Exercise: Calculation of tube-side pressure drop for a 1-2 heat exchanger

Gonzalo Mantilla

PROBLEM STATEMENT:

Suppose you are designing a power plant wherein feedwater must be fed through a 1-2 heat exchanger as part of a vaporization process to then be fed into a turbine - for the power demands, the water flow rate must be 27.12 kg/s through the heat exchanger, and the pressure must be maintained at 10.69 MPa. The water is being fed through the tube-side at the specified pressure, with inlet and outlet water temperatures of 45.81 and 315.9 centigrade, respectively. From steam tables, it is known that the inlet and outlet water enthalpy is 201.1 and 1437 kJ/kg, respectively. Solar salt must be used as the heat transfer fluid (HTF) feeding through shell-side - however, it has a very high melting point (238 centigrade), and so the outlet HTF temperature must be kept at 300 centigrade to save on heating costs for thermal energy storage (thermal stability is up to 600 centigrade). For solar salt as the HTF, the following thermochemical data is available [1]:

ρ (kg/m³) = 2090 - 0.63T (°C)

C_P (kJ/kg-K) = 1.5404 + (3.0924*10^-5)T (°C)

From standard policy, you can assume that the maximum water flow velocity for the water going into the exchanger is 3 m/s [2]. ASME standards dictate that water service conditions exceeding 250 psi and 220°F must use seamless carbon-steel (also called wrought iron) and have at least schedule 80 thickness (pg. 32) [3]. You will assume water thermochemical properties are independent of pressure, and that they remain the same as in atmospheric pressure conditions, can be found through interpolation of tabulated values [4, 5]. Roughness of common piping materials are also tabulated [6]. For heat exchanger water in forced liquid flow, the overall heat transfer coefficient can be estimated to be between 900 and 2500 W/m²-K [7]. As part of your design, you must ensure that the tube-side pressure drop and tube length are balanced, such that pressure of the water in the outlet is maintained without needing too use much tubing length - this appears to be mostly dependent on the number of tubes. Using the listed resources, and design equations reviewed in class, determine the pressure drop on the tube-side and the tubing length and plot against number of tubes available.

[1]: https://www.mdpi.com/1996-1073/14/22/7486

[2]: https://www.engineeringtoolbox.com/flow-velocity-water-pipes-d_385.html

[3]: https://engstandards.lanl.gov/esm/pressure_safety/Section%20REF-3-R0.pdf

[4]: https://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html

[5]: https://www.engineeringtoolbox.com/water-dynamic-kinematic-viscosity-d_596.html

[6]: https://www.engineeringtoolbox.com/surface-roughness-ventilation-ducts-d_209.html

[7]: https://www.engineeringtoolbox.com/overall-heat-transfer-coefficient-d_434.html

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SOLUTION:

The first step to create the plot is determining the exchanger heat duty. Given the water flow rates and enthalpies, the calculation is simple:

$\dot{Q} = \dot{m} C p Δ T = \dot{m} Δ H = U A (L M T D)$

Doing this results in Q = 33511.8 kW. However, now it must be recognized that the heat duty must match the temperature change of the HTF - with the equations given, it is possible to estimate properties at the bulk temperature, such that an iterative equation can be created:

$T_{H T F} = \frac{| T_{i n} - T_{o u t} |}{2}$ , $ρ = 2090 \frac{k g}{m^{3}} - 0.63 C^{- 1} T_{H T F}$ , $C_{P} = 1.5404 \frac{k J}{k g * C} + (3.0924 * 1 0^{- 5} \frac{k J}{k g * C^{2}}) * T_{H T F}$ , $Δ H_{H T F} = C_{P} * Δ T$ , ${\dot{Q}}_{H T F} = {\dot{m}}_{H T F} * Δ H_{H T F}$ , $\dot{Q} - {\dot{Q}}_{H T F} = 0$

The last equation can be solved using Excel solver by manipulating the outlet HTF mass flow rate and temperature - it should be noted that inlet HTF temperature must NOT exceed 600 centigrade, as it was previously mentioned that thermal stability is maintained until this point (this is set as a constraint in the program). A range of solutions is possible, but only some will be acceptable for heat exchanger design. To determine this, the log-mean temperature difference (LMTD) correction factor (F) must be determined. This is defined as the geometric correction factor applied to the LMTD, since the LMTD is calculated as though we were modeling a 1-1 heat exchanger. For F, the variables in question which must be defined are the hourly heat-capacity ratio (Z) and the heating effectiveness (η_H) - from these, F is determined as follows:

$Z = \frac{T_{H T F, i n} - T_{H T F, o u t}}{T_{w a t e r, o u t} - T_{w a t e r, i n}}$ , $η_{H} = \frac{T_{w a t e r, o u t} - T_{w a t e r, i n}}{T_{H T F, i n} - T_{w a t e r, i n}}$ , $B_{1} = (Z^{2} + 1)^{0.5}$ , $F_{N} = \frac{B_{1}}{Z - 1} \log_{10} (\frac{1 - η_{H}}{1 - η_{H} Z})$ , $F_{D} = l o g_{10} (\frac{(2 / η_{H}) - 1 - Z + B_{1}}{(2 / η_{H}) - 1 - Z - B_{1}})$ , $F = \frac{F_{N}}{F_{D}}$

