Nuclear Power Plants

In nuclear power plants, the nuclear reactor must be isolated, for safety reasons, from the turbines that convert the heat to mechanical or electrical energy. A heat transfer medium is required that can absorb and transfer large quantities of heat at a temperature of several hundred degrees Celsius, so it can be used to boil water in a steam turbine. Steam (water vapor) itself is an excellent heat transfer medium (see Example below), but it reacts with metal parts at high temperatures to produce hydrogen. This was the cause of the chemical explosion that destroyed the Chernobyl power plant in 1986 ^[1].

Figure \(\PageIndex{1}\) Chernobyl Nuclear Plant

Figure \(\PageIndex{2}\) Steam Turbine Rotor^[3]

The heat transfer medium must be stable at high temperatures, have a low vapor pressure to reduce danger of pump seal failures, have good heat conductivity, low corrosiveness, have a low enough melting point so that it doesn't solidify in steam turbines at <400 ºC, and have high "heat capacity" (see below) so that large amounts of heat can be stored without changing the temperature drastically. Surprisingly, molten salt mixtures are leading candidates^[4]. They have much lower melting points than the pure salts. Below is a table of candidates for Next Generation Nuclear Plant (NGNP) to the Nuclear Hydrogen Initiative (NHI) hydrogen-production plant. Suppliers provide extensive data for commercially available mixtures like HITEC(potassium nitrate, sodium nitrite and sodium nitrate).

Table \(\PageIndex{1}\): Summary of the properties of candidate coolants for the NGNP/NHI heat-transfer loop (Heat-transfer properties at 700ºC)^[5]

Salt	Melting point (ºC)	900ºC vapor pressure (mm Hg)	ρ, density (g/cm^3)	ρ*Cp,volumetric heat capacity (cal/cm^3-ºC)	viscosity (cP)	k,thermal conductivity (W/m-K)
LiF-NaF-KF	454	~ 0.7	2.02	0.91	2.9	0.92
NaF-ZrF4	500	5	3.14	0.88	5.1	0.49
KF-ZrF4	390	1.2	2.80	0.70	< 5.1	0.45
LiF-NaF-ZrF4	436	~ 5	2.92	0.86	6.9	0.53
LiCl-KCl	355	5.8	1.52	0.435	1.15	0.42
LiCl-RbCl	313	--	1.88	0.40	1.30	0.36
NaCl-MgCl2	445	< 2.5	1.68	0.44	1.36	0.50
KCl-MgCl2	426	< 2.0	1.66	0.46	1.40	0.40
NaF-NaBF4	385	9500	1.75	0.63	0.90	0.40
KF-KBF4	460	100	1.70	0.53	0.90	0.38
RbF-RbF4	442	< 100	2.21	0.48	0.90	0.28

When heat energy is absorbed by matter, it causes a rise in temperature proportional to the quantity of heat energy supplied (Provided that no chemical changes or phase changes take place). If q is the quantity of heat supplied and the temperature rises from T₁ to T₂ then

\[q = C * (T_{2} – T_{1})\]

or

\[q = C * (\Delta T)\]

where the constant of proportionality C is called the heat capacity of the sample. The sign of q in this case is + because the sample has absorbed heat (the change was endothermic), and (ΔT) is defined in the conventional way.

If we add heat to any homogenous sample of matter of variable mass, such as a pure substance or a solution, the quantity of heat needed to raise its temperature is proportional to the mass as well as to the rise in temperature. That is,

\[q = C * m * (T_2 – T_1)\]

OR

\[q = C * m * (\Delta T)\]

The new proportionality constant C is the heat capacity per unit mass. It is called the specific heat capacity (or sometimes the specific heat), where the word specific means “per unit mass.”

The heat capacity per unit volume may be used in engineering applications where the volume of heat transfer medium required may be critical. Some values are given in the table above. In the table below, the specific heat capacity of helium is given (5.19 J/(g·K), but since the density of He at 1 Atm and 0 ^oC is 0.18 g/L, this is a volume heat capacity of only 0.00093 J/cm³-ºC or 0.00022 cal/cm³-ºC. Compared to the other entries in the table above, helium absorbes only about 1/1000 as much heat per cm³ per degree of temperature increase, so it appears at first to be a poor heat exchange medium. It is used in some nuclear plants, however, because of it's negligible toxicity and zero corrosiveness.

