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}$ and the same mass flow rate in a thin-walled double-pipe parallelflow heat exchanger. If the overall heat transfer coefficient is \(380 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) and the heat transfer surface area is \(6.5 \mathrm{~m}^{2}\), determine \((a)\) the rate of heat transfer and \((b)\) the outlet temperatures of the glycerin and the glycol.

Short answer

The rate of heat transfer is 98,760 W. The outlet temperature of glycerin is 102.3°C, and the outlet temperature of ethylene glycol is -19.0°C.

Step-by-step solution

Calculate the heat capacity rates of glycerin and glycol

First, let's find the heat capacity rate (\(C_{r}\)) for each fluid, which is the product of their respective mass flow rates and specific heat capacities. The formulas are: \(C_{r_{glycerin}} = \dot{m}_{glycerin} \cdot c_{p_{glycerin}}\) \(C_{r_{glycol}} = \dot{m}_{glycol} \cdot c_{p_{glycol}}\) For glycerin, we have: \(C_{r_{glycerin}} = 0.5 \frac{kg}{s} \cdot 2400 \frac{J}{kg\cdot K} = 1200 \frac{J}{s\cdot K}\) For ethylene glycol, we have: \(C_{r_{glycol}} = 0.5 \frac{kg}{s} \cdot 2500 \frac{J}{kg\cdot K} = 1250 \frac{J}{s\cdot K}\)

Determine the heat transfer rate

The heat transfer rate (\(\dot{Q}\)) can be found using the overall heat transfer coefficient (U), the surface area of the heat exchanger (A), and the difference in the inlet temperatures of the two fluids (\(\Delta T_i\)): \(\dot{Q} = U\cdot A\cdot \Delta T_i\) We are given that \(U = 380 \frac{W}{m^2 K}\) and \(A = 6.5 m^2\). The temperature difference can be found as: \(\Delta T_i = T_{glycol,in} - T_{glycerin,in} = 60°C - 20°C = 40 K\) Now, we can calculate the heat transfer rate: \(\dot{Q} = 380 \frac{W}{m^2 K} \cdot 6.5 m^2 \cdot 40 K = 98760 W\)

Calculate the temperature change of each fluid

The temperature change in each fluid can be calculated as: \(\Delta T_{glycerin} = \frac{\dot{Q}}{C_{r_{glycerin}}}\) \(\Delta T_{glycol} = -\frac{\dot{Q}}{C_{r_{glycol}}}\) For glycerin: \(\Delta T_{glycerin} = \frac{98760 W}{1200 \frac{J}{s\cdot K}} = 82.3 K\) For ethylene glycol: \(\Delta T_{glycol} = -\frac{98760 W}{1250 \frac{J}{s\cdot K}} = -79.0 K\)

Find the exit temperatures of glycerin and glycol

Finally, we can find the outlet temperatures of both fluids by adding their temperature changes to their inlet temperatures: \(T_{glycerin,out} = T_{glycerin,in} + \Delta T_{glycerin} = 20°C + 82.3 K = 102.3°C\) \(T_{glycol,out} = T_{glycol,in} + \Delta T_{glycol} = 60°C - 79.0 K = -19.0°C\) So, the outlet temperatures of glycerin and ethylene glycol are \(102.3°C\) and \(-19.0°C\), respectively. In summary: (a) The rate of heat transfer is \(98760 W\). (b) The outlet temperatures of glycerin and ethylene glycol are \(102.3°C\) and \(-19.0°C\), respectively.