Contents

## What is Volume

**Volume** is a basic **physical quantity**. **Volume** is a derived quantity and it expresses the t**hree dimensional extent** of an **object**. Volume is often quantified numerically using the SI derived unit, the **cubic metre**. For example, the volume inside a **sphere** (that is the volume of a ball) is derived to be** V = 4/3πr ^{3}**, where r is the radius of the sphere. As another example, volume of a cube is equal to side times side times side. Since each side of a square is the same, it can simply be the length of one side

**cubed**.

If a square has one side of 3 metres, the volume would be 3 metres times 3 metres times 3 metres, or 27 cubic metres.

## What is Specific Volume

**Specific volume** is an** intensive variable**, whereas volume is an extensive variable. The standard unit for specific volume in the SI system is cubic meters per kilogram (m^{3}/kg). The standard unit in the English system is cubic feet per pound mass (ft^{3}/lbm).

The density (ρ) of a substance is the reciprocal of its** specific volume** (ν).

**ρ = m/V = 1/ρ**

**Density** is defined as the **mass per unit volume**. It is also an **intensive property**, which is mathematically defined as mass divided by volume:

**ρ = m/V**

## Volume of an Atom and Nucleus

**The atom** consist of a small but massive **nucleus** surrounded by a cloud of rapidly moving **electrons**. The nucleus is composed of **protons and ****neutrons**. Typical nuclear radii are of the order 10^{−14} m. Assuming spherical shape, nuclear radii can be calculated according to following formula:

r = r_{0} . A^{1/3}

where r_{0} = 1.2 x 10^{-15 }m = 1.2 fm

If we use this approximation, we therefore expect the volume of the nucleus to be of the order of 4/3πr^{3} or 7,23 ×10^{−45 }m^{3} for hydrogen nuclei or 1721×10^{−45} m^{3} for ^{238}U nuclei. These are volumes of nuclei and atomic nuclei (protons and neutrons) contains of about **99.95%** of mass of atom.

## Is an atom an empty space?

**The volume of an atom** is about **15 orders of magnitude** **larger** than the volume of a nucleus. For **uranium atom**, the **Van der Waals radius** is about **186 pm = 1.86 ×10 ^{−10} m**. The Van der Waals radius, r

_{w}, of an atom is the radius of an imaginary hard sphere representing the distance of closest approach for another atom. Assuming spherical shape, the uranium atom have volume of about

**26.9 ×10**. But this “huge” space is occupied primarily by electrons, because the

^{−30}m^{3}**nucleus**occupies only about

**1721×10**of space. These electrons together weigh only a fraction (let say 0.05%) of entire atom.

^{−45}m^{3}It may seem, that the space and in fact the matter is **empty**, **but it is not**. Due to the **quantum nature of electrons**, the electrons are not point particles, they are smeared out over the whole atom. The classical description cannot be used to describe things on the atomic scale. On the atomic scale, physicists have found that quantum mechanics describes things very well on that scale. Particle locations in quantum mechanics are not at an exact position, they are described by a **probability density function**. Therefore the space in an atom (between electrons and an atomic nucleus) is not empty, but it is filled by a probability density function of electrons (usually known as “**electron cloud**“).

## Volume of Coolant in Reactor Coolant System

In typical modern **pressurized water reactors** (PWRs), the **Reactor Coolant System** (RCS), shown in the figure, consists of:

- the
**reactor vessel**, which contains the nuclear fuel - four parallel heat transfer
**loops**connected to a reactor vessel. - each loop contains a
**main coolant pump**and a**steam generator**. - the system includes a
**pressurizer**and its auxiliary systems

All RCS components are located inside the containment building.

