EGR 334 Thermodynamics: Homework 06 Prob 3: 5 Determine the phase of phases in a system consisting of H 2O at the following conditions and sketch p-v and T-v diagrams showing the location of each state. a) p = 80 psi T = 312.07 deg F. b) p = 80 psi T = 400 deg F. c) p = 360 psi T = 400 deg F. d) p = 70 psi T = 320 deg F. e) 14.7 psi T = 10 deg F. ----------------------------------------------------------------------------------------------------------------------------a) p = 80 psi T = 312.07 deg F. from Table A-3E: p = 80 and T = 312.07 are saturation values. Therefore this is somewhere along the liquidvapor line and exists as two phases. Quality is not given, so the spec. volume cannot cannot be determined. determined. ------------------------------------------------------------------------------------------------------------------------------b) p = 80 psi T = 400 deg F. from Table A-3E at p = 80 note that that T=400 > 312.07 (saturation (saturation pressure) pressure) so the substance substance is vapor or superheated. from Table A-4E p = 80 and T = 400 gives v = 6.22 ft3 / lbm ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ -c) p = 360 psi T = 400 deg F. From table A-2E: at T = 400 note that that p=360 > 247.1 (saturation (saturation pressure) pressure) so that the substance is is condensed liquid. from table A-5E p=500 psi and T = 400 is as close as you can get to p = 360 and 400. therefore use v as approximately equal to 0.0186 ft 3 /lbm. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ --------d) p = 70 psi T = 320 deg F. From table A-2E: at T=320 note that p= 70 < 89.60 (saturation pressure) pressure) so the substance is vapor or superheated. from table A-4E at p = 60 at p = 80 T v T v 300 7.26 312.1 5.47 350 7.82 350 5.80 Using double interpolation: at p=60: at p = 80
v320deg
7.26 7.26 (320 300) 00)
7.82 7.82 7.26 7.26
v320deg
350 300
= 7.48 then at 70 psi:
v
5.47 5.47 (320 (320 312. 312.1 1)
5.80 5.80 5.47 5.47 350 350 312. 312.1 1
= 5.54
7.48 (70 60)
5.54 5.54 7.48 7.48 80 60
= 6.51 ft3 /lbm.
---------------------------------------------------------------------------------------------------------------------------------------e) 14.7 psi T = 10 deg F. We know that water freezes at at 32 deg F., so at 14.7 psi and 10 deg, the water is ice. At 10 deg the sat. spec. volume 3 of ice is 0.01744 ft /lbm.
p
(c)
T
(a)
80
400 320
(b)
(c)
(b) (d)
(a)
(d) (e)
(e) v
v
EGR 334 Thermodynamics: Homework 06 Prob 3:7 the following table lists temperature and specific volume of water vapor at two pressures. p= 1.0 MPa p=1.5 MPa T (deg C) v (m3 /kg) T (deg C) 200 0.2060 200 240 0.2275 240 280 0.2480 280
v (m3 /kg) 0.1325 0.1483 0.1627
Data encounter in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate a) the specific volume at T=240 deg C, p = 1.25 MPa in m3 /kg b) the temperature at p = 1.5 MPa, v = 0.155 m 3 /kg in deg C. c) the specific volume at T =220 deg C, p = 1.4 MPa in m3 /kg ----------------------------------------------------------------------------------------------------------------------------a) the specific volume at T=240 deg C, p = 1.25 MPa in m3 /kg
v
v
va
( p pa )
vb
va
pb
pa
0.2275 (1.25 1.0)
0.1483 0.2275 1.5 1.0
0.1879 m3 /kg
b) the temperature at p = 1.5 MPa, v = 0.155 m 3 /kg in deg C.
T
T
Ta
(v va )
Tb
T a
vb
va
240 (220 200)
280 240 240 200
258.6 oC
c) the specific volume at T =220 deg C, p = 1.4 MPa in m3 /kg: need to do a double linear interpolation: then at p = 1.0
vaT 220
vaT 220
since 220 is halfway between 200 and 240
v200 v240
2 0.2060 0.2275 2
at p = 1.5
vbT 220
v200 v240
2 vbT 220
0.2168
then:
v
v
va
( p pa )
vb
va
pb
pa
0.2168 (1.4 1.)
