Visual aids are useful for all of the following reasons except

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage

ivolga24 [154]

Answer:

a. 46.15%

b. 261.73 kg/s

c. 54.79 kW/K

Explanation:

a. State 1

The parameters given are;

T₁ = 560°C

P₁ = 12 MPa = 120 bar

Therefore;

h₁ = 3507.41 kJ/kg, s₁ = 6.6864 kJ/(kg·K)

State 2

p₂ = 1 MPa = 10 bar

s₂ = s₁ = 6.6864 kJ/(kg·K)

h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52

= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg

State 3

p₃ = 6 kPa = 0.06 bar

s₃ = s₁ = 6.6864 kJ/(kg·K)

sg = 8.3291 kJ/(kg·K)

sf = 0.52087 kJ/(kg·K)

x = s₃/sfg = (6.6864- 0.52087)/(8.3291 - 0.52087) = 0.7896

(h₃ - 151.494)/2415.17 = 0.7896

∴ h₃ = 2058.56 kJ/kg

State 4

Saturated liquid state

p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)

State 5

Open feed-water heater

p₅ = p₂ = 1 MPa = 10 bar = 1000000 Pa

s₄ = s₅ = 0.52087 kJ/(kg·K)

h₅ = h₄ + work done by the pump on the saturated liquid

∴ h₅ = h₄ + v₄ × (p₅ - p₄)

h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg

Step 6

Saturated liquid state

p₆ = 1 MPa = 10 bar

h₆ = 762.683 kJ/kg

s₆ = 2.1384 kJ/(kg·K)

v₆ = 0.00112723 m³/kg

Step 7

p₇ = p₁ = 12 MPa = 120 bar

s₇ = s₆ = 2.1384 kJ/(kg·K)

h₇ = h₆ + v₆ × (p₇ - p₆)

h₇ = 762.683 + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg

The fraction of flow extracted at the second stage, y, is given as follows

$y = \dfrac{762.683 - 152.4944113 }{2824.01 - 152.4944113 } = 0.2284$

The turbine control volume is given as follows;

$\dfrac{\dot{W_t}}{\dot{m_{1}}} = \left (h_{1} - h_{2} \right ) + \left (1 - y \right )\left (h_{2} - h_{3} \right )$

= (3507.41 - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg

For the pumps, we have;

$\dfrac{\dot{W_p}}{\dot{m_{1}}} = \left (h_{7} - h_{6} \right ) + \left (1 - y \right )\left (h_{5} - h_{4} \right )$

= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 - 151.494)

= 13.17 kJ/kg

For the working fluid that flows through the steam generator, we have;

$\dfrac{\dot{Q_{in}}}{\dot{m_{1}}} = \left (h_{1} - h_{7} \right )$

= 3507.41 - 775.08253 = 2732.32747 kJ/kg

The thermal efficiency, η, is given as follows;

$\eta = \dfrac{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}{\dfrac{\dot{Q_{in}}}{\dot{m_{1}}}}$

η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%

(762.683 - 152.4944113)/(2824.01 - 152.4944113)

b. The mass flow rate, $\dot{m_{1}}$ , into the first turbine stage is given as follows;

$\dot{m_{1}} = \dfrac{\dot{W_{cycle}}}{\dfrac{\dot{W_t}}{\dot{m_{1}}} -\dfrac{\dot{W_p}}{\dot{m_{1}}}}$

$\dot{m_{1}}$ = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s

c. From the entropy rate balance of the steady state form, we have;

$\dot{\sigma }_{cv} = \sum_{e}^{}\dot{m}_{e}s_{e} - \sum_{i}^{}\dot{m}_{i}s_{i} = \dot{m}_{6}s_{6} - \dot{m}_{2}s_{2} - \dot{m}_{5}s_{5}$

$\dot{\sigma }_{cv} = \dot{m}_{6} \left [s_{6} - ys_{2} - (1 - y)s_{5} \right ]$

= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K

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3 years ago

The Hubble space telescope is a bus-sized satellite powered by huge ___ panels

Artemon [7]

Answer:

Solar

Explanation:

The Hubble space telescope is a bus-sized satellite powered by huge solar panels.

4 0

3 years ago

Read 2 more answers

The files provided in the code editor to the right contain syntax and/or logic errors. In each case, determine and fix the probl

Yanka [14]

Question Continuation

public class DebugOne3{

public static void main(String args){

System.out.print1n("Over the river");

System.out.pr1ntln("and through the woods");

system.out.println("to Grandmother's house we go");

}

Answer:

Line 2: Invalid Syntax for the main method. The main method takes the form

public static void main (String [] args) { }

or

public static void main (String args []) { }

Line 3: The syntax to print is wrong.

To print on a new line, make use of System.out.println(".."); not System.out.print1n();

Line 4:

To print on a new line, make use of System.out.println(".."); not System.out.pr1ntln();

Line 5:

The case of "system" is wrong.

The correct case is sentence case, "System.out.println" without the quotes

The correct program goes, this:

public class DebugOne3{

public static void main(String [] args){

System.out.println("Over the river");

System.out.println("and through the woods");

System.out.println("to Grandmother's house we go");

}

Explanation:

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When an electrical signal travels through a conductive wire, it produces an electromagnetic (EM) field. Likewise, when an EM fie

PolarNik [594]

Answer:

A. True

Explanation:

When an electromagnetic field wave strikes a conductor, say a wire, it induces an alternating current that is proportional to the wave in the conductor. This is a reversal of generating electromagnetic wave from accelerating a charged particle. This phenomenon is used in radio antena for receiving radio wave signals and also use in medicine for body scanning.

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Bryan a project manager and his team have been assigned a new project. The team members have already started working on their as

Vedmedyk [2.9K]

Answer:

there teams management is good

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