At normal operation, there is a compressed liquid water inside the reactor vessel, loops and steam generators. The pressure is maintained at approximately **16MPa**. At this pressure water boils at approximately **350°C** (662°F). Inlet temperature of the water is about** 290°C** (554°F). The water (coolant) is heated in the reactor core to approximately** 325°C** (617°F) as the water flows through the core. As it can be seen, the reactor contains approximately 25°C subcooled coolant (distance from the saturation).This high pressure is maintained by the pressurizer, a separate vessel that is connected to the primary circuit (hot leg) and partially filled with water (partially with saturated steam) which is heated to the **saturation temperature** (boiling point) for the desired pressure by submerged **electrical heaters**. Temperature in the pressurizer can be maintained at 350 °C. At normal conditions, about **60%** of volume of pressurizer occupies the **compressed water** and about **40%** of volume occupies the **saturated steam**.

Volumes of typical PWR are in the following table.

**It is an illustrative example, following data do not correspond to any reactor design.**

It must be noted the **volume of coolant** significantly **changes** with the **temperature** of the coolant. The** total mass** of the coolant remains always the same, a change in water volume is not a change in water inventory. The reactor coolant volume changes with temperature because of **changes in density**. Most substances **expand w****hen heated** and **contract when cooled**. However, the amount of expansion or contraction varies, depending on the material. This phenomenon is known as **thermal expansion**. The change in volume of a material which undergoes a temperature change is given by following relation:

where ∆T is the change in temperature, V is the original volume, ∆V is the change in volume, and **α**** _{V}** is the

**coefficient of volume expansion**.

The** volumetric thermal expansion coefficient** for water is **not constant** over the temperature range and increases with the temperature (**especially at 300°C**), therefore the change in density is **not linear** with temperature (as indicated in the figure).

See also: Steam Tables

**At normal conditions** the total volume of coolant in the reactor coolant system is almost constant. On the other hand, **during transient load conditions** the **volume can significantly change**. These changes are naturally reflected in a change in pressurizer water level. When the average temperature of reactor coolant goes gradually down, the total water volume is also decreasing, which lowers the pressurizer level. On a gradual load pick-up, the increase in reactor coolant average temperature causes the total water volume to expand, which raises the pressurizer level. These effects must be controlled by pressurizer level control system.

## Control Volume – Control Volume Analysis

A **control volume** is a fixed region in space chosen for the thermodynamic study of mass and energy balances for flowing systems. The boundary of the **control volume** may be a real or imaginary **envelope**. The control surface is the boundary of the control volume.

A control volume analysis can be used for example to determine the rate of change of momentum for a fluid. In this analysis, we will consider a streamtube (**control volume**) as we did for the Bernoulli equation. In this** control volume** any change in momentum of the fluid within a control volume is due to the action of external forces on the fluid within the volume.

See also: Momentum Formula

As can be seen from the picture the **control volume method** can be used to analyze the law of conservation of momentum in fluid. **Control volume** is an **imaginary surface** enclosing a volume of interest. The control volume can be fixed or moving, and it can be rigid or deformable. In order to determine all forces acting on the surfaces of the control volume we have to solve the conservation laws in this control volume.

## Choosing a Control Volume

A control volume can be selected as any arbitrary volume through which fluid flows. This volume can be static, moving, and even deforming during flow. In order to solve any problem we have to solve basic conservation laws in this volume. It is very important to know all relative flow velocities to the control surface and therefore it is very important to define exactly the boundaries of the control volume during an analysis.

## Example: Water Jet Striking a Stationary Plate

**A stationary plate** (e.g. blade of a watermill) is used to deflect water flow at a velocity of **1 m/s** and at an angle of **90°**. It occurs at atmospheric pressure and the mass flow rate is equal to **Q =1 m ^{3}/s**.

**Calculate the pressure force.****Calculate the body force.****Calculate the total force.****Calculate the resultant force.**

**Solution**

- The
**pressure force**is zero as the pressure at both the inlet and the outlets to the control volume are atmospheric. - As the control volume is small we can ignore the
**body force**due to the weight of gravity. **F**= ρ.Q.(w_{x}_{1x}– w_{2x}) = 1000 . 1 . (1 – 0) =**1000 N**

**F**=_{y}**0**

**F**=**(1000, 0)**- The
**resultant force**on the plane is the same magnitude but in the opposite direction as the total force**F**(friction and weight are neglected).

The water jet exerts on the plate the force of **1000 N** in the x-direction.