Question

Glycerin \(\left(c_{p}=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(20^{\circ} \mathrm{C}\) and \(0.5 \mathrm{~kg} / \mathrm{s}\) is to be heated by ethylene glycol $\left(c_{p}=2500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)60^{\circ} \mathrm{C}$ in a thin-walled double-pipe parallel-flow heat exchanger. The temperature difference between the two fluids is \(15^{\circ} \mathrm{C}\) at the outlet of the heat exchanger. If the overall heat transfer coefficient is $240 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and the heat transfer surface area is \)3.2 \mathrm{~m}^{2}$, determine \((a)\) the rate of heat transfer, \((b)\) the outlet temperature of the glycerin, and \((c)\) the mass flow rate of the ethylene glycol.

Short answer

Question: Determine the rate of heat transfer (a), the outlet temperature of glycerin (b), and the mass flow rate of ethylene glycol (c) in a parallel-flow heat exchanger given the properties and flow rates of glycerin and ethylene glycol. Answer: (a) The rate of heat transfer is 18932.16 W. (b) The outlet temperature of glycerin is 43.14°C. (c) The mass flow rate of ethylene glycol is 0.381 kg/s.

Step-by-step solution

Overall heat transfer equation

For a heat exchanger, the overall heat transfer is given by: \(Q = U \cdot A \cdot \Delta T_{lm}\) Where: \(Q\) is the rate of heat transfer, \(U\) is the overall heat transfer coefficient, \(A\) is the heat transfer surface area, \(\Delta T_{lm}\) is the logarithmic mean temperature difference. Now, we are given \(U = 240 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(A = 3.2 \mathrm{~m}^{2}\). We need to find \(\Delta T_{lm}\) to calculate the rate of heat transfer \((a)\).

Calculate logarithmic mean temperature difference

Since it's a parallel-flow heat exchanger, we have: \(\Delta T_{1} = T_{h1} - T_{c1} = 60 - 20 = 40^{\circ} \mathrm{C}\) \(\Delta T_{2} = T_{h2} - T_{c2}\) From the problem statement, \(T_{h2} - T_{c2} = 15^{\circ} \mathrm{C}\). Now using the formula for logarithmic mean temperature difference: \(\Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\) \(\Delta T_{lm} = \frac{40 - 15}{ln(\frac{40}{15})} = 24.71^{\circ} \mathrm{C}\)

Calculate the rate of heat transfer (a)

Now that we have the logarithmic mean temperature difference, we can find the heat transfer rate: \(Q = U \cdot A \cdot \Delta T_{lm}\) \(Q = 240 \cdot 3.2 \cdot 24.71 = 18932.16 \mathrm{~W}\)

Calculate the outlet temperature of glycerin (b)

We are given the mass flow rate of glycerin, \(m_{c} = 0.5 \mathrm{~kg} / \mathrm{s}\) and its specific heat capacity, \(c_{p} = 2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). We can now find the outlet temperature of glycerin using the equation: \(Q = m_{c} \cdot c_{p} \cdot (T_{c2} - T_{c1})\) Solving for \(T_{c2}\): \(T_{c2} = \frac{Q}{m_{c} \cdot c_{p}} + T_{c1} = \frac{18932.16}{0.5 \cdot 2400} + 20 = 43.14^{\circ} \mathrm{C}\)

Calculate the mass flow rate of ethylene glycol (c)

We are given the specific heat capacity of ethylene glycol, \(c_{p} = 2500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). We can now find the mass flow rate of ethylene glycol, \(m_{h}\), using the equation: \(Q = m_{h} \cdot c_{p} \cdot (T_{h1} - T_{h2})\) First, we find \(T_{h2}\) using the relationship: \(T_{h2} = T_{c2} + 15 = 43.14 + 15 = 58.14^{\circ} \mathrm{C}\) Now, solving for \(m_{h}\): \(m_{h} = \frac{Q}{c_{p} \cdot (T_{h1} - T_{h2})} = \frac{18932.16}{2500 \cdot (60 - 58.14)} = 0.381 \mathrm{~kg} / \mathrm{s}\) In conclusion, we have: (a) The rate of heat transfer is \(18932.16 \mathrm{~W}\). (b) The outlet temperature of the glycerin is \(43.14^{\circ} \mathrm{C}\). (c) The mass flow rate of the ethylene glycol is \(0.381 \mathrm{~kg} / \mathrm{s}\).