity after its last tube bank, but its height is zero.

Consider how the performance of the above economizer is solved. The economizer is as-sumed to be a parallel-flow heat exchanger which inside and outside inlet enthalpies and pressures are known. First, unknown outlet enthalpies and pressures are guessed. Then, ini-tial guesses for intermediate enthalpies and pressures can also be calculated if linear enthalpy and pressure profiles inside the heat exchanger for both fluids are assumed. Thus, a similar pressure drop is assumed also through cavities than through tube rows, even if the height of the cavity is zero. Also, overall average enthalpies end pressures for the whole heat ex-changer are calculated which are used in pressure loss calculations. Tube surface and fouling layer surface temperatures in both sides of the tube shown in figure 36 for each tube row are also guessed and logarithmic mean temperature for all those layers are calculated. The sur-face temperature of the outside fouling layer is needed for radiation calculations and the average tube wall temperature is needed to determine the thermal conductivity of the tube wall.

In the solution procedure, small cavities are solved first one by one proceeding in a direction of the outside fluid. In the case of the first small cavity, the total absorption fraction of the last tube bank of the superheater and the outside fouling layer surface temperature of the last tube row of the superheater are brought into the calculation as an input. If there is a large cavity instead of a heat exchanger next to the small cavity as is illustrated in figure 57, radi-ation from the small cavity to the large cavity is not calculated because there is no tube temperature that would receive the radiation.

Figure 57. Large cavity next to a small cavity.

When the energy released from the outside fluid in the first cavity has been solved, the new outlet enthalpy is calculated for the first cavity. At this point, it would be possible to calculate the new average enthalpy for the first cavity by using the new outlet enthalpy and solve the cavity again. This iteration could be continued until the energy balance of the first cavity matches. However, this is not done but the first cavity is solved just once at this point. After the first cavity, the second cavity is solved, and the new outlet enthalpy of the second cavity is calculated. The calculation model also calculates the third cavity but because in this case, its height is zero, the energy released from the outside fluid in the third cavity is basically just set to be zero.

When cavities are solved and thus radiations that they send for adjacent tube banks are known, it is time to solve the first tube bank in a direction of the outside fluid. Like cavities, also tube banks are not iterated but calculated just once through. After enthalpies are solved for every tube row using the appropriate matrix equation, also new tube surface and fouling layer surface temperatures are solved for every tube row. New temperatures are calculated from the following equations which are based on the thermal circuit:

𝜙_𝑐 − 𝜙_𝑒𝑥 =^{𝑇𝑜−𝑇𝑓,𝑜}

𝑅_{𝑜,𝑡𝑜𝑡} (7.7)

𝜙_𝑐 − 𝜙_𝑒𝑥 =^{𝑇𝑓,𝑜−𝑇𝑡,𝑜}

𝑅_𝑓,𝑜 (7.8)

𝜙_𝑐 − 𝜙_𝑒𝑥 =^{𝑇𝑡,𝑜−𝑇𝑡,𝑖}

𝑅𝑤 (7.9)

𝜙_𝑐 − 𝜙_𝑒𝑥 =^{𝑇𝑡,𝑖−𝑇𝑓,𝑖}

𝑅𝑓,𝑖 (7.10)

In equations 7.7 – 7.10, external radiation must be reduced from the heat to the cold fluid because external radiation is not driven by temperature differences in question. For example, the surface temperature of the outside fouling layer is obtained from equation 7.7 as follows:

𝑇_𝑓,𝑜 = 𝑇_𝑜− (𝜙_𝑐− 𝜙_𝑒𝑥) ⋅ 𝑅_{𝑜,𝑡𝑜𝑡} (7.11)

If the fouling layer thermal resistance is not directly known, it can be calculated from the fouling factor. Connections of fouling factors and fouling resistances can be seen from equa-tions 5.72 and 5.73 and those are as follows:

1

ℎ𝑜,𝑐+ℎ𝑜,𝑟+ 𝑅_𝑓,𝑜𝐴_𝑠,𝑜 = ¹

𝑓𝑜(ℎ𝑜,𝑐+ℎ𝑜,𝑟) (7.12)

𝑅_𝑓,𝑖𝐴_𝑠,𝑜+ ^𝑑𝑜

ℎ_𝑖,𝑐𝑑_𝑖 = ^𝑑𝑜

𝑓_𝑖ℎ_𝑖,𝑐𝑑_𝑖 (7.13)

From equations 7.12 and 7.13, foul layers thermal resistances can be solved as follows:

𝑅_𝑓,𝑜 = ( ¹

𝑓_𝑜(ℎ_𝑜,𝑐+ℎ_𝑜,𝑟)− ¹

ℎ_𝑜,𝑐+ℎ_𝑜,𝑟) /𝐴_𝑠,𝑜 (7.14)

𝑅_𝑓,𝑖 = ( ^𝑑𝑜

𝑓𝑖ℎ𝑖,𝑐𝑑𝑖 − ^𝑑𝑜

ℎ𝑖,𝑐𝑑𝑖) /𝐴_𝑠,𝑜 (7.15)

In some situations, heat transfer may have been corrected by a fouling factor greater than one that would lead to a negative foul layer resistance. In those situations, the foul layer thermal resistance is set to be zero, and outside/inside resistance is calculated with the foul-ing factor. For example, if the inside foulfoul-ing factor is greater than one, the inside convective thermal resistance is calculated as follows:

𝑅_𝑖,𝑐 = ¹

𝑓_𝑖ℎ_𝑖,𝑐𝐴_𝑠,𝑖 (7.16)

After surface temperatures are solved, new logarithmic mean temperatures for foul layers and tube wall can be calculated. For example, the mean temperature for the tube wall is calculated as follows:

𝑇_𝑡,𝑚= ^{𝑇𝑡,𝑜−𝑇𝑡,𝑖}

ln (^{𝑇𝑡,𝑜}

𝑇𝑡,𝑖) (7.17)

When the first tube bank is solved, inside outlet enthalpies of the first tube bank are updated for the inlets of the second tube bank. This can be done because parts of the tubes which go through the second small cavity and connect the outlet of the first tube bank and the inlet of the second tube bank are assumed to be unheated. It is also assumed that there are no pressure losses in those tube parts. After the first tube bank, the second tube bank is solved. External radiation from the first cavity of the air heater is brought into the calculation of the second tube bank as an input.

After cavities and tube banks are solved, pressure losses for the whole heat exchanger are calculated using overall average fluid properties. Then outlet pressures are updated in this case and linear pressure losses through the heat exchanger are once again assumed to get new intermediate pressures. Now all temperatures, enthalpies, and pressures have been up-dated once and the new values can be compared to the old ones. The calculation procedure is repeated until the values no longer change significantly.

In situations where inside outlet enthalpies are known instead of inside inlet enthalpies, the enthalpy profile obtained from the iteration needs to be corrected as was mentioned earlier.

That happens so that the average is taken from the inside inlet enthalpies and this average enthalpy is set for all inside inlets. Then the calculation procedure described in this chapter is repeated so that inside outlets are solved. Then the obtained average inside outlet enthalpy is compared to the known inside outlet enthalpy. If the outlet enthalpy is for example 100 kJ/kg bigger than the target enthalpy, the inlet enthalpy is reduced by 100 kJ/kg and the calculation procedure is repeated. This guessing procedure is continued until the calculated outlet enthalpy matches with the known outlet enthalpy on wanted accuracy.