Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 4, Problem 41P
To determine

The temperature and total enthalpy at initial and final states in the container.

The temperature and total enthalpy at initial and final states when the heating is completed.

Expert Solution & Answer
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Explanation of Solution

Given:

The mass of a refrigerant R-134a (m) is 10 kg.

The volume of a rigid container (ν) is 14 L.

The final pressure of the container is 600 kPa.

Calculation:

Since the process is a constant volume, calculate the specific volume.

  v1=v2

Here, specific volume at states 1 and 2 are v1andv2 respectively.

  v1=νm=14L10kg=14L10kg(1m31000L)=0.0014m3/kg

Refer to Table A-12, obtain the values of below variables as in Table (I).

Pressure, kPa (x)Temperature, °C (y)
280–1.25
300?
3202.46

Write the formula of interpolation method of two variables.

  y2=(x2x1)(y3y1)(x3x1)+y1

  y2=(300280)(2.46(1.25))(320280)+(1.25)=0.605

Here, the variables denote by x and y are pressure and temperature.

Thus, the initial temperature (T1) is 0.61°C._

The initial state represents the mixture, so temperature to be considered as the saturation temperature at given pressure.

Similarly, calculate the values of vf, vg, hf, and hfg using interpolation method at pressure of 300 kPa as 0.0007735m3/kg, 0.067776m3/kg, 52.71kJ/kg, and 198.17 kJ/kg respectively.

Calculate the dryness fraction at state 1.

  x1=v1vfvgvf

Here, specific volume at saturated water is vf and specific volume at saturated vapor is vg.

Substitute 0.0014m3/kg for v1, 0.0007735m3/kg for vf, and 0.067776m3/kg for vg in Equation .

  x1=0.0014m3/kg0.0007735m3/kg0.067776m3/kg0.0007735m3/kg=0.009351

Calculate the specific enthalpy at state 1.

  h1=hf+x1(hfg)

Here, specific enthalpy at saturated liquid is hf and specific enthalpy at evaporation is hfg.

Substitute 52.71kJ/kg for hf, 0.009351 for x1, and 198.17 kJ/kg for hfg in Equation.

  h1=52.71kJ/kg+(0.009351)(198.17kJ/kg)=54.56kJ/kg

Calculate the total enthalpy at state 1.

  H1=mh1

Here, mass of the refrigerant R-134a is m.

Substitute 10 kg for m and 54.56 kJ/kg for h1 in equation (IV).

  H1=10kg(54.56kJ/kg)=545.6kJ

Thus, total enthalpy final states in the container is 545.6kJ_.

Similarly, calculate for x2, h2, and H2 respectively.

  x2=v2vfvgvf        (I)

Here, dryness fraction at state 2 is x2 and specific volume at state 2 is v2.

Calculate the specific enthalpy at state 2.

  h2=hf+x2(hfg)        (II)

Calculate the total enthalpy at state 2.

  H2=mh2        (III)

Similarly, calculate the values of vf, vg, hf, and hfg using interpolation method at pressure of 300 kPa as 0.0008198m3/kg, 0.034335m3/kg, 81.50kJ/kg, and 180.95kJ/kg respectively.

Repeat the above steps for T2,x2,h2,andH2 by substituting the values of vf, vg, hf, and hfg in equations (I), (II), and (III).

  T2=21.55°Cx2=0.01731h2=84.64kJ/kgH2=846.4kJ

Thus, the temperature and total enthalpy at initial and final states in the container are 21.55°C_ and 846.4kJ_ respectively.

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Chapter 4 Solutions

Fundamentals of Thermal-Fluid Sciences

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