20 -250 Pressure [kPa] 100 80 60 40 Problem 1 R717 is refrigerant grade Ammonia (NH3). Ammonia is at a temperature of 130°C and a pressure of 3 MPa. Determine the specific volume of the ammonia based on: • a) Ideal gas equation; b) The generalized compressibility chart; c) The p-h diagram below indicating (marking) the position on the diagram clearly (Hint: Use the Zoom function of View.); and d) The superheat ammonia table (See: BrightSpace / Content / Thermodynamic Property Tables). e) Determine the error involved in cases (a), (b), & (c) using case (d) as the reference. 10000 8000 s [kJ/(kg K)] v [m³/kg] 6000 T [°C] 4000 2000 1000 800 600 400 200 $=-0.25 -50 40 -30 20 -10 10 10 20 30 40 50 50 09 70 80 06 8. 100 110 120 v=0.006- v= 0.008- v=0.010 v= 0.015 v=0.020- v= 0.030 v= 0.040 v= 0.060 v= 0.0807 v=0.100- v= 0.150 0.200 V= 0.300 v= 0,400 y= 0.600 v=0.800 v= 1.000 va 1.500 v 2.000 v= 3.000 v=4.000 x=0.10 0.25 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75 6.25 6.75 0 250 500 750 1000 1250 Enthalpy [kJ/kg] 60 50 710 0.008 0.009 120. 0.010 110 100 -90 80 70/ 30 60 50 40 = 4.75 s=5.00 =5.25 s=5.50 0.015 0.020 0.030 0.040 0.050 0.060 0.070 0.080 0.090 0.100 0.150 0.200 0.300 -20 S=7.00 -7.25 -10 =6.75 s=6.50 s=6.25 s=6.00 S=5.75 -40 -30 =7.50 S=7.75 S=8.00 8.25 R717 Ammonia -60-40-20 0 20 40 60 80 100120 140 160 180 200 220 1500 1750 2000 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.500 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.00 15.00 20.00
20 -250 Pressure [kPa] 100 80 60 40 Problem 1 R717 is refrigerant grade Ammonia (NH3). Ammonia is at a temperature of 130°C and a pressure of 3 MPa. Determine the specific volume of the ammonia based on: • a) Ideal gas equation; b) The generalized compressibility chart; c) The p-h diagram below indicating (marking) the position on the diagram clearly (Hint: Use the Zoom function of View.); and d) The superheat ammonia table (See: BrightSpace / Content / Thermodynamic Property Tables). e) Determine the error involved in cases (a), (b), & (c) using case (d) as the reference. 10000 8000 s [kJ/(kg K)] v [m³/kg] 6000 T [°C] 4000 2000 1000 800 600 400 200 $=-0.25 -50 40 -30 20 -10 10 10 20 30 40 50 50 09 70 80 06 8. 100 110 120 v=0.006- v= 0.008- v=0.010 v= 0.015 v=0.020- v= 0.030 v= 0.040 v= 0.060 v= 0.0807 v=0.100- v= 0.150 0.200 V= 0.300 v= 0,400 y= 0.600 v=0.800 v= 1.000 va 1.500 v 2.000 v= 3.000 v=4.000 x=0.10 0.25 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75 6.25 6.75 0 250 500 750 1000 1250 Enthalpy [kJ/kg] 60 50 710 0.008 0.009 120. 0.010 110 100 -90 80 70/ 30 60 50 40 = 4.75 s=5.00 =5.25 s=5.50 0.015 0.020 0.030 0.040 0.050 0.060 0.070 0.080 0.090 0.100 0.150 0.200 0.300 -20 S=7.00 -7.25 -10 =6.75 s=6.50 s=6.25 s=6.00 S=5.75 -40 -30 =7.50 S=7.75 S=8.00 8.25 R717 Ammonia -60-40-20 0 20 40 60 80 100120 140 160 180 200 220 1500 1750 2000 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.500 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.00 15.00 20.00
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
Solve step by step
![20
-250
Pressure [kPa]
100
80
60
40
Problem 1
R717 is refrigerant grade Ammonia (NH3). Ammonia is at a temperature of 130°C and a pressure of 3 MPa.
