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goldenfox [79]
2 years ago
7

[3] (25%) Superheated steam at 9 MPa and 480oC leaves the steam generator of a vapor power plant. Heat transfer and frictional e

ffects in the line connecting the steam generator and the turbine reduce the pressure and temperature at the turbine inlet to 8.6 MPa and 440oC, respectively. The pressure at the exit of the turbine is 10 kPa, and the turbine operates adiabatically. Liquid leaves the condenser at 8 kPa, 35 oC. The pressure is increased to 9.6 MPa across the pump. The turbine and pump isentropic efficiencies are 85%. The mass flow rate of steam is 80 kg/s. Determine (a) the net power output, in kW. (b) the thermal efficiency (c) the rate of heat transfer from the line connecting the steam generator and the turbine, in kW. (d) the mass flow rate of the condenser cooling water, in kg/s, if the cooling water enters at 15oC and exits at 35oC with negligible pressure change.
Engineering
1 answer:
aleksklad [387]2 years ago
8 0
The condenser is a device which is used thermal power plants to change the phase of the steam to liquid by releasing the latent heat to the surroundings.

The condenser in the Rankine cycle rejects heat to the surrounding or the flowing fluid at the constant pressure
You might be interested in
A(n) is a detailed, structured diagram or drawing.
monitta

Answer:

Schematics

Explanation:

A schematic is a detailed structured diagram or drawing. It employs illustrations to help the viewer understand detailed information on the machine or object being described. Its main aim is not to help the observer know what the object looks like physically. It is rather aimed at helping the viewer know how the machine works. This is achieved by only including key and important details to the drawing.

It is most times used in the blueprint and user guides of machines and gadgets used in the home to help users know how these things work so that they can do little fixings should there be such needs.

6 0
3 years ago
A well-insulated rigid vessel contains 3 kg of saturated liquid water at 40oC. The vessel also contains an electrical resistor t
user100 [1]

Answer:

The final temperature is 111.66°C

Explanation:

The given conditions :-

i) Well insulated means no heat loss.

ii) Rigid vessels means volume remains same.

iii) Initial temperature ( T₁ ) = 40°C. = 273 + 40 = 313 K

iv ) Mass of water in vessel = 3 kg.

v) current drawn by resistor ( i ) = 10 ampere.

vi) Voltage applied ( V ) = 50 volts.

vii) The time for which resistor operating ( t ) = 30 minute = 30 * 60 = 1800 seconds.

Now we have to calculate heat developed by resistor in vessel.

Q = V * i * t  = 50 * 10 * 1800 = 900,000 J = 900 KJ.

Since it is a rigid container so the work done is zero.

Q = du    ( du - change in internal energy)

Q = m * C * dT      ( C = 4.186 KJ/KgK )

Q = 3 * 4.186 * (T₂ - T₁ )

900 = 12.558 * ( T₂ - 313 )

T₂ - 313 = 71.6674

T₂ = 384.6674 K

T = 384.6674 - 273 = 111.66°C

So the final temperature is 111.66°C.

3 0
3 years ago
Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)
sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c)  cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) =  cos(π0.4n)

so

w₀ = π0.3

2π/w₀  = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀  = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

6 0
3 years ago
The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.91 and 0.09, respectively. Is this a hypoe
WINSTONCH [101]

Answer:

hypoeutectoid

Explanation:

ferrite:  pure form of iron

cementite: It is iron carbide with 93.3% iron and 6.67% carbon

hypoeutectoid: Eutectoid steel with carbon fraction less than 0.8%

hypereutectoid: Eutectoid Steel with carbon content more than 0.8%

For the mentioned iron-carbide alloy,

% of carbon in iron-carbide alloy= percentage of cementite × percentage of carbon in cementite

% of carbon in iron-carbide alloy= 0.09× 0.0667

                                                     = 0.6%

so the alloy is hypoeutectoid

4 0
4 years ago
A steam turbine in a power plant receives 5 kg/s steam at 3000 kPa, 500°C. Twenty percent of the flow is extracted at 1000 kPa t
MrMuchimi

Answer:

The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW

Explanation:

If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.

In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg

For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg

For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg

For the power output, we have to multiply the steam flow with the enthalpic jump.

The addition of the 2 jumps is the total power output.

4 0
3 years ago
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