Question 6.85: A heat exchanger is to cool ethylene glycol (cp = 2.56 kJ/kg......
A heat exchanger is to cool ethylene glycol \left(c_{p}=\right. 2.56 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C} ) flowing at a rate of 2 \mathrm{~kg} / \mathrm{s} from 80^{\circ} \mathrm{C} to 40^{\circ} \mathrm{C} by water \left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right) that enters at 20^{\circ} \mathrm{C} and leaves at 55^{\circ} \mathrm{C}. Determine (a) the rate of heat transfer and (b) the mass flow rate of water.
Ethylene glycol is cooled by water in a heat exchanger. The rate of heat transfer in the heat exchanger and the mass flow rate of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. \mathrm{3} Changes in the kinetic and potential energies of fluid streams are negligible. \mathrm{4} Fluid properties are constant.
Properties The specific heats of water and ethylene glycol are given to be 4.18 and 2.56 \mathrm{~kJ} / \mathrm{kg} .{ }^{\circ} \mathrm{C}, respectively.
Analysis (a) We take the ethylene glycol tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as
\begin{aligned} & \underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c} \text { Rate of net energy transfer } \\ \text { by heat, work, and mass } \end{array}}=\underbrace{\Delta \dot{E}_{\text {system }} {}^{\nearrow 0 \text { (steady) }}}_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc. energies } \end{array}}=0 \\ & \dot{E}_{\text {in }}=\dot{E}_{\text {out }} \\ & \dot{m} h_{1}=\dot{Q}_{\text {out }}+\dot{m} h_{2} \quad(\text { since } \Delta \mathrm{ke} \cong \Delta \mathrm{pe} \cong 0) \\ & \dot{Q}_{\text {out }}=\dot{m} c_{p}\left(T_{1}-T_{2}\right) \end{aligned}
Then the rate of heat transfer becomes
\dot{Q}=\left[\dot{m} c_{p}\left(T_{\text {in }}-T_{\text {out }}\right)\right]_{\text {glycol }}=(2 \mathrm{~kg} / \mathrm{s})\left(2.56 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(80^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}\right)=\mathrm{2 0 4 . 8} \mathrm{~ k W}
(b) The rate of heat transfer from glycol must be equal to the rate of heat transfer to the water. Then,
\dot{Q}=\left[\dot{m} c_{p}\left(T_{\text {out }}-T_{\text {in }}\right)\right]_{\text {water }} \longrightarrow \dot{m}_{\text {water }}=\frac{\dot{Q}}{c_{p}\left(T_{\text {out }}-T_{\text {in }}\right)}=\frac{204.8 \mathrm{~kJ} / \mathrm{s}}{\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(55^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}\right)}=\mathrm{1 . 4} \mathrm{~ k g} / \mathrm{s}
