Question

The lowest flow in the Menominee River in July is about 40 m3/s. If the river temperature is 18°C and a power plant discharges 2 m3/s of cooling water at 80°C, what is the final river temperature after the cooling water and river have mixed?

Answer 1

$21^{\circ} \mathrm{C}$ is the final river temperature after the cooling water and river have mixed.

Explanation:

Final river temperature after mixing using energy energy balance equation

$\Delta E_{\text {River }}+\Delta E_{\text {Cooling water}}=0 \ldots \ldots(1)$

$\Delta E_{\text {River }}=Q_{\text {River}} \rho C_{P}\left(T-T_{\text {River }}\right)$

$\Delta E_{\text {cooling water }}=Q_{\text {cooling water }} \rho C_{P}\left(T-T_{\text {Cooling water }}\right)$

$Q_{\text {River}} \rho C_{P}\left(T-T_{\text {River }}\right)+Q_{\text {Cooling water }} \rho C_{P}\left(T-T_{\text {Cooling water }}\right)=0 \ldots \ldots .(2)$

Where,$C_{P}$ is specific heat at constant pressure,$\Delta T$ change in temperature,$Q_{\text {River}}$ is flow in the river,$Q_{\text {cooling water }}$is the flow of cooling water from plant,$T$ final required temperature after mixing cooling water and river water,and,$\rho$ is density of water.

From Diagram

While substituting,

  $40 \mathrm{m}^{3} / \mathrm{s} \text { for } Q_{\text {River }}$

$2 \mathrm{m}^{3} / \mathrm{s} \quad for\quad Q_{\text {cooling water }}$

$80^{\circ} \mathrm{C} \text { for } T_{\text {cooling water }}$ and

$18^{\circ} \mathrm{C} \text { for } T_{\mathrm{River}}$

$Q_{\text {River}} \rho C_{P}\left(T-T_{\text {River }}\right)+Q_{\text {Cooling water }} \rho C_{P}\left(T-T_{\text {Cooling water }}\right)=0$

$40 \times \rho C_{P}(T-18)+2 \times \rho C_{P}(T-40)=0$

$40 \times(T-18)+2 \times(T-40)=0$

$T=21^{\circ} \mathrm{C}$

Finally, the temperature after the mixing of cool water and river is $21^{\circ} \mathrm{C}$ .