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3 years ago

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100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t

he volume is one-half its original size. This is done such that the pressure of the R-134a does not change. Determine the final temperature and the change in the total internal energy of the R-134a.

1 answer:

LekaFEV [45]3 years ago

5 0

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.

Using the given mass m, molar mass M, we can get the following equation:

pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

U=-1.848*10^6 J

Note, that the energy was taken away from the system.

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Explanation:

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3 years ago

An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces

Allushta [10]

Answer:

(a) The amount of heat transferred to the air, $q_{out}$ is 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

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Process 1 to 2 is isentropic compression, therefore;

$T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}} \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K$

From;

$\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }$

We have;

$p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa$

Process 2 to 3 is reversible constant volume heating, therefore;

$\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}$

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

$T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86 \times 729.21}{2190.43} = 1458.42 \, K$

Process 3 to 4 is isentropic expansion, therefore;

$T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}} \right )^{k-1}$

$1458.42= T_{4} \times \left (9.2 \right )^{0.4}$

$T_4 = \dfrac{1458.42}{(9.2)^{0.4}} = 600.3 \, K$

$q_{out} = m \times c_v \times (T_4 - T_1) = 0.718 \times (600.3 - 300.15) = 215.5077 \, kJ/kg$

The amount of heat transferred to the air, $q_{out}$ = 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is found as follows;

$W_{net} = q_{in} - q_{out}$

$q_{in} = m \times c_v \times (T_3 - T_2) = 0.718 \times (1458.42 - 729.21) = 523.574 \, kJ/kg$

$\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg$

(c) The thermal efficiency is given by the relation;

$\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100= \dfrac{308.07}{523.574} \times 100= 58.8\%$

(d) From the general gas equation, we have;

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The Mean Effective Pressure, MEP, is given as follows;

$MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa$

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Explanation:

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A sketch of the problem is uploaded along this answer.

Now

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3 years ago

A steam power plant operates on the reheat Rankine cycle. Steam enters the highpressure turbine at 12.5 MPa and 550°C at a rate

gayaneshka [121]

Answer:

A) condenser pressure = 9.73 kPa,

B) 10242 kw

C) 36.9%

Explanation:

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temperature of steam = 550⁰c

flow rate of steam = 7.7 kg/s

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reheated steam temperature = 450⁰c

isentropic efficiency of turbine( nt ) = 85% = 0.85

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From steam tables

at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg, S3 = 6.6317 kJ/kgK

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nt (0.85) = $\frac{h3-h4}{h3-h4s}$ = $\frac{3476.5 - h4}{3476.5 - 2948.1}$

making h4 subject of the equation

h4 = 3476.5 - 0.85 (3476.5 - 2948.1)

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h5 = 3358.2 kj/kg, s5 = 7.2815 kj/kgk

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p6 = 9.73 kPa, h6 = 2463.3 kj/kg

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solution is attached below

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8 0

3 years ago