Ask question

Ask question

All categories

3 years ago

15

Why is the reactor coolant water kept contained within the primary loop instead of allowing it to mix with the feed water and le

ave through the cooling tower? it is contaminated and radio active?

2 answers:

WITCHER [35]3 years ago

5 0

Answer:

Yes, It is contaminated and radio active

Explanation:

The coolant in the primary loop is actually in contact with the reactor fuel. So it will have some trace amount of the reactor fuel and all the other radioactive by products like cobalt.

Also the neutrons in the reactor will convert the hydrogen in the water into radioactive tritium. and the oxygen into radioactive nitrogen.

This is why the primary loop water is kept isolated from the feed water.

uysha [10]3 years ago

4 0

For the answer to the question above, I the answer is yes, It is <u><em>both </em></u><span><u><em>contaminated and </em></u><u><em>radioactive</em></u></span><u><em> at the same time</em></u>. That's why they keep the water spinning.I hope my answer helped you. Have a nice day!

You might be interested in

Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

3 years ago

A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

$I_1=7\times (0.25)^2=0.437\ kgm^2$

The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

l = r

$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

$I=0.437+0.15=0.587\ kgm^2$

So, the moment of inertia of the wagon wheel for rotation about its axis is $0.587\ kgm^2$ . Hence, this is the required solution.

4 0

3 years ago

Suppose the initial kinetic energy and final potential energy in an experiment are both zero. What can you conclude?

puteri [66]

Answer:

That an item is neither moving nor staying still in a position that is building up energy.

Explanation:

3 0

3 years ago

Read 2 more answers

Suppose that a constant force is applied to an object. Newton's Second Law of Motion states that the acceleration of the object

omeli [17]

<span>(9 kg)(5 m/s^2) = M(3 m/s^2)
</span><span>that the acceleration of the object varies inversely with its mass.</span>

4 0

3 years ago

A balloon is launched at 2.5 m every second, how far did it travel in 10 seconds?

Elina [12.6K]

Answer:

25 meters

Explanation:

1 sec is 2.5, 1x10 so 2.5x10=25

7 0

2 years ago