Heat Transport
Basic Equations and Applications
Environmental Hydraulics
Heat Exchange
Important for circulation in a receiving water.
Determines the rate at which artificially added heat is transferred to the atmosphere
Examples:
• annual temperature variation and stratification in a lake
• evaporation
• discharge of cooling water from fossil and nuclear power plants
1

Example of a Power Plant
Cooling power: Q o
Δ ρ c : specific heat (4.19 · 10 3 J/kg o C)
Examples of Cooling Water Systems
Open system
Only open cooling water systems in Sweden
Receiving waters: The Baltic Sea,
Öresund, Kattegatt
Savannah
River
2

Examples of Cooling Water Systems
Closed system:
Chernobyl
Nuclear Power Plant, Daya Bay, China
3

Öresundsverket, Malmö
Plant outline
Annual production:
3 TWhr electric power
1 TWhr heat
Öresundsverket, Malmö
Plant for production of electric power and district heating.
Advanced cooling system:
Pipe diameter 2.2 m
Handle a flow of 3 – 6 m 3 /s
Water intake in the harbor area
360 m long outfall in the harbor area
Excess temperature 10 deg
4

Heat Exchange Mechanisms
Heat exchange at a water surface:
Components in Heat Budget
Φ s
= incoming short wave solar radiation (0.17
μ m < λ <
3.8
μ m); instantaneous: 0 – 1000 W/m 2 ; daily average
Φ sr
=
60 – 300 W/m 2 reflected short wave solar radiation ; daily average 7
- 20 W/m 2
Φ a
= incoming long-wave atmospheric radiation (3.8
μ m <
λ < 80 μ m); 200 – 450 W/m 2
Φ ar
Φ br
Φ e
=
= reflected atmospheric radiation 3 % of Φ a emitted long wave radiation from the water surface
250 – 500 W/m 2
= heat exchange due to evaporation or condensation ,
; naturally occuring (density driven) or forced (winds)
Φ c
0 – 1000 W/m 2
= heat exchange due to conduction ; 70 – 200 W/m 2
5

Net heat flow to a water surface:
Φ n
= ( Φ s
Φ sr
+ Φ a
Φ ar
) – ( Φ br
+ Φ e
+ Φ c
) = Φ m
– Φ v
(difference between positive and negative flows)
Φ m depends on meteorological conditions
Φ v depends on water surface temperature, air temperature, air moisture, and wind speed
Equilibrium Temperature
Equilibrium temperature ( T
E
): for given meteorological conditions, the temperature that corresponds to a neat heat flow of zero at the water surface
⇒ The temperature that the water surface will approach under constant meteorological conditions
T
S
< T
E
: heating up
T
S
> T
E
: cooling down
6

Example: Temperature Variation and Heat
Balance for Lake Cayuga
For small deviations between T
S and T
E
:
Φ n
= K ( T
E
– T
S
)
K : heat exchange coefficient
Approximate expression for K :
K
= + +
+
W
2
2
)
=
T
β
=
T s
+
T d
2
+
T
β
+
0.000156
T
β
2
7

A water mass is supplied with Φ x cooling water discharge) extra heat (e.g.,
=> increase in surface water temperature to T
S
1 to allow for release of extra heat to the atmosphere x
(
E
−
T s
1
)
K T s
1 −
T s
)
= Δ s
Example: Estimate Pond Size for Cooling Water
Calculate the pond surface water area that is required for emitting the excess heat from a 600 MW nuclear power plant, with an efficiency of 33 % (i.e., 33 % of the produced heat energy in the power plant is transformed into electric energy). The maximum allowable excess temperature in the pond is 2 0 C : a summer day, when the air temperature is 25 0 C, the relative humidity 40%, and the water temperature 15 0 C an autumn day, when the air temperature is 10 0 C, the relative humidity 70 %, and the water temperature 15 0 C.
The wind velocity is assumed to be 5 m/s at both occasions.
8

Determine dew point temperature from a Mollier diagram.
⇒ T d
= 10.5 0 C and T d
= 5.0 0 C
T
β
=
2
=
12.7
0
C and T
β
=
2
=
10.0
0
C
=> K = 28.3 W/m 2 0 C and K = 26.0 W/m 2 0 C
The amount of heat power in the cooling water is: 1800 · 0.67 ≈
1200 MW (the total heat production in the power plant is 1800
MW, out of which 600 MW is electric energy).
Mollier Diagram for
Humid Air
9

If 1200 MW heat power should be transferred from the pond to the atmosphere, the following pond area A will be required:
A
=
⋅ 6
= ⋅ 6 =
21.2 km
2
A
=
⋅ 6
= ⋅ 6 =
23.1 km 2
Example: Discharge of Cooling Water to River
Steady state conditions prevail with the river flow rate Q and the cooling water flow rate Q
0
. The discharged cooling water has got an excess temperature Δ T
0
. Assume that the water velocity U does not change in the river flow direction
( x -direction).
10

A short distance downstream of the discharge point x =0 well-mixed conditions are obtained, and the excess temperature Δ T due to the cooling water discharge will only depend on the x -coordinate.
Δ T will asymptotically approach zero in the downstream direction due to heat exchange (emittance) with the atmosphere. The cooling of the river water can be described according to the following relationship (the excess temperature Δ T can be considered as a pollutant concentration c ): u d (
Δ
T ) dx
=
E x d
2
(
Δ
T )
−
KB dx
2 ρ cA
Δ
T
(steady-state AD equation with advection, dispersion, and sink term)
Solution to governing equation:
Q
0
Δ
T
0
Q 1
1
+ α exp
⎝
⎛
⎜ u
2 E x x
(
1
−
1
+ α
) ⎞
⎠ where:
α =
4 KE x
ρ
A c u
B
2
If a << 1 =>
Q
0
Δ
T
0
Q exp
⎛
⎝
−
BK
ρ cQ x
⎞
⎠
11

Numerical Example
Assume that:
Q = 400 m 3 /s
Q
0
Δ T
0
= 80 m
= 10 0
B = 400 m
3
C
/s
W
2
= 10 m/s
Water surface temperature ( T s
) 20 0 C
Air temperature 25 0 C
Relative humidity of 40 %.
The Mollier diagram gives that T d
= 10.6 0 C.
We assume that T s gives: is constant = 20 0 C when calculating T
β
, which
T
β
=
10.6
+
20
2
=
15.3
0
C
=> β = 0.111 and K = 3.7 + (0.0613 + 0.111) (70 + 3.5 · 10 2 ) ≈ 76
This yields:
400
⋅ exp
−
⋅ ⋅ 3 ⋅
⋅ x 2 exp
(
− ⋅ −
6 ⋅ x
)
12

The excess temperature will be 2 0 C close to the cooling water discharge (implying a complete mixing across the entire river cross section).
10 km further downstream the excess temperature will be:
2 exp
(
− ⋅ −
6 ⋅
10
4
)
≈
1.6
0
C
13