Determine the specific volume of the ammonia based on:
•
a) Ideal gas equation;
b) The generalized compressibility chart;
c) The p-h diagram below indicating (marking) the position on the diagram clearly (Hint: Use the Zoom
function of View.); and
d) The superheat ammonia table (See: BrightSpace / Content / Thermodynamic Property Tables).
e) Determine the error involved in cases (a), (b), & (c) using case (d) as the reference.
10000
8000
s [kJ/(kg K)]
v [m³/kg]
6000
T [°C]
4000
2000
1000
800
600
400
200
$=-0.25
-50
40
-30
20
-10
10
10
20
30
40
50
50
09
70
80
06
8.
100
110
120
v=0.006-
v= 0.008-
v=0.010
v= 0.015
v=0.020-
v= 0.030
v= 0.040
v= 0.060
v= 0.0807
v=0.100-
v= 0.150
0.200
V= 0.300
v= 0,400
y= 0.600
v=0.800
v= 1.000
va 1.500
v 2.000
v= 3.000
v=4.000
x=0.10
0.25
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
0.75
1.25
1.75
2.25
2.75
3.25
3.75
4.25
4.75 5.25
5.75
6.25
6.75
0
250
500
750
1000
1250
Enthalpy [kJ/kg]
60
50
710
0.008
0.009
120.
0.010
110
100
-90
80
70/
30
60
50
40
= 4.75
s=5.00
=5.25
s=5.50
0.015
0.020
0.030
0.040
0.050
0.060
0.070
0.080
0.090
0.100
0.150
0.200
0.300
-20
S=7.00
-7.25
-10
=6.75
s=6.50
s=6.25
s=6.00
S=5.75
-40
-30
=7.50
S=7.75
S=8.00
8.25
R717 Ammonia
-60-40-20 0 20 40 60 80 100120 140 160 180 200 220
1500
1750
2000
0.400
0.500
0.600
0.700
0.800
0.900
1.000
1.500
2.000
3.000
4.000
5.000
6.000
7.000
8.000
9.000
10.00
15.00
20.00](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8f7b4131-1469-43f0-8481-774e91aa2dd0%2Fbeda638f-40b1-408d-b837-c32961de8f11%2Fz5iva8k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:20
-250
Pressure [kPa]
100
80
60
40
Problem 1
R717 is refrigerant grade Ammonia (NH3). Ammonia is at a temperature of 130°C and a pressure of 3 MPa.
Determine the specific volume of the ammonia based on:
•
a) Ideal gas equation;
b) The generalized compressibility chart;
c) The p-h diagram below indicating (marking) the position on the diagram clearly (Hint: Use the Zoom
function of View.); and
d) The superheat ammonia table (See: BrightSpace / Content / Thermodynamic Property Tables).
e) Determine the error involved in cases (a), (b), & (c) using case (d) as the reference.
10000
8000
s [kJ/(kg K)]
v [m³/kg]
6000
T [°C]
4000
2000
1000
800
600
400
200
$=-0.25
-50
40
-30
20
-10
10
10
20
30
40
50
50
09
70
80
06
8.
100
110
120
v=0.006-
v= 0.008-
v=0.010
v= 0.015
v=0.020-
v= 0.030
v= 0.040
v= 0.060
v= 0.0807
v=0.100-
v= 0.150
0.200
V= 0.300
v= 0,400
y= 0.600
v=0.800
v= 1.000
va 1.500
v 2.000
v= 3.000
v=4.000
x=0.10
0.25
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
0.75
1.25
1.75
2.25
2.75
3.25
3.75
4.25
4.75 5.25
5.75
6.25
6.75
0
250
500
750
1000
1250
Enthalpy [kJ/kg]
60
50
710
0.008
0.009
120.
0.010
110
100
-90
80
70/
30
60
50
40
= 4.75
s=5.00
=5.25
s=5.50
0.015
0.020
0.030
0.040
0.050
0.060
0.070
0.080
0.090
0.100
0.150
0.200
0.300
-20
S=7.00
-7.25
-10
=6.75
s=6.50
s=6.25
s=6.00
S=5.75
-40
-30
=7.50
S=7.75
S=8.00
8.25
R717 Ammonia
-60-40-20 0 20 40 60 80 100120 140 160 180 200 220
1500
1750
2000
0.400
0.500
0.600
0.700
0.800
0.900
1.000
1.500
2.000
3.000
4.000
5.000
6.000
7.000
8.000
9.000
10.00
15.00
20.00
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