Specific heat capacities provide a convenient way of determining the heat added to, or removed from, material by measuring its mass and temperature change. As mentioned [|previously], James Joule established the connection between heat energy and the intensive property temperature, by measuring the temperature change in water caused by the energy released by a falling mass. In an ideal experiment, a 1.00 kg mass falling 10.0 m would release 98.0 J of energy. If the mass drove a propeller immersed in 0.100 liter (100 g) of water in an insulated container, its temperature would rise by 0.234^oC. This allows us to calculate the specific heat capacity of water:

98 J = C × 100 g × 0.234 ^oC

C = 4.184 J/g^oC

At 15°C, the precise value for the specific heat of water is 4.184 J K^–1 g^–1, and at other temperatures it varies from 4.178 to 4.218 J K^–1 g^–1. Note that the specific heat has units of g (not the base unit kg), and that since the Centigrade and kelvin scales have identical graduations, either ^oC or K may be used.

Example \(\PageIndex{1}\): Heat Absorbed

Compare the heat absorbed by steam and other heat transfer media:

a. How much heat is required to raise the temperature of 1 kg of steam from 475.0 ^oC to 675.0 ^oC, given that the specific heat capacity of water vapor is 2.080 J K^–1 g^–1?

Solution:

\(\text{q} = \text{C} × \text{m} × (\text{T}_2–\text{T}_1)\)

\(\text{q} = \text{2.080} \dfrac{J}{g ^\circ C} × \text{1000 g} × \text{(675.0 - 475.0)}\)

\(\text{q} = \text{416,000 J or 416 kJ}\)

 

b. How much heat is required to raise the temperature of 1 kg of LiF/NaF/KF salt mixture from 475.0 ^oC to 675.0 ^oC? C = 0.91 cal/cm³-ºC x (4.18 J /cal) x ( 1/ 2.02 g/cm³) = 1.88 J/gºC

Solution

\(\text{q} = \text{C} × \text{m} × (\text{T}_2–\text{T}_1)\)

\(\text{q} = \text{1.88} \dfrac{J}{g ^\circ C} × \text{1000 g} × \text{(675.0 - 475.0)}\)

\(\text{q} = \text{376,000 J or 376 kJ}\)

c. Helium has been proposed as a secondary heat exchanger in nuclear plants. How much heat is required to raise the temperature of 1 kg of helium from 475.0 ^oC to 675.0 ^oC?

Solution

\(\text{q} = \text{C} × \text{m} × (\text{T}_2–\text{T}_1)\)

\(\text{q} = \text{5.19} \dfrac{J}{g ^\circ C} × \text{1000 g} × \text{(675.0 - 475.0)}\)

\(\text{q} = \text{1,038,000 J or 1,038 kJ}\)

The volume of helium or steam is large (depending on the temperature and pressure). At STP it's be thousands of liters, compared to around 0.5 L for the NaF/KF/LiF salt mixture.

Table \(\PageIndex{2}\) Specific heat capacities (25 °C unless otherwise noted)

Substance	phase	Cp(see below) J/(g·K)
air, (Sea level, dry, 0 °C)	gas	1.0035
argon	gas	0.5203
carbon dioxide	gas	0.839
helium	gas	5.19
hydrogen	gas	14.30
methane	gas	2.191
neon	gas	1.0301
oxygen	gas	0.918
water at 100 °C (steam)	gas	2.080
water at 100 °C	liquid	4.184
ethanol	liquid	2.44
water at -10 °C (ice))	solid	2.05
copper	solid	0.385
gold	solid	0.129
iron	solid	0.450
lead	solid	0.127

Electrical Energy Conversion

The most convenient way to supply a known quantity of heat energy to a sample is to use an electrical coil. The heat supplied is the product of the applied potential V, the current I flowing through the coil, and the time t during which the current flows:

\[\text{q} = \text{V} × \text{I} × \text{t}\]