0.1404 0.2168 1.5 1.0
0.15568 m3 /kg:
0.1325 0.1483 2
0.1404
EGR 334 Thermodynamics: Homework 06 Prob 3:10 For H20 determine the specified property at the indicated state. Locate the state on a sketch on the T -v diagram. a) p=300 kPa, v = 0.5 m3 /kg, find T in deg C. b) p = 28 MPa, T = 200 deg C, find v in m3 /kg c) p = 1 MPa, T = 405 deg C, Find v in m3 /kg d) T = 100 deg C, x = 60 %, find v in m3 /kg. ----------------------------------------------------------------------------------------------------------------------------a) p=300 kPa = 3 bar, v = 0.5 m3 /kg , find T in deg C. Table A-3 at p = 300 and vapor.
vf = 0.0010732 < 0.5 < 0.6058 = vg (sat. vapor)…so the substance is mixture of liquid
Therefore the temperature is the saturation temp for p=300 which is T = 133.6 deg C. ----------------------------------------------------------------------------------------------------------------------------- ----------b) = 28 MPa = 280 bar T = 200 deg C, find v in m3 /kg Table A-2 at T = 200 using table A-5:
p=280 > 15.54 (sat pressure) … so the substance is compressed liquid. p = 250 bar p = 300 bar T= 200 T = 200 v = 0.0011344 v = 0.0011302
interpolating:
v
0.0011344 (280 250)
0.0011302 0.0011344 300 250
0.0011319 m3 /kg
---------------------------------------------------------------------------------------------------------------------------------------- c) p = 1 MPa= 10 bar T = 405 deg C, Find v in m3 /kg Table A-3 at 10 bar Using table A-4: at
T= 405 > 179.9 (sat temp) so that the substance is vapor or superheated. p = 10 bar T v 400 0.3066 440 0.3257
interpolating:
v
0.3066 (405 400)
0.3257 0.3066 440 400
0.3090 m3 /kg
----------------------------------------------------------------------------------------------------------------------------- -------d) T = 100 deg C, x = 60 %, find v in m3 /kg. Table A-2 at T = 100 using the quality:
v
v f
x ( vg
vf =0.0010435
and vg = 1.673
v f )
0.0010435 0.6(1.673 0.0010435) 1.004 m3 /kg. T
(c) (a) (b) (d) v
EGR 334 Thermodynamics: Homework 06 Prob 3: 29 Ammonia contained in a piston cylinder assembly initially saturated vapor at 0 deg F, undergoes an isothermal process during which its volume (a) double and ( b) reduces by a half. For each case, fix the final state by giving 2 the quality or pressure in lb /in as appropriate. Locate the initial and final states on sketch of the p-v and T-v f diagrams. ----------------------------------------------------------------------------------------------------------------------------For ammonia the tables A-13E, A-14E, and A-15E will be used. a) Double the volume: State 1: saturated vapor at 0 deg F. Process 1-2: isothermal (T= constant) and v 2 = 2 v1 ( volume doubles) St1ate 2: substance becomes vapor…Need to use the superheated table: A-15E. saturated vapor give p1 = 30.416 psi and v1 = vg = 9.1100 ft3 /lbm
At State 1: T 1 = 0 degrees
T2=0 and v2 = 2v1 = 2(9.11)= 18.2200 ft 3 /lbm Using the super heated table A-15E
At state 2:
at T=0
p = 14 psi v = 20.289
p = 16 psi T=0 v = 17.701
interpolating:
p
14 (18.220 20.289)
16 14 17.701 20.289
15.60psi
----------------------------------------------------------------------------------------------------------------------------- ----b) half the volume: State 1: saturated vapor at 0 deg F. Process 1-2: isothermal (T= constant) and v 2 = 0.5 v1 ( volume is halved) St1ate 2: substance becomes liquid-vapor mixture…. saturated vapor give p 1 = 30.416 psi and v1 = vg = 9.1100 ft3 /lbm
At State 1: T 1 = 0 degrees
T2=0 and v2 = 0.5v1 = 0.5(9.11)= 4.555 ft3 /lbm
At state 2: using
vf = 0.02419
and vg = 9.1100
then to find the quality:
x
v vg v f
vg
4.555 0.02419 9.1100 0.02419
.499 49.9%
since the substance is a two phase mixture, the pressure is the saturation pressure for T = 0 deg or p = 30.416 psi
p
T
2b
1
1 2a
v
2a
2b
v
