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2 years ago

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Water vapor at 6 MPa, 500°C enters a turbine operating at steady state and expands to 20 kPa. The mass flow rate is 3 kg/s, and

the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible.
Determine:

(a) the isentropic turbine efficiency and

(b) the rate of entropy production within the turbine, in kW/K.

1 answer:

Blababa [14]2 years ago

6 0

Answer:

a)75.8%

b)2.517KW/K

Explanation:

Hello!

To solve this problem follow the steps below

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 such as pressure and temperature.

1. use thermodynamic tables to find the following variables.

a.enthalpy and entropy at the turbine entrance

h1=Enthalpy(Water;T=500;P=6000)

=3422KJ/kg

s1=Entropy(Water;T=500;P=6000)

=6.881KJ/kgK

b. enthalpy and ideal entropy at the turbine outlet

h2i=Enthalpy(Water;s=6.881;P=20)

=2267KJ/kg

s2i=s1=6.881KJ/kgK

2. uses the output power and the first law of thermodynamics to find the real enthalpy at the turbine's output

W=m(h1-h2)

h2=h1-W/m

h2r=3422-2626/3=2546.6KJ/kg

3.

find efficiency with the following equation

$eficiency=\frac{h1-h2r}{h1-h2i}$

$\frac{h1-h2r}{h1-h2i}=\frac{3422-2546.6}{3422-2267} =0.758=75.8%$

4.

find the real entropy at the turbine exit

s2=Entropy(Water;h=2546,6;P=20)=7.72KJ/kgK

5.Finally find the entropy generated, using the following equation

ΔS=m(s2-s1)=(3kg/s)(7.72 KJ/kgK-6.881 KJ/kgK)=2.517KW/K

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A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta

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Answer:

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Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0$

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$\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}$

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Inlet (Superheated Steam)

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$P = 1\,MPa$

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The work output is:

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3 years ago

Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a

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Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

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Total thermal resistance

$R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}$

Now by putting the values

$R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}$

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We know that

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$Q=\dfrac{\Delta T}{R_{th}}$

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So heat transfer per unit volume is 39.15 W/m²

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Answer:

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Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

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Method (b)

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Method (c)

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Step2

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Step3

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Answer:

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