Chapter 15, Problem 63GP

(a)

To determine

The average temperature increase in the river water.

Answer to Problem 63GP

The average temperature increase in the river water is $5.3°\text{C}$ .

Explanation of Solution

Given info:

The output of the power plant is, $W=980\text{MW}=980\text{MW}\times \frac{{10}^6\text{W}}{1\text{MW}}=9.80\times {10}^8\text{W}\times \frac{1\text{J}/\text{s}}{1\text{W}}=9.80\times {10}^8\text{J}/\text{s}$ .

The temperature of the steam is, $T_h=625\text{K}$ .

The temperature of the river water is, $T_l=285\text{K}=\left(285\text{K}-273\right)°\text{C}=12°\text{C}$ .

The flow rate of the river is, $V=37{\text{m}}^3/\text{s}$ .

Formula Used:

The expression to calculate the heat transferred to river water is,

  $Q=W\left(\frac{T_l}{T_h-T_l}\right)$

The expression to calculate theaverage temperature increase in the river water is,

  $\begin{array}{c}Q=\rho Vc\Delta T\\ \Delta T=\frac{Q}{\rho Vc}\end{array}$

Here,

• $\rho$ is the density of the water and its value is $1000\text{kg}/{\text{m}}^3$ .
• $c$ is the specific heat of the water and its value is $4186\text{J}/\text{kg}\cdot °\text{C}$ .

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}Q=\left(9.80\times {10}^6\text{J}/\text{s}\right)\left(\frac{285\text{K}}{625\text{K}-285\text{K}}\right)\\ =8.21\times {10}^8\text{J}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}\Delta T=\frac{8.21\times {10}^8\text{J}/\text{s}}{\left(1000\text{kg}/{\text{m}}^3\right)\left(37{\text{m}}^3/\text{s}\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)}\\ =5.3°\text{C}\end{array}$

Conclusion:

Thus, the average temperature increase in the river water is $5.3°\text{C}$ .

(b)

To determine

The entropy increase per kilogram of the downstream river water.

Answer to Problem 63GP

The entropy increase per kilogram of the downstream river water is $1514.39\text{J}/\text{kg}\cdot °\text{C}$ .

Explanation of Solution

Formula Used:

The expression to calculate the entropy increase per kilogram of the downstream river water is,

  $\Delta s=\frac{c\Delta T}{T_l+0.5\Delta T}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}\Delta s=\frac{\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)\left(5.3°\text{C}\right)}{12°\text{C}+0.5\left(5.3°\text{C}\right)}\\ =1514.39\text{J}/\text{kg}\cdot °\text{C}\end{array}$

Conclusion:

Thus, the entropy increase per kilogram of the downstream river water is $1514.39\text{J}/\text{kg}\cdot °\text{C}$ .