If the SI units volt for applied potential, ampere for current, and second time are used, the energy is obtained in joules. This is because the volt is defined as one joule per ampere per second:

\(\text{1 volt} × \text{1 ampere} × \text{1 second} = \text{1} \dfrac{J}{A s} × \text{1 A} × \text{1 s} = \text{1 J}\)

Example \(\PageIndex{2}\): Heat Energy

An electrical heating coil, 230 cm³ of water, and a thermometer are all placed in a polystyrene coffee cup. A potential difference of 6.23 V is applied to the coil, producing a current of 0.482 A which is allowed to pass for 483 s. If the temperature rises by 1.53 K, find the heat capacity of the contents of the coffee cup. Assume that the polystyrene cup is such a good insulator that no heat energy is lost from it.

Solution The heat energy supplied by the heating coil is given by

\(\text{q} = \text{V} × \text{I} × \text{t} = \text{6.23 V} × \text{0.482 A } × \text{483 s} = \text{1450 V A s} = \text{1450 J}\)

However,

\(\text{q} = \text{C} × \text{(T}_2 – \text{T}_1)\)

Since the temperatue rises, T₂ > T₁ and the temperature change ΔT is positive:

\(\text{1450 J} = \text{C} × \text{1.53 K}\)

so that

\(\text{C} = \dfrac{1450 J}{1.53 K} = \text{948} \dfrac{J}{K}\)

Note

Note: The heat capacity found applies to the complete contents of the cup-water, coil, and thermometer taken together, not just the water.

As discussed in other sections, an older, non-SI energy unit, the calorie, was defined as the heat energy required to raise the temperature of 1 g H₂O from 14.5 to 15.5°C. Thus at 15°C the specific heat capacity of water is 1.00 cal K^–1 g^–1. This value is accurate to three significant figures between about 4 and 90°C.

If the sample of matter we are heating is a pure substance, then the quantity of heat needed to raise its temperature is proportional to the amount of substance. The heat capacity per unit amount of substance is called the molar heat capacity, symbol C_m. Thus the quantity of heat needed to raise the temperature of an amount of substance n from T₁ to T₂ is given by

\[\text{q} = \text{C} × \text{n} × (\text{T}_2 – \text{T}_1)\label{6}\]

The molar heat capacity is usually given a subscript to indicate whether the substance has been heated at constant pressure (C_p)or in a closed container at constant volume (C_V).

Example \(\PageIndex{3}\): Heat Supply

A sample of neon gas (0.854 mol) is heated in a closed container by means of an electrical heating coil. A potential of 5.26 V was applied to the coil causing a current of 0.336 A to pass for 30.0 s. The temperature of the gas was found to rise by 4.98 K. Find the molar heat capacity of the neon gas, assuming no heat losses.

Solution

The heat supplied by the heating coil is given by

\(\text{q} = \text{V} × \text{I} × \text{t}\)

= \(\text{5.26 V} × \text{0.336 A} × \text{30.0 s}\)

= \(\text{53.0 V A s}\)

= \(\text{53.0 J}\)

Rearranging Eq. \(\ref{6}\), we then have

\(\text(C)_m = \dfrac{q}{n(T_2-T_1)} = \dfrac{53.0 J}{0.854 mol * 4.98 K} = \text{12.47} \dfrac{J}{mol*k}\)

However, since the process occurs at constant volume, we should write

\(\text{C}_V = \text{12.47} \dfrac{J}{mol*K}\)

From ChemPRIME: 15.1: Heat Capacities

References

1. ↑ en.Wikipedia.org/wiki/File:Chernobyl_Disaster.jpg
2. ↑ en.Wikipedia.org/wiki/File:Chernobyl_Disaster.jpg
3. ↑ http://en.Wikipedia.org/wiki/File:Chernobyl_Disaster.jpg
4. ↑ inlportal.inl.gov/portal/ser...9_htl_salt.pdf
5. ↑ inlportal.inl.gov/portal/ser...9_htl_salt.pdf

Contributors and Attributions

• Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.