**fuel assembly**

**pressure drop**is formed by fuel bundle frictional drop (dependent on relative roughness of fuel rods, reynolds number, hydraulic diameter etc.) and other pressure drops of structural elements (top and bottom nozzle, spacing grids or mixing grids).

In general, it is not so simple to calculate pressure drops in fuel assemblies (especially the spacing grids) and it belongs to key **know-how** of certain fuel manufacturer. Mostly, pressure drops are measured in **experimental hydraulic loops**, rather than calculated.

Engineers use the **pressure loss coefficient**, **PLC**. It is noted K or **ξ** (pronounced “xi”). This coefficient characterizes pressure loss of a certain hydraulic system or of a part of a hydraulic system. It can be easily measured in hydraulic loops. The pressure loss coefficient can be defined or measured for both straight pipes and especially for** local (minor) losses**.

Using data from below mentioned example, the **pressure loss coefficient** (only frictional from straight pipe) is equal to **ξ = f _{D}L/D_{H} = 4.9**. But the overall pressure loss coefficient (including spacing grids, top and bottom nozzles etc.) is usually about three times higher. This PLC (

**ξ = 4.9**) causes that the pressure drop is of the order of (using the previous inputs)

**Δp**= 4.9 x 714 x 5

_{friction}^{2}/ 2 =

**43.7 kPa**(without spacing grids, top and bottom nozzles). About three times higher real PLC means about three times higher

**Δp**will be.

_{fuel }The overall reactor pressure loss, **Δp _{reactor}**, must include:

- downcomer and reactor bottom
- lower support plate
- fuel assembly including spacing grids, top and bottom nozzles and other structural components –
**Δp**_{fuel} - upper guide structure assembly

In result the overall reactor pressure loss – **Δp _{reactor}** is usually of the order of hundreds kPa (let say 300 – 400 kPa) for design parameters.

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

See also: Fluid Acceleration – Pressure Loss

## Example: Frictional Pressure Loss – Fuel Bundle

Calculate the **frictional pressure loss **of a **single fuel rod** inside a reactor core at normal operation (design flow rate). Assume that this fuel rod is part of a fuel bundle with the rectangular fuel lattice and this fuel bundle does not contain spacing grids. Its height is **h = 4m** and the core flow velocity is constant and equal to **V**_{core}** = 5 m/s.**

Assume that:

- the outer diameter of the cladding is:
**d = 2 x r**_{Zr,1}**= 9,3 mm** - the pitch of fuel pins is:
**p = 13 mm** - the relative roughness is
**ε/D = 5×10**^{-4} - the fluid density is:
**ρ = 714 kg/m**^{3} - the core flow velocity is constant and equal to
**V**_{core}**= 5 m/s** - the average temperature of reactor coolant is:
**T**_{bulk}**= 296°C**

## Calculation of the Reynolds number

To calculate the Reynolds number, we have to know:

- the outer diameter of the cladding is:
**d = 2 x r**_{Zr,1}**= 9,3 mm**(to calculate the hydraulic diameter) - the pitch of fuel pins is:
**p = 13 mm**(to calculate the hydraulic diameter) - the dynamic viscosity of saturated water at 300°C is:
**μ = 0.0000859 N.s/m**^{2} - the fluid density is:
**ρ = 714 kg/m**^{3}

**The hydraulic diameter, D**** _{h}**, is a commonly used term when handling flow in

**non-circular tubes and channels**. The

**hydraulic diameter of the fuel channel**,

*D*

*, is equal to*

_{h}**13,85 mm**.

See also: Hydraulic Diameter

The **Reynolds number **inside the fuel channel is then equal to:

This fully satisfies the **turbulent conditions**.[/lgc_column]l