me 3310 spring 19 hw3 solution - wright state universityphilippe.sucosky/downloads/me3310... · me...

ME 3310 – Spring 19 HW3 solution

Problem 1 System 1

Water at 4 MPaP = , 30.030 m /kg =

We start by sketching the isobar 4 MPaP = on the T − diagram:

Using the saturated property table with pressure entry, the saturation temperature at that

pressure is: 250.35 CsatT = .

From the table, we can also obtain the values of the specific volume of the saturated liquid (

30.01252 m /kgf = ) and saturated vapor (30.049779 m /kgg = ).

Based on the specific volume given in the problem: f g . Therefore, the state of water is:

2-phase mixture

The temperature is directly obtained as the saturation temperature at 4 MPa:

250.35 CsatT T= =

System 2

Water at 0.225 MPaP = , 150 CT =

We start by sketching the isobar 0.225 MPaP = on the T − diagram:



pressure is: 123.97 CsatT = .

Since 150 C > satT T= , the state of the water is:

superheated vapor

The specific volume can be obtained from the superheated vapor table.


However, the properties are not available at the given operating pressure 0.225 MPaP = .

Therefore, interpolation is needed. We assume that the specific volume varies linearly with pressure between 0.2 and 0.3 MPa:

An expression for 0.225 MPa can be generated by expressing the slope of the line in two ways:

0.3 MPa 0.2 MPa 0.225 MPa 0.2 MPa

0.3 0.2 0.225 0.2

− −=

− −

( )0.225 MPa 0.2 MPa 0.3 MPa 0.2 MPa

0.225 0.2

0.3 0.2

−= + −

−

Numerical application: ( )0.225 MPa

0.225 0.20.95986 0.63402 0.95986

0.3 0.2

−= + −

−

3

0.225 MPa 0.8784 m /kg =

System 3

R-134a at 0.14 MPaP = , 20 CT = −

We start by sketching the isobar 0.14 MPaP = on the T − diagram:



pressure is: 18.77 CsatT = − .

Since 20 C < satT T= − , the state of the R-134a is:

compressed liquid

The compressed liquid properties are not provided in the textbook. Since the properties of compressed liquid do not different significantly from those of the saturated liquid, the specific volume of the gas can be approximated as the specific volume of the saturated liquid at the same temperature:

Using the saturated property table at a temperature 20 CT = − :

30.0007361 m /kgf =

Summary

System P (MPa) T (°C) (m3/kg) State

Water 4.0 250.35 0.030 2-phase mixture Water 0.225 150 0.8784 superheated vapor Refrigerant-134a 0.14 -20 0.0007361 compressed liquid


Problem 2 1- Initial state characterization

The volume of the tank is 5 L=V . Initially, the pressure of R-134a is 600 kPaP = and the

temperature is 30 CT = .

Using the saturated property table for R-134a (pressure entry), we obtain the value of the saturation temperature at that pressure:

@600 kPa 21.55 CsatT =

Since @600 kPasatT T , the refrigerant is in the

superheated state.

2- Initial mass calculation The mass can be obtained from the

definition of the specific volume: m =V ,

where the specific volume value is extracted from the superheated property table:


30.035984 m /kg =

Therefore: 35 10

0.035984m

−= 0.13895 kgm =

3- Final state characterization

At the end of the recharging process, the pressure of R-134a is 150 kPaP = and the

temperature is 0 CT = .

Using the saturated property table (pressure entry):

The working pressure is not available in the table. Properties are obtained by interpolation

between 140 kPaP = and 160 kPaP = . In particular, for the saturation temperature:

@140 kPa @160 kPa

@150 kPa

18.77 15.6017.185 C

2 2

sat sat

sat

T TT

+ − −= = = −

Since @150 kPasatT T , the refrigerant is in the superheated state.

4- Final mass

calculation We use the superheated property table to obtain the specific volume in the final state. Since properties are not available at the target pressure of 150 kPa, we consider the properties at 140 and 180 kPa:


We perform the interpolation assuming a linear relationship between specific volume and pressure between 140 and 180 kPa:

An expression for 0.15 MPa can be generated by expressing the slope of the line in two ways:

0.18 MPa 0.14 MPa 0.15 MPa 0.14 MPa

0.18 0.14 0.15 0.14

− −=

− −

( )0.15 MPa 0.14 MPa 0.18 MPa 0.14 MPa

0.15 0.14

0.18 0.14

−= + −

−

Numerical application: ( )0.15 MPa

0.15 0.140.15263 0.11722 0.15263

0.18 0.14

−= + −

−

3

0.225 MPa 0.1437775 m /kg =

The mass is calculated as:

35 10

0.1437775m

−= =V

0.03478 kgm =

Problem 3 1- Ideal gas method IG law:

P RT =

Since the heating process is isobaric (constant pressure): constantT P

R= =

Therefore:

1 2

1 2

T T

= 2

2 1

1

T T

=

Since the volume has increased by 60%: 2 11.6 = 2 11.6 T T=


Numerical application: 2 1.6 (100 273.15)T = + 2 597.04 KT =

2- Compressibility factor method To determine the compressibility factor at the initial state, we first calculate the reduced

temperature 1rT and reduced pressure 1rP using the critical point properties obtained from

Table A-1:

11

102.23

4.48r

cr

PP

P= = = and 1

1

373.151.22

305.5r

cr

TT

T= = =

The compressibility factor 1Z at the initial state can now be obtained using the generalized

compressibility chart (see construction lines on top of next page):

1 0.58Z =

On the chart, we can also obtain the value of the reduced specific volume at the initial state:

1 0.32r =

To determine the compressibility factor at the final state, we first calculate the reduced pressure

2rP and reduced specific volume 2r :

2 1 2.23r rP P= = and 2 11.6 0.51r r = =

We then use the generalized compressibility chart to obtain the compressibility factor in the final state:

2 0.78Z =


Construction lines for initial state

Construction lines for final state


We then use the definition of the compressibility factor to calculate the final temperature:

2 22

2

PZ

RT

= 22 2 2

2

2 2

r cr

cr

TP PT

Z R Z P

= =

Numerical application: 2

10 0.51 305.5

0.78 4.48T

= 2 445.9 CT =

3- Discussion The compressibility factor is substantially different from 1 in both the initial and final state. Therefore, the ideal gas behavior is not satisfied by this gas under these conditions. Provided that the lines on the compressibility chart are constructed with sufficient accuracy, the results obtained with the second method are the most accurate. Problem 4 1- IG method IG law:

P RT = RT

P =

The gas constant is obtained from Table A-1:

Numerical application: 6

461.5 (450 273.15)

3.5 10

+=

30.093525 m /kg =


2- Compressibility factor method We first calculate the reduced pressure and reduced temperature:

3.50.16

22.06r

cr

PP

P= = = and

723.151.12

647.1r

cr

TT

T= = =

Construction lines:

The chart yields:

0.96Z =

From the definition of the compressibility factor:

ideal

Z

= idealZ =

Numerical application: 0.96 0.093525 = 30.089784 m /kg =

3- Property table method We use the superheated water table to determine the specific volume:


We obtain: 30.09198 m /kg =

Error calculation:

- Method 1: 3 1

3

0.09198 0.093525

0.09198

method method

method

error

− −= = 1.7%error =

- Method 2: 3 2

3

0.09198 0.089784

0.09198

method method

method

error

− −= = 2.4%error =

me 3310 spring 19 hw3 solution - wright state universityphilippe.sucosky/downloads/me3310... · me...

Documents