Question

A shell-and-tube (two shell passes) heat exchanger is to heat $0.5 \mathrm{~kg} / \mathrm{s}\( of water \)\left(c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ by geothermal brine flowing through the shell passes. The heated water is then fed into commercial warewashing equipment. The National Sanitation Foundation (NSF) standard for commercial warewashing equipment (ANSI/NSF 3) requires that the final rinse water temperature be between 82 and \(90^{\circ} \mathrm{C}\). The geothermal brine enters and exits the heat exchanger at 98 and \(90^{\circ} \mathrm{C}\), respectively. The water flows through a thin-walled tube inside the shell passes. The tube diameter is \(25 \mathrm{~mm}\), and the tube length per pass is 4 \(\mathrm{m}\). The corresponding convection heat transfer coefficients on the outer and inner tube surfaces are 450 and $2700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The estimated fouling factor caused by the accumulation of deposit from the geothermal brine is \(0.0002\) \(\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If the water enters the heat exchanger at \(20^{\circ} \mathrm{C}\), determine the number of tube passes required inside each shell pass to heat the water to \(86^{\circ} \mathrm{C}\) so that it complies with the ANSI/ NSF 3 standard.

Short answer

Answer: Approximately 46 tube passes are required per shell pass.

Step-by-step solution

Determine the mass flow rate of water

Given that the mass flow rate of the water is \(0.5\,\mathrm{kg/s}\).

Calculate the heat transfer in the water

Given that the specific heat capacity of water is \(4200\,\mathrm{J/kg\cdot K}\). The temperature of water increases from \(20^{\circ}\mathrm{C}\) to \(86^{\circ}\mathrm{C}\). The equation for heat transfer, \(Q\), can be written as follows: \(Q = (\dot m\,c_p\Delta T)_{water}\) \(Q = 0.5\,\mathrm{kg/s} \times 4200\,\mathrm{J/kg \cdot K} \times (86 - 20)\) \(Q = 138600\,\mathrm{W}\)

Calculate the overall heat transfer coefficient

The overall heat transfer coefficient, \(U\), is given by: \(1 / U = 1 / h_{i} + 1 / h_{o} + R_{f}\) where \(h_i\) and \(h_o\) are the convection heat transfer coefficients on the inner and outer surfaces of the tube, and \(R_f\) is the fouling factor. \(1 / U = 1 / 2700 + 1 / 450 + 0.0002\) Solving for \(U\), we get: \(U = 399.6\,\mathrm{W/m^2 \cdot K}\)

Calculate the Log Mean Temperature Difference (LMTD)

As this is a two-shell pass heat exchanger, we need to consider the temperature changes happening within the shell. The temperatures are given as: - Hot fluid inlet temp: \(T_{Hi} = 98^\circ\mathrm{C}\) - Hot fluid outlet temp: \(T_{Ho} = 90^\circ\mathrm{C}\) - Cold fluid inlet temp: \(T_{Ci} = 20^\circ\mathrm{C}\) - Cold fluid outlet temp: \(T_{Co} = 86^\circ\mathrm{C}\) LMTD = \(\frac{(ΔT_{1ab}-ΔT_{2ab})}{\ln(\frac{ΔT_{1ab}}{ΔT_{2ab}})}\) \(\Delta T_{1ab} = T_{Hi} - T_{Co}\) \(\Delta T_{1ab} = 98 - 86 = 12\,\mathrm{K}\) \(\Delta T_{2ab} = T_{Ho} - T_{Ci}\) \(\Delta T_{2ab} = 90 - 20 = 70\,\mathrm{K}\) Substituting the values in the LMTD equation: LMTD = \(\frac{(12-70)}{\ln(\frac{12}{70})}\) LMTD = \(24.399\,\mathrm{K}\)

Calculate the required heat transfer area

Using the heat transfer equation \(Q = UA\Delta T\), we can calculate the required heat transfer area \(A\): \(A = \frac{Q}{U \cdot \text{LMTD}}\) \(A = \frac{138600}{399.6 \times 24.399}\) \(A = 14.352\,\mathrm{m^2}\)

Calculate the heat transfer area per tube pass

Using the tube diameter and length information given, we can calculate the heat transfer area per tube pass: \(A_t = \pi D L = \pi \times 0.025 \times 4\) \(A_t = 0.314\,\mathrm{m^2}\) per tube pass

Calculate the number of tube passes required per shell pass

Finally, we can determine the number of tube passes required per shell pass by dividing the total heat transfer area by the heat transfer area per tube pass: Number of tube passes per shell pass = \(\frac{A}{A_t}\) Number of tube passes per shell pass = \(\frac{14.352}{0.314}\) Number of tube passes per shell pass = \(45.71\) Since it's not possible to have a fraction of a tube pass, we can round up to the nearest whole number: The number of tube passes required per shell pass is approximately 46.