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2 years ago

What is electrical energy used for?

Engineering

2 answers:

Stolb23 [73]2 years ago

8 0

Answer:

Electrical energy is used throughout the world to power devices, appliances and methods of transportation utilized in daily life. To make things operate, electrical energy must be emitted from energy sources such as power plants, to enable an object to consume the power it needs to function.

Explanation:

FromTheMoon [43]2 years ago

6 0

Answer:

to power devices appliances and some methods of transportation

Explanation:

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Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago

A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis

kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

$V_1=0.3m^3,T_1=298 K$

$V_2=0.1m^3,T_1=1273 K$

Mass of air=1 kg

For polytropic process $pv^n=C$ ,n is the polytropic constant.

$Tv^{n-1}=C$

$T_1v^{n-1}_1=T_2v^{n-1}_2$

$298\times .3^{n-1}_1=1273\times .1^{n-1}_2$

n=2.32

Work in polytropic process given as

w= $\dfrac{P_1V_1-P_2V_2}{n-1}$

w= $mR\dfrac{T_1-T_2}{n-1}$

Now by putting the values

w= $1\times 0.287\dfrac{289-1273}{2.32-1}$

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0

2 years ago

What would happen to a plane if the weight force becomes greater than the lift force?

mrs_skeptik [129]

Answer:

When the lift is greater than the weight, the aircraft gains altitude. ... Drag must be overcome for the aircraft to move, and movement is essential to obtain lift. To overcome drag and move the aircraft forward, another force is essential. This force is thrust.

Explanation:

Hope this helps!

8 0

2 years ago

Read 2 more answers

a vertical cylindrical container is being cooled in ambient air at 25 °C with no air circulation. if the initial temperature of

Sloan [31]

Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period. = 4.93 w/m²k

Explanation:

check attachement for answer explanation