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2 years ago

What is an advantage of a nuclear-fission reactor?.

Engineering

1 answer:

Kobotan [32]2 years ago

4 0

Answer:

Nuclear fission is almost 8,000 times more efficient than traditional fossil fuels at producing energy. That's a lot of energy packed into a small space. Nuclear energy is more efficient, which means it uses less fuel to power the plant and produces less waste.

advantages:
-produces no polluting gases
-does not contribute to global warming
-very low fuel costs
-Low fuel quantity reduces mining and transportation effects on environment
-High technology research required benefits other industries
-Power station has very long lifetime

Disadvantages:
-Waste is radioactive and safe disposal is very difficult and expensive
-Local thermal pollution from wastewater affects marine life
-Large-scale accidents can be catastrophic
-Public perception of nuclear power is negative
-Costs of building and safely decommissioning are very high
-Cannot react quickly to changes in electricity demand

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Technician A says that the use of methanol in internal combustion engines has declined over the years. Technician B says that th

enyata [817]

Answer:

Both Technician A and B are correct

Explanation:

Methanol is used in internal combustion engines. However, the use of methanol in internal combustion engines has decreased lately even though it was thought to lead to cleaner emissions than gasoline. Methanol internal combustion engines produce formaldehyde which is also an environmental pollutant. Also, the cost of methanol is slightly higher than that of good quality gasoline.

MTBE replaced tetraethyllead as a gasoline additive because the former lead to the emission of particulate lead from automobile exhausts which is a serious environmental pollutant. The use of MTBE has declined over the years due to environmental concerns. It has been banned because it has been found to be a significant groundwater pollutant if gasoline containing MTBE is spilled or leaked at gas stations.

4 0

3 years ago

1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-

Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * $\frac{79.25}{21.2}$

= 0.0214 * 3.738

= 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0

3 years ago

Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati

Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0

3 years ago

technician A says that voltage is electric pressure .Technician B says the amperage is electrical flow who is right

Dmitrij [34]

Answer:

I would say technician A

Explanation:

Definition of voltage- An electromotive force or potential difference expressed in volts.

electromotive force is pressure so technician A would make sense.

5 0

3 years ago

Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected e

Lady_Fox [76]

Answer:

a) 19 or select the closest answer

b) 5%

Explanation:

from the voltage divide rule

$V_{in} = V_0 * \frac{R_2}{R_2 + R_f}$

$\frac{V_0}{V_{in}} = \frac{R_2 + R_f}{R_2} => 1 + \frac{R_f}{R_2} = 20$

$\frac{R_f}{R_2} = 19$

Select the nearest answer

obtained gain = $\frac{V_0}{V_{in}} = 1 + \frac{36}{2} = 19$

Expected gain = $\frac{V_0}{V_{in}} = 20$

∴ error = | $\frac{19 - 20}{20}$ | × 100

= 1/20 × 100

= 5%

6 0

3 years ago