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Andrew [12]
4 years ago
12

Consider a nuclear power plant that produces 1200 MW of power and has a conversion efficiency of 34 percent (that is, for each u

nit of fuel energy used, the plant produces 0.34 units of electrical energy. Assuming continuous operation, determine the amount of nuclear fuel consumed by the plant per year.
Engineering
1 answer:
KIM [24]4 years ago
7 0

Answer with Explanation:

The relation between power and energy is

Energy=Power\times Time

Since the nuclear reactor operates at 1200 MW throughout the year thus the energy produced in 1 year equals

E=1200\times 10^{6}\times 3600\times 24\times 365=3.784\times 10^{16}

Now from the energy mass equivalence we have

E=mass\times c^2

where

'c' is the speed of light in free space

Thus equating both the above values we get

3.784\times 10^{16}=mass\times (3\times 10^{8})^{2}\\\\\therefore mass=\frac{3.784\times 10^{16}}{9\times 10^{16}}=0.42kg

Since it is given that 1 kg of mass is 34% effective thus the mass reuired for the reactor is

mass_{req}=\frac{mass}{\eta }=\frac{0.43}{0.34}=1.235

Thus 1.235 kg of nuclear fuel is reuired for operation.

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An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ide
storchak [24]

Answer:

Explanation:

Given that

Mass of 1 = m_1

Mass of 2 = m_2

Temperature in  1 = T_1

Temperature in 2 = T_2

Pressure remains i  the group apartment

The closed system and energy balance is

E_{in}-E_{out}=\Delta E_{system}

The kinetic energy and potential energy are negligible

since it is insulated tank ,there wont be eat transfer from the system

And there is no work involved

\Delta U = 0

Let the final temperature be final temperature

m_1+c_v(T_3-T_1)+m_2c_v(T_3+T_2)=0\\\\m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)---(i)

Using mass balance

m_3+m_2+m_1

from eqn i

m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)m_1T_3-m_1T_1=m_2T_2-m_2T_3\\\\m_1T_3+m_2T_3=m_2T_2+m_1T_1\\\\(m_1+m_2)T_3=m_1T_1+m_2T_2\\\\(m_1)T_3=m_1T_1+m_2T_2\\\\T_3=\frac{m_1T_1_m_2T_2}{m_3}

Therefore the final temperature can be express as

\large \boxed {T_3=\frac{m_1}{m_3} T_1+\frac{m_2}{m_3}T_2 }

8 0
3 years ago
A thermodynamicist claims to have developed a heat pump with a COP of 1.7 when operating with thermal energy reservoirs at 273 K
Vinil7 [7]

Answer:

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Explanation:

3 0
3 years ago
Please calculate the current for the circuit below
Aleks04 [339]

Answer:

The answer is "I = 0.0085106383 \ A"

Explanation:

Given:

R= 470  \ \Omega \\\\V= 4 \ v

Formula:

\to V=IR\\\\\to I = \frac{V}{R}\\\\

      = \frac{4}{470}\\\\ = 0.0085106383 \ A

3 0
3 years ago
In contrast to the leading-trailing drum brake system, the duo-servo drum brake system will:
prohojiy [21]

Leading/trailing shoe type drum brake This is called the servo effect (self-boosting effect) which realizes the powerful braking forces of drum brakes. ... This is because drum brakes generate the same braking force in either direction. Generally, this type is used for the rear brakes of passenger cars.

6 0
4 years ago
Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from
Greeley [361]

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet (h_{1}=40.09 Btu/lb)

liquid specific enthalpy at the exit (h_{2}=40.94 Btu/lb)

initial elevation (z_1=0ft)

final elevation (z_2=100ft)

acceleration due to gravity (g) = 32.174 ft/s²

W_{cv} = 3 Btu/s

The energy balance equation is given as:

Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0

Since  kinetic energy effects are negligible, the equation becomes:

Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0

Substituting values:

Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

8 0
3 years ago
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