A small PWR plant operates at a power of 485 MWt. The core, which is approximately 75.4 in. in diameter and 91.9 in. high, consists of a square lattice of 23,142 fuel tubes of thickness 0.021 in. and inner diameter of 0.298 in. on a 0.422-in. pitch. The tubes are filled with 3.40 w/o-enriched The core is cooled by water, which enters at the bottom at and passes through the core at a rate of at 2,015 psia. Compute (a) the average temperature of the water leaving the core; (b) the average power density in kW/liter; (c) the maximum heat production rate, assuming the reactor core is bare.

Since the core is cooled by water, which enters at the bottom at 496°F, the heat input ([tex]h_{in}[/tex]) would be equal to 483.02 Btu/lb based on the steam table.

Given the following data:

Power = 485 MWt to Btu/hr = 1.635 × 10⁹ Btu/hr.

Diameter = 75.4 in.

Height = 91.9 in.

Thickness = 0.021 in.

Inner diameter = 0.298 in.

Pitch = 0.422-in.

Work done = 34 × 10⁶ lb/hr.

Substituting the given parameters into the formula, we have;

[tex]h_{out}= \frac{q}{w} + h_{in}\\\\h_{out}= \frac{1.635 \times 10^9}{34 \times 10^6} + 483.02\\\\h_{out}= 48.09 + 483.02\\\\h_{out}= 531.11\;Btu/lb[/tex]

How to calculate the average power density.

First of all, we would determine the area of this core by using this formula:

Area = πd²/2

Area = 3.142 × (0.298/2)²

Area = 0.0698 in².

Mathematically, the average power density is given by:

[tex]q_{av}=\frac{q}{W} =\frac{q}{\pi A^2H \times 23142} \\\\q_{av}=\frac{485 \times 10^3}{3.142 \times 0.0698^2 \times 91.9 \times 0.01639 \times 23142} \\\\q_{av}=199.5\;kW/liter.[/tex]

How to calculate maximum heat production rate.

In this scenario, we would assume the reactor core is bare. Thus, the heat is given by:

[tex]q(0) = \frac{2.32PE_1}{HE_R} \\\\q(0) =\frac{2.32 \times 485 \times 180}{23142 \times 200} \\\\q(0) =0.044 \;MW = 1.5 \times 10^8 \;Btu/hr^2[/tex]

For the maximum heat production rate, we have:

[tex]q_{max}=\frac{q(0)}{2HA^2} \\\\q_{max}=\frac{1.5 \times 10^8}{2 \times (\frac{75.4}{2} \times \frac{1}{12}) \times (\frac{0.298}{2} \times \frac{1}{12})} \\\\q_{max}=\frac{1.5 \times 10^8}{2 \times 3.1417 \times 0.00124} \\\\q_{max}=\frac{1.5 \times 10^5}{0.007791416}\\\\q_{max}=1.55 \times 10^8 \;Btu/hr.ft^3[/tex]

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