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e-lub [12.9K]
4 years ago
10

An adiabatic air compressor is to be powered by a direct coupled adiabatic steam turbine that is also driving a generator. Steam

enters the turbine at 12.5 MPa and 500-degree C at a rate of 25 Kg/s and exits at 10 kPa and a quality (or dryness factor) of 0.92. Air enters the compressor at 98 kPa and 295 K at a rate of 10 Kg/s and exits at 1 MPa and 620 K. determine the net power delivered to the generator by the turbine.
Engineering
1 answer:
Allisa [31]4 years ago
5 0

Answer:

the net power delivered to the generator by the turbine. is 26486Kw

Explanation:

Hello!

To solve this problem we must use the following steps,

1)First we apply the first law of thermodynamics in the turbine that states that the energy must always be conserved, then we find in the energy balance that the power generated by the turbine must be equal to the power consumed by the compressor and the power consumed by the generator

Wt=Wc+Wg

Wg=Wt-Wc

where

Wt= turbine power

Wc=compresor power

Wg=generator power

2) we find the power generated by the turbine, for this we apply the first law of thermodynamics and find the enthalpies in the two states using the thermodynamic tables.

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties

h1=  enthalpy at the entrance at 12.5Mpa and 500C

h1=3342KJ/kg

h2=enthalpy at the exit at 10kpa and quality=0.92

h2=2151KJ/kg

Wt=m(h1-h2)

Wt=25(3342-2151)=29775Kw

2) Now we do the same as we did in step two for the compressor remembering that to calculate the power the following equation is used considering that it operates with air

Wc=m Cp (T2-T1)

Where

m=mass flow=10kg/S

Cp=specific heat for air=1.012KJ/Kg

T2= oulet temperature=620K

T1=inlet temperature=295K

solving

Wc=10(1.012)(620-295)=3289Kw

finally we apply the equation found in step 1 to find the power in the generator

Wg=Wt-Wc =29775Kw-3289Kw=26486Kw

the net power delivered to the generator by the turbine. is 26486Kw

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Answer:

The answer is not in the options. It is one-fourth.

Explanation:

As of 2017, it was recorded that nuclear power supplies 25% of electricity in Europe. That's 1/4 of the total electrical power supply.

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3 years ago
Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra
VashaNatasha [74]

Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

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2) struct array{

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6 0
3 years ago
Who works alongside and assists the engineers?
nika2105 [10]

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5 0
3 years ago
A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug
Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

\rho = \frac{m}{V}

Where

m = Mass

V = Volume

For state one we know that

\rho_1 = \frac{m_1}{V}

m_1 = \rho_1 V

m_1 = 1.18*1

m_1 = 1.18Kg

For state two we have to

\rho_2 = \frac{m_2}{V}

m_2 = \rho_2 V

m_1 = 7.2*1

m_1 = 7.2Kg

Therefore the total change of mass would be

\Delta m = m_2-m_1

\Delta m = 7.2-1.18

\Delta m = 6.02Kg

Therefore the mass of air that has entered to the tank is 6.02Kg

5 0
3 years ago
a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

6 0
3 years ago
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