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2 years ago

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Steam enters an adiabatic turbine at 400◦C, 2 MPa pressure. The turbine has an isentropic efficiency of 0.9. The exit pressure i

s set so that the actual work output from the turbine is 700 kJ/kg. Determine the required exit pressure and the actual exit temperature and quality, if saturated.

1 answer:

pychu [463]2 years ago

8 0

Answer:

Explanation:

Find attached the solution

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Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea

Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = $h_{f4}$ + x₄× $h_{fg}$ = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ = $s_{f4}$ + x₄× $s_{fg}$ = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, $\dot X_{dest}$ , is given as follows;

$\dot X_{dest}$ = T₀ × $\dot S_{gen}$ = T₀ × $\dot m$ × (s₄ + s₂ - s₁ - s₃)

$\dot X_{dest}$ = T₀ × $\dot W$ ×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ $\dot X_{dest}$ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, $\dot W_{rev}$ = $\dot W_{}$ + $\dot X_{dest}$ ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

6 0

3 years ago

The high electrical conductivity of copper is an important design factor that helps improve the energy efficiency of electric mo

ludmilkaskok [199]

Answer:

B

Explanation:

This is a two sample t-test and not a matched pair t-test

null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors

alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.

So, option D is rejected

The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.

Option A and C are also rejected

5 0

2 years ago

How did studying pagodas help engineers create earthquake-proof structures in modern society?A. Engineers learned that the desig

Mila [183]

D cuz i took the test

7 0

2 years ago

For welding the most important reason to use jigs and fixtures in a welding shop is to

diamong [38]

Answer:

Reduce manufacturing costs.

Explanation:

Hope This Helps

Have A Great Day

7 0

2 years ago

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

zvonat [6]

Answer:

R= 1.25

Explanation:

As given the local heat transfer,

$Nu_x = 0.035 Re^{0.8}_x Pr^{1/3}$

But we know as well that,

$Nu=\frac{hx}{k}\\h=\frac{Nuk}{x}$

Replacing the values

$h_x=Nu_x \frac{k}{x}\\h_x= 0.035Re^{0.8}_xPr^{1/3} \frac{k}{x}$

Reynolds number is define as,

$Re_x = \frac{Vx}{\upsilon}$

Where V is the velocity of the fluid and \upsilon is the Kinematic viscosity

Then replacing we have

$h_x=0.035(\frac{Vx}{\upsilon})^{0.8}Pr^{1/3}kx^{-1}$

$h_x=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}kx^{0.8-1}$

$h_x=Ax^{-0.2}$

<em>*Note that A is just a 'summary' of all of that constat there.</em>

<em>That is $A=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}k$ </em>

Therefore at x=L the local convection heat transfer coefficient is

$h_{x=L}=AL^{-0.2}$

Definen that we need to find the average convection heat transfer coefficient in the entire plate lenght, so

$h=\frac{1}{L}\int\limit^L_0 h_x dx\\h=\frac{1}{L}\int\limit^L_0 AL^{-0.2}dx\\h=\frac{A}{0.8L}L^{0.8}\\h=1.25AL^{-0.2}$

The ratio of the average heat transfer coefficient over the entire plate to the local convection heat transfer coefficient is

$R = \frac{h}{h_L}\\R= \frac{1.25Al^{-0.2}}{AL^{-0.2}}\\R= 1.25$

3 0

2 years ago