A shell-and-tube heat exchanger is to be designed to cool down the petroleum- based organic vapor available at a flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) and at a saturation temperature of \(75^{\circ} \mathrm{C}\). The cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) used for its condensation is supplied at a rate of \(25 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(15^{\circ} \mathrm{C}\). The cold water flows through copper tubes with an outside diameter of \(20 \mathrm{~mm}\), a thickness of \(2 \mathrm{~mm}\), and a length of \(5 \mathrm{~m}\). The overall heat transfer coefficient is assumed to be \(550 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the latent heat of vaporization of the organic vapor may be taken to be \(580 \mathrm{~kJ} / \mathrm{kg}\). Assuming negligible thermal resistance due to pipe wall thickness, determine the number of tubes required.

Step by step solution

01

Calculate the heat transfer needed for condensation

First, we need to calculate the amount of heat transfer required to condense the organic vapor. We'll use the flow rate and latent heat of vaporization values for this calculation: $$Q = \dot{m}_{vapor} \cdot L_{vapor}$$ where: \(Q\) - Heat transfer (W) \(\dot{m}_{vapor}\) - Flow rate of organic vapor (5 kg/s) \(L_{vapor}\) - Latent heat of vaporization of the organic vapor (580 kJ/kg) $$Q = 5 \frac{kg}{s} \cdot 580000 \frac{J}{kg} = 2900000 \mathrm{~W}$$

02

Calculate the heat transfer rate on the water side

Next, we need to determine the heat transfer rate on the water side. We'll use the flow rate, specific heat capacity, and temperature difference values for this calculation: $$\dot{Q} = \dot{m}_{water} \cdot c_{p} \cdot \Delta T$$ where: \(\dot{Q}\) - Heat transfer rate (W) \(\dot{m}_{water}\) - Flow rate of cold water (25 kg/s) \(c_{p}\) - Specific heat capacity of water (4187 J/kg·K) \(\Delta T\) - Temperature difference (75°C - 15°C = 60°C) $$\dot{Q} = 25 \frac{kg}{s} \cdot 4187 \frac{J}{kg \cdot K} \cdot 60 K = 6304500 \mathrm{~W}$$

03

Calculate the heat transfer coefficient for the system

Now, we need to determine the heat transfer coefficient for the system. The overall heat transfer coefficient was given in the problem statement: $$U = 550 \frac{W}{m^{2} \cdot K}$$

04

Calculate the convective heat transfer surface area

We need to calculate the convective heat transfer surface area for the pipes with the given diameter and length: $$A = N \cdot \pi d \cdot L$$ where: \(A\) - Convective heat transfer surface area (m²) \(N\) - Number of tubes \(d\) - Diameter of the tubes (20 mm = 0.02 m) \(L\) - Length of the tubes (5 m)

05

Calculate the number of tubes needed

The heat transfer equation, assuming negligible thermal resistance due to pipe wall thickness, is: $$Q = U \cdot A \cdot \Delta T$$ Rearrange the equation to find the number of tubes: $$N = \frac{Q}{\pi d \cdot L \cdot U \cdot \Delta T}$$ Now substitute the values: $$N = \frac{2900000 \mathrm{~W}}{\pi (0.02 \mathrm{~m}) (5 \mathrm{~m}) (550 \frac{W}{m^{2} \cdot K}) (60 \mathrm{~K})} = 16.65$$ Since we can't have a fraction of a tube, we'll round up to the next whole number: $$N = 17$$ There are 17 tubes required for the shell-and-tube heat exchanger.

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