All categories

3 years ago

Explain the two advantages and the two disadvantages of fission as an energy source.

Engineering

1 answer:

yawa3891 [41]3 years ago

6 0

Answer with Explanation:

1) The advantages of fission energy are:

a) Higher concentration of energy : Concentration of energy or the energy density is defined as the amount of energy that is produced by burning a unit mass of the fuel. The nuclear energy obtained by fission has the highest energy density among all the other natural sources of energy such as coal,gas,e.t.c.

b) Cheap source of energy : The cost at which the energy is produced by a nuclear reactor after it is operational is the lowest among all the other sources of energy such as coal, solar,e.t.c

2) The disadvantages of fission energy are:

a) Highly dangerous residue: The fuel that is left unspent is highly radioactive and thus is very dangerous. Usually the residual material is taken deep into the earth for it's disposal.

b) It has high initial costs of design and development: The cost to design a nuclear reactor and to built one after it is designed is the most among all other types of energy sources and requires highly skilled personnel for operation.

You might be interested in

Give me top 5 British snacks

tia_tia [17]

Answer:

1. Mini Cheddars

2. Sausage Roll

3. Monster Munch

4. Cheese Twists

5. Flapjacks

6 0

2 years ago

Lagest organs of the human body

aliya0001 [1]

Answer:

skin is the largest organ

3 0

3 years ago

Read 2 more answers

The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter

Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

( $p_t$ )₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ $[$ α₂/( $p_t$ )₂ $]^{1/2$ ------- let this be equation 2

where ( $p_t$ )₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ from equation for (σf)₂ in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ = 2(6( σ₀ )₁) $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

divide both sides by 2(σ₀)₁

$[$ α₁/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]$ = 36 $[$ 0.84/( $p_t$ )₂ $]$

1 / ( $p_t$ )₁ = 30.24 / ( $p_t$ )₂

( $p_t$ )₂ = 30.24( $p_t$ )₁

( $p_t$ )₂/( $p_t$ )₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0

3 years ago

According to the Bureau of Labor Statistics, which occupation employed 4.2 million people – more than any other occupation – in

Ludmilka [50]

Answer:

Retail Salespersons

Explanation:

The Bureau of Labor Statistics or the BLS in short is the federal unit or agency of the United States Labor Department. It is the main stats finding agency of the United States federal government in the field of statistics and labor economics.

According to a publication published by the Labor Statistics Bureau, the retail salesperson sector of occupation employed the most people in the U.S. in the year 2010. It provided employment to about 4.2 million people.

5 0

3 years ago

A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol

insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

$W_{f}=\gamma(L)(B)(D)$

Where $W_{f}$ is the weight of footing, γ is the unit weight of concrete, L is the length of footing is the width of footing, and D is the depth of footing

Substitute $2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3}$ for γ in the equation

$\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}$

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

$\sigma_{z p}^{\prime}=\gamma(D+B)-u$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0$ for u in the equation

$\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}$

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

$\sigma_{z D}^{\prime}=\gamma D$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3}\) for $\gamma, 0.5 m$ for $D$ and 0 for \(u$ in the equation.

$\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}$

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

$I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}$

Here $I_{e p}$ is the influence factor at the midpoint of soil layer $\sigma_{z^{p}}^{\prime}$ is initial vertical stress, $\sigma_{z^{p}}^{\prime}$ is vertical effective stress, and $Q$ is bearing pressure

Substitute $36 kPa for $\sigma_{z p}^{\prime}, 228.47$ kPa for $q,$ and 9 kPa for $\sigma_{z D}^{\prime}$$ in the equation $\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}$

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0

4 years ago