QuestionA power station gets rid of 600 MW of surplus energy into a nearby river. The river is 20 m wide, 3.0 m deep and flows at $1.2 \mathrm{~m} \mathrm{~s}^{-1}$ . Calculate the temperature rise.

Studdy Solution

STEP 1

1. The power station releases 600 MW of energy into the river.
2. The river is 20 meters wide, 3.0 meters deep, and flows at a speed of $1.2 \, \mathrm{m/s}$ .
3. We need to calculate the temperature rise of the river water.

STEP 2

1. Calculate the volume flow rate of the river.
2. Calculate the mass flow rate of the river.
3. Use the energy balance to find the temperature rise.

STEP 3

Calculate the volume flow rate of the river.
The volume flow rate $Q$ is given by the product of the cross-sectional area $A$ of the river and the flow velocity $v$ .
$A = \text{width} \times \text{depth} = 20 \, \mathrm{m} \times 3.0 \, \mathrm{m} = 60 \, \mathrm{m}^2$
$Q = A \times v = 60 \, \mathrm{m}^2 \times 1.2 \, \mathrm{m/s} = 72 \, \mathrm{m}^3/s$

STEP 4

Calculate the mass flow rate of the river.
The mass flow rate $\dot{m}$ is given by the product of the volume flow rate $Q$ and the density $\rho$ of water (assuming $\rho = 1000 \, \mathrm{kg/m}^3$ ).
$\dot{m} = Q \times \rho = 72 \, \mathrm{m}^3/s \times 1000 \, \mathrm{kg/m}^3 = 72000 \, \mathrm{kg/s}$

STEP 5

Use the energy balance to find the temperature rise.
The energy added to the river is $600 \, \mathrm{MW} = 600 \times 10^6 \, \mathrm{J/s}$ .
The temperature rise $\Delta T$ can be calculated using the formula:
$P = \dot{m} \times c \times \Delta T$
Where $P$ is the power added, $\dot{m}$ is the mass flow rate, $c$ is the specific heat capacity of water (approximately $4186 \, \mathrm{J/(kg \cdot °C)}$ ), and $\Delta T$ is the temperature rise.
Rearranging for $\Delta T$ :
$\Delta T = \frac{P}{\dot{m} \times c} = \frac{600 \times 10^6 \, \mathrm{J/s}}{72000 \, \mathrm{kg/s} \times 4186 \, \mathrm{J/(kg \cdot °C)}}$
$\Delta T \approx \frac{600 \times 10^6}{72000 \times 4186}$
$\Delta T \approx 1.99 \, °C$
The temperature rise of the river water is approximately:
$\boxed{1.99 \, °C}$

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