Question

A shell-and-tube heat exchanger with one shell pass and 14 tube passes is used to heat water in the tubes with geothermal steam condensing at $120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)$ on the shell side. The tubes are thin-walled and have a diameter of \(2.4 \mathrm{~cm}\) and a length of \(3.2 \mathrm{~m}\) per pass. Water \(\left(c_{p}=4180\right.\) \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) ) enters the tubes at $18^{\circ} \mathrm{C}\( at a rate of \)6.2 \mathrm{~kg} / \mathrm{s}$. If the temperature difference between the two fluids at the exit is \(46^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer, \((b)\) the rate of condensation of steam, and \((c)\) the overall heat transfer coefficient.

Short answer

Based on the given parameters of a shell-and-tube heat exchanger and the step-by-step calculations: (a) The rate of heat transfer is 1.448 x 10^6 W. (b) The rate of condensation of steam is 0.657 kg/s. (c) The overall heat transfer coefficient is 623.9 W/m^2.K.

Step-by-step solution

Calculate the temperature change of water

In order to determine the rate of heat transfer, we first need to find the temperature change of the water. The initial temperature of water is given as \(18^{\circ}\mathrm{C}\); the temperature difference between the two fluids at the exit is given as \(46^{\circ}\mathrm{C}\). The geothermal steam condenses at \(120^{\circ}C\), so we can calculate the final temperature of the water as follows: Final temperature of water = \(120^{\circ}\mathrm{C}\) - \(46^{\circ}\mathrm{C}\) = \(74^{\circ}\mathrm{C}\) Temperature change of water = Final temperature - Initial temperature = \(74^{\circ}\mathrm{C}\) - \(18^{\circ}\mathrm{C}\) = \(56^{\circ}\mathrm{C}\).

Determine the rate of heat transfer (a)

Now we can find the rate of heat transfer using the formula: Rate of heat transfer = mass flow rate of water × specific heat of water × temperature change \(q = \dot{m}_{water} \times c_p \times \Delta T\) Let's plug in the values: \(q = (6.2\, \mathrm{kg/s})\times(4180\, \mathrm{J/kg \cdot K})\times(56^{\circ}\mathrm{C})\) \(q = 1.448\times10^6\, \mathrm{W}\)

Calculate the rate of condensation of steam (b)

The rate of condensation of steam can be calculated as follows: Rate of condensation = Rate of heat transfer / Latent heat of vaporization \(\dot{m}_{steam} = \frac{q}{h_{fg}}\) Plug in the given heat transfer rate and latent heat of vaporization: \(\dot{m}_{steam} = \frac{1.448\times10^6\, \mathrm{W}}{2203\, \mathrm{kJ/kg}} = \frac{1.448\times10^6\, \mathrm{W}}{2.203\times10^6\, \mathrm{J/kg}}\) \(\dot{m}_{steam} = 0.657\, \mathrm{kg/s}\)

Calculate the overall heat transfer coefficient (c)

Now, we can calculate the overall heat transfer coefficient using the formula: \(U=\frac{q}{A \cdot \Delta T_{lm}}\) First, we need to determine the log mean temperature difference (\(\Delta T_{lm}\)). Using the temperatures of water and steam at the inlet and exit, we can calculate \(\Delta T_{lm}\) as follows: \(\Delta T_1 = T_{inlet, steam} - T_{inlet,water} = 120^{\circ}\mathrm{C} - 18^{\circ}\mathrm{C} = 102^{\circ}\mathrm{C}\) \(\Delta T_2 = T_{exit, steam} - T_{exit,water} = 120^{\circ}\mathrm{C} - 74^{\circ}\mathrm{C} = 46^{\circ}\mathrm{C}\) \(\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)} = \frac{102^{\circ}\mathrm{C} - 46^{\circ}\mathrm{C}}{\ln(102^{\circ}\mathrm{C}/46^{\circ}\mathrm{C})} = 69.76^{\circ}\mathrm{C}\) Now, let's calculate the area of the heat exchange surface: \(A = N \times \pi d L\) \(N\) is the number of tube passes: \(N = 14\) \(d\) is the diameter of the tubes: \(d = 2.4\times10^{-2}\, \mathrm{m}\) \(L\) is the length per pass of the tubes: \(L = 3.2\, \mathrm{m}\) \(A = 14 \times \pi (2.4\times10^{-2}\, \mathrm{m}) (3.2\, \mathrm{m}) = 3.368\, \mathrm{m^2}\) Finally, let's calculate the overall heat transfer coefficient: \(U = \frac{1.448\times10^6\, \mathrm{W}}{3.368\, \mathrm{m^2} \times 69.76^{\circ}\mathrm{C}} = 623.9\, \mathrm{W/m^2\cdot K}\) To summarize the results: (a) The rate of heat transfer is \(1.448\times10^6\, \mathrm{W}\). (b) The rate of condensation of steam is \(0.657\, \mathrm{kg/s}\). (c) The overall heat transfer coefficient is \(623.9\, \mathrm{W/m^2\cdot K}\).