It is common practice to reject heat exchanger design with F < 0.75, as it indicates the countercurrent LMTD driving force is too small for design parameters to be met - this is why only some solutions to the iterative equation above are accepted, as it heavily depends on the inlet temperature of the HTF. In fact, one can set a cell equal to F in an Excel sheet with all the aforementioned equations and add F > 0.75 as a constraint. In the solved example, our solution has inlet HTF temperature is 464.217 centigrade, and HTF mass flowrate is 131.471 kg/s. Thus, our F = 0.76 approximately. Now the LMTD must be calculated, which is defined accordingly for countercurrent flow with the correction factor:

$Δ T_{1} = T_{H T F, i n} - T_{w a t e r, o u t}$ , $Δ T_{2} = T_{H T F, o u t} - T_{w a t e r, i n}$ , $L M T D = \frac{F (Δ T_{2} - Δ T_{1})}{\ln [(Δ T_{2} / Δ T_{1})]}$

This leads to an LMTD of 149.23 K, which will prove useful later. Now, the limits of our process are considered. With a maximum water flow rate of 3 m/s, the density must be found before volumetric flow rates and piping dimensions can be determined. Evaluating at the bulk mean water temperature (identical to HTF evaluation), interpolation yields ρ = 886.1 kg/m³. Thus, the minimum tube internal diameter for the heat-exchanger is defined as follows:

$\dot{V} = \frac{\dot{m}}{ρ}$ , $A_{i} = \frac{\dot{V}}{v_{m a x}}$ , $d_{i n} = \sqrt (\frac{4 A_{i}}{π})$

Inserting all values, a diameter of ~114 mm is considered. At this point, the number of tubes (N_T) becomes a variable, as the number of tubes per pass is a direct variable in the volumetric flow-rate (and thus diameter) of each tube - in this example, we will consider tube counts of 75, 150, 300, 450, 600, and 900 for our plotting. With 2 tube passes (n_pass = 2), the real tube internal diameter is considered as such:

${\dot{V}}_{t u b e} = \frac{\dot{V}}{(N_{T} / n_{p a s s})}$ , $A_{i n, t u b e} = \frac{A_{i}}{(N_{T} / n_{p a s s})}$ , $d_{i n, t u b e} = \sqrt (\frac{4 A_{i n, t u b e}}{π})$

The reasoning behind these operations is that the number of tubes per pass represents the "real" count of tubes (i.e. uninterrupted lines of flow) rather than the somewhat misleading terminology of "number of tubes". This can be illustrated by following the flow arrows of the tube-side fluid throughout the 1-2 heat exchanger pictured below - notice that N_T counts the number of repeated horizontal flow lines across the whole exchanger, while the number of tubes per pass represents how many continuous lines of flow can be corresponded one-to-one with a tube in each pass (if still confused, simply count the number of pipes at the tube-side inlet plenum - notice it is equal to the number of tubes at the outlet plenum).

Recall that carbon steel was selected as the material for the piping - absolute roughness values are tabulated at ε = 0.09 * 10^-3 m. The pressure drop is indirectly related to this value due to the Colebrook-White equation to determine the Fanning friction factor, which is directly related to the pressure drop. The Colebrook-White equation requires computation of the Reynold's number of tube-side flow, both of which are computed as such:

$\frac{1}{\sqrt f_{F}} = - 4.0 \log_{10} [(\frac{ϵ}{d_{i n, t u b e}}) (\frac{1}{3.7}) + \frac{1.255}{R e \sqrt f_{F}}]$ , $R e = \frac{ρ v_{m a x} d_{i n, t u b e}}{μ}$

At the bulk mean water temperature, viscosity is interpolated at μ = 0.0001496 N*s/m². Thus, the Colebrook-White equation is solved for iteratively and the Fanning friction factor is obtained. Thus, the pressure drop along a tube (and thus, all tubes) is written as follows, with there notably being one more unspecified variable - the length (L):

$\frac{| Δ P_{t u b e} |}{ρ g} = n_{p a s s} (4 f_{F} \frac{L}{d_{i n, t u b e}} + 4) (\frac{v_{m a x}^{2}}{2 g})$

Recall that we previously defined an overall heat transfer coefficient range of U = 900 - 2500 W/m²-K for heat exchanger applications of forced water flow. While the defined U may make the assumption that the heat transfer area along the inside and outside pipe of the tube may be equal, the requirement of schedule 80 thickness causes the assumption to be dubious at best. Therefore, we rely on the inside overall heat transfer coefficient (U_i) for the calculation, and refer back to our heat duty equation while defining the inside heat transfer area (A_i) to then define the tube length (L):

$\dot{Q} = U_{i} A_{i} (L M T D)$ , $A_{i} = N_{T} (π d_{i n, t u b e}) L$

With the tube length defined, the pressure drop can finally be solved for, and the results can be plotted - if performed correctly, the graphs should resemble the example below (one for the lower boundary of U_i, and one for the upper boundary of U_i):

https://drive.google.com/file/d/1TR1vHK02WZxKn_DBct4nnuYRiNys108U/view?usp=sharing

https://drive.google.com/file/d/1g2Me84LVrNSCcUpsdSnGMrRzLeuBA8PL/view?usp=sharing

Thus, the problem is solved. The plotting was for the purpose of demonstration - note a true balance of pressure drop and length would have to account for the costs of materials and the energy balance differences from a higher inlet pressure to account for the tube-side drop.