Question

Question: Suppose a power plant delivers energy at 880 MW using steam turbines. The steam goes into the turbines superheated at 625 K and deposits its unused heat in river water at 285 K. Assume that the turbine operates as an ideal Carnot engine. (a) If the river flow rate is \({\bf{37}}\;{{{{\bf{m}}^{\bf{3}}}} \mathord{\left/{\vphantom {{{{\bf{m}}^{\bf{3}}}} {\bf{s}}}} \right.} {\bf{s}}}\), estimate the average temperature increase of the river water immediately downstream from the power plant. (b) What is the entropy increase per kilogram of the downstream river water in \({{\bf{J}} \mathord{\left/{\vphantom {{\bf{J}} {{\bf{kg}} \cdot {\bf{k}}}}} \right.} {{\bf{kg}} \cdot {\bf{k}}}}\)?

Short answer

(a) The average temperature increase of the river water is \(4.7\;{\rm{K}}\).

(b)The entropy increase per kilogram of the downstream river water is\(69.46\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.} {{\rm{kg}} \cdot {\rm{K}}}}\).

Step-by-step solution

Understanding entropy

The entropy is the measure of the disorder in a particular thermodynamic system. It is related to the spread of energy among the available molecular energy levels.

Given information

Given data:

The power is \(P = 880\;{\rm{MW}}\).

The lower temperature is \({T_{\rm{L}}} = 285\;{\rm{K}}\).

The higher temperature is \({T_{\rm{H}}} = 625\;{\rm{K}}\).

The volume flow rate is \(\frac{V}{t} = 37\;{{{{\rm{m}}^{\rm{3}}}} \mathord{\left/{\vphantom {{{{\rm{m}}^{\rm{3}}}} {\rm{s}}}} \right.} {\rm{s}}}\).

Evaluation of the average temperature increase of the river water

(a)

The expression for the efficiency of the heat engine is

\(e = 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\). … (i)

The expression for the efficiency of a heat engine in terms of work done \(\left( W \right)\) and heat given to the system \(\left( {{Q_{\rm{H}}}} \right)\) is

\(e = \frac{W}{{{Q_{\rm{H}}}}}\). … (ii)

The work done by the system is equal to the difference in heat given to the system \(\left( {{Q_{\rm{H}}}} \right)\) and heat coming out from the system \(\left( {{Q_{\rm{L}}}} \right)\). Therefore,

\({Q_{\rm{H}}} = W + {Q_{\rm{L}}}\). … (iii)

Substitute the value of equation (iii) in equation (ii).

\(\begin{aligned}{c}e &= \frac{W}{{W + {Q_{\rm{L}}}}}\\W + {Q_{\rm{L}}} &= \frac{W}{e}\\{Q_{\rm{L}}} &= W\left( {\frac{1}{e} - 1} \right)\end{aligned}\)

Substitute the value of equation (i) in the above equation.

\(\begin{aligned}{c}{Q_{\rm{L}}} &= W\left( {\frac{1}{{\left( {\frac{{{T_{\rm{H}}} - {T_{\rm{L}}}}}{{{T_{\rm{H}}}}}} \right)}} - 1} \right)\\{Q_{\rm{L}}} &= W\left( {\frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}} \right)\end{aligned}\)

Express the above equation in terms of heat rejection rate by dividing the above equation by \(t\).

\(\begin{aligned}{c}\frac{{{Q_{\rm{L}}}}}{t} &= \frac{W}{t}\left( {\frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}} \right)\\\frac{{{Q_{\rm{L}}}}}{t} &= P\left( {\frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}} \right)\end{aligned}\) … (iv)

The expression for the heat energy given to the power plant is

\({Q_{\rm{L}}} = mc\Delta T\).

This equation can also be written as shown below

\(\begin{aligned}{c}m &= \frac{{\left( {\frac{{{Q_{\rm{L}}}}}{t}} \right)t}}{{c\Delta T}}\\\rho V &= \frac{{\left( {\frac{{{Q_{\rm{L}}}}}{t}} \right)t}}{{c\Delta T}}\\V &= \frac{{\left( {\frac{{{Q_{\rm{L}}}}}{t}} \right)t}}{{\rho c\Delta T}}\\\frac{{{Q_{\rm{L}}}}}{t} &= \frac{{\rho Vc\Delta T}}{t}\end{aligned}\). … (v)

Here, \(\rho \) is the density of the water with the value \(1 \times {10^3}\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.} {{{\rm{m}}^{\rm{3}}}}}\), and \(c\) is the specific heat of the water with the value \(4186\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.} {{\rm{kg}} \cdot {\rm{K}}}}\).

Equate equations (iv) and (v).

\(\begin{aligned}{c}P\left( {\frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}} \right) &= \frac{{\rho Vc\Delta T}}{t}\\\Delta T &= \frac{P}{{\rho \left( {\frac{V}{t}} \right)c}}\left( {\frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}} \right)\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\Delta T &= \frac{{\left( {880\;{\rm{MW}}} \right)\left( {\frac{{{{10}^6}\;{\rm{W}}}}{{1\;{\rm{MW}}}}} \right)}}{{\left( {1 \times {{10}^3}\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.} {{{\rm{m}}^{\rm{3}}}}}} \right)\left( {37\;{{{{\rm{m}}^{\rm{3}}}} \mathord{\left/{\vphantom {{{{\rm{m}}^{\rm{3}}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\left( {4186\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.} {{\rm{kg}} \cdot {\rm{K}}}}} \right)}}\left( {\frac{{285\;{\rm{K}}}}{{625\;{\rm{K}} - 285\;{\rm{K}}}}} \right)\\\Delta T &= 4.7\;{\rm{K}}\end{aligned}\)

Thus, the average temperature increase of the river water is \(4.7\;{\rm{K}}\).

Evaluation of the entropy increase per kilogram of the downstream river water

(b)

The expression for the increase in entropy is

\(\Delta S = \frac{Q}{{\Delta T}}\). … (vi)

The expression for the heat given to the system is

\(Q = mc\Delta T\). … (vii)

The expression for the average temperature of the system is

\(\Delta T = {T_{\rm{L}}} + \frac{1}{2}\Delta T\). … (viii)

Substitute the values of equations (vii) and (viii) in equation (vi).

\(\begin{aligned}{c}\Delta S &= \frac{{mc\Delta T}}{{{T_{\rm{L}}} + \frac{1}{2}\Delta T}}\\\frac{{\Delta S}}{m} &= \frac{{c\Delta T}}{{{T_{\rm{L}}} + \frac{1}{2}\Delta T}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\frac{{\Delta S}}{m} &= \frac{{\left( {4186\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.} {{\rm{kg}} \cdot {\rm{K}}}}} \right)\left( {4.7\;{\rm{K}}} \right)}}{{\left( {285\;{\rm{K}}} \right) + \frac{1}{2}\left( {4.7\;{\rm{K}}} \right)}}\\\frac{{\Delta S}}{m} &= 69.46\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right. } {{\rm{kg}} \cdot {\rm{K}}}}\end{aligned}\)

Thus, the entropy increase per kilogram of the downstream river water is\(69.46\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.} {{\rm{kg}} \cdot {\rm{K}}}}\).