Ocean Power Plant

## **Problem**

A proposed ocean power plant will utilize the temperature difference between surface seawater and seawater at a depth of 100 meters. Assume the surface temperature is 25° Celsius and the temperature at the 100-meter depth is 3° Celsius.

**Part 1:** What is the Carnot efficiency of the plant?

**Part 2:** If the power plant generates useful energy at the rate of 100 megawatts while operating with the efficiency found in part 1, at what rate is heat given off to the surroundings?

**Part 3:** The diagram below represents the Carnot cycle for a simple reversible (Carnot) engine in which a fixed amount of gas, originally at pressure po and volume Vo follows the path ABCDA.

In the chart below, for each part of the cycle indicate with +, -, or 0 whether the heat transferred

Q

{\displaystyle Q}

 and temperature change

Δ
T

{\displaystyle \Delta T}

 are positive, negative, or zero, respectively. (

Q

{\displaystyle Q}

 is positive when heat is added to the gas, and

Δ
T

{\displaystyle \Delta T}

 is positive when the temperature of the gas increases.)

Q

{\displaystyle \;Q\;}

Δ
T

{\displaystyle \Delta T}

A
B

{\displaystyle AB}

B
C

{\displaystyle BC}

C
D

{\displaystyle CD}

D
A

{\displaystyle DA}

## **Solution**

**Part 1**

The Carnot efficiency formula is

# e

T
H

−

T
C

T
H

{\displaystyle e = \frac{T\_H - T\_C}{T\_H}}

where the temperatures are mesaured in the Kelvin scale. Convert the hot temperature and cold temperature as follows:

25
# 273.15
298.15

K

3
# 273.15
276.15

K

{\displaystyle \begin{align}
25 + 273.15 = 298.15\, K \\
3 + 273.15 = 276.15 \, K
\end{align}}

Substituting the temeperatures into the Carnot engine formula yields:

# e

298.15
−
276.15

298.15

=
0.07379

{\displaystyle e = \frac{298.15 - 276.15}{298.15} = 0.07379}

or 7.38% efficiency.

**Part 2**

The power plant generates useful energy at the rate of 100 megawatts. Using the formula

# e

P

o
u
t

P

i
n

{\displaystyle e = \frac{P\_{out}}{P\_{in}}}

we have

# 0.07379

100

M
W

P

i
n

{\displaystyle 0.07379 = \frac{100 \, MW}{P\_{in}}}

P

i
n

=

100

M
W

0.07379

=
1355

M
W

{\displaystyle P\_{in} = \frac{100 \, MW}{0.07379} = 1355 \, MW}

The rate of the heat given off to the surroundings is equal to

P

i
n

−

P

o
u
t

{\displaystyle P\_{in} - P\_{out} }

.

P

i
n

−

P

o
u
t

=
1355

M
W
−
100

M
# W
1255

M
W

{\displaystyle P\_{in} - P\_{out} = 1355 \, MW - 100 \, MW = 1255 \, MW }

**Part 3**

AB is isothermal so

Δ
# T
0

{\displaystyle \Delta T = 0 }

. It is an expansion so the work is negative and because

# Q
−
W

{\displaystyle Q = -W }

, we get

Q
>
0

{\displaystyle Q > 0 }

.

BC is adiabatic so

# Q
0

{\displaystyle Q = 0}

. Since the temperature drops, we get

Δ
T
<
0

{\displaystyle \Delta T < 0 }

.

CD is isothermal so

Δ
# T
0

{\displaystyle \Delta T = 0 }

. It is a compression so the work is positive and because

# Q
−
W

{\displaystyle Q = -W }

, we get

Q
<
0

{\displaystyle Q < 0 }

.

BC is adiabatic so

# Q
0

{\displaystyle Q = 0}

. Since the temperature rises so

Δ
T
>
0

{\displaystyle \Delta T > 0 }

.

Q

{\displaystyle \;Q\;}

Δ
T

{\displaystyle \Delta T}

A
B

{\displaystyle AB}

{\displaystyle +}

0

{\displaystyle 0}

B
C

{\displaystyle BC}

0

{\displaystyle 0}

−

{\displaystyle -}

C
D

{\displaystyle CD}

−

{\displaystyle -}

0

{\displaystyle 0}

D
A

{\displaystyle DA}

0

{\displaystyle 0}

{\displaystyle +}