(sp 1) - purdue engineeringย ยท (sp 1) given: in a closed rigid ... (saturated vapor)" p[1] =...
TRANSCRIPT
![Page 1: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/1.jpg)
(SP 1)
Given:
In a closed rigid tank,
State 1: ๐1,๐๐๐ = 1 ๐๐, ๐1,๐ = 0.05 ๐๐
P1= 0.0381 kPa, T1= -30oC
State 2: the liquid โ vapor equilibrium line, either saturated liquid or saturated vapor
Find:
(a) The volume of the tank
(b) P1 in bar
(c) a saturated liquid or a saturated vapor
(d) T2
(e) P-T and P-V diagrams
System sketch:
State 1 State 2
DLLL
Assumptions:
- Water vapor and ice are in equilibrium at state 1
- Pure saturated liquid or pure saturated vapor at state 2
Solution:
(a)
Vtank = ๐1๐ฃ1 = ๐1,๐๐๐๐ฃ1,๐๐๐ + ๐1,๐๐ฃ1,๐
1 kg ice
0.05 kg water vapor
P1= 0.0381 kPa
T1= -30oC
1.05 kg of
Saturated liquid
or
Saturated vapor Q
![Page 2: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/2.jpg)
From Table A-6,
๐ฃ1,๐๐๐ = 1.0858 ร 10โ3 ๐3
๐๐
๐ฃ1,๐ = 2943 ๐3
๐๐
Then, the equation we wrote can be,
Vtank = (1 ๐๐)(1.0858 ร 10โ3 ๐3
๐๐) + (0.05 ๐๐) (2943
๐3
๐๐) = 147.15 ๐3
The volume of the tank is ๐๐๐. ๐๐ ๐๐
(b)
P1 = (0.0381 kPa)(1 ๐๐๐
105๐๐) (
1000 ๐๐
1 ๐๐๐) = 0.000381 ๐๐๐
(c)
๐ฃ2 = ๐ฃ1 =๐๐ก๐๐๐
๐1=
๐๐ก๐๐๐
๐2=
147.15 ๐3
1.05 ๐๐= 140.14
๐3
๐๐
which is greater than ๐ฃ๐๐๐๐ก , 0.0031๐3
๐๐. Then, the state is a saturated vapor.
(d)
From Table A-2,
๐ฃ๐ = 147.120 ๐3
๐๐ at 5 oC
๐ฃ๐ = 137.734 ๐3
๐๐ at 6 oC
Interpolate to know temperature at ๐ฃ๐ is 140.14 ๐3
๐๐.
This provides T2 = 5.744 oC.
(e)
![Page 3: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/3.jpg)
![Page 4: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/4.jpg)
(SP 2)
Given:
Two-phase R-134a
State 1: T1= 50oF
State 2: superheated vapor at P2= 4 psia
Find:
(a) min and max of T2
(b) x1,min
(c) P-h diagram of x1=0.9
(d) plot x1 vs T2
System sketch:
Assumptions:
1. Steady-state, Steady-flow
2. โ๐พ๐ธ = โ๐๐ธ = 0
3. Adiabatic
4. No work
Basic equations:
![Page 5: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/5.jpg)
Solution:
Apply our assumptions into the energy balance equation,
The equation above is now,
๏ฟฝ๏ฟฝ1โ1 = ๏ฟฝ๏ฟฝ2โ2
Because of ๏ฟฝ๏ฟฝ1 = ๏ฟฝ๏ฟฝ2,
โ1 = โ2
(a)
Using the properties we know, establish an equation with the above equation. ๐ฅ1 ๐๐๐ ๐2 are
independent value and dependent value.
โ1(๐1, ๐ฅ1) = โ2(๐2, ๐2)
Use EES to solve (EES code is attached).
T2,max= 29.5 oF for x1=1 and T2,min= -60 oF for x2=1
Maximum T2 when x1 occurs at 1.0 (sat. vapor at the throttle inlet). Minimum T2 when x2
occurs at 1.0 (sat. vapor at the throttle outlet).
โต 1 โต 3 โต 4 โต 2 โต 2
![Page 6: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/6.jpg)
(b)
Minimum x1 occurs for T2= T2,min. See the attached EES code.
x1,min= 0.804
(c) and (d)
EES code is attached.
![Page 7: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/7.jpg)
"Specified inlet temperature and outlet pressure" T[1] = 50 [F] P[2] = 4 [psia] "Maximum outlet temperature where device works is when X[1] = 1 (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) h_1_max = enthalpy(R134a,T=T[1],x=1.0) h_2_max = h_1_max T_2_max = temperature(R134a,P=P[2],h=h_2_max) "Minimum outlet temperature where device works is when X[2] = 1 (saturated vapor)" T_2_min = temperature(R134a,P=P[2],x=1.0) h_2_min = enthalpy(R134a,P=P[2],x=1.0) h_1_min = h_2_min X_1_min = quality(R134a,P=P[1],h=h_1_min) "Data for process line on P-h diagram with X[1] = 0.9" x[1] = 0.9 h[1] = enthalpy(R134a,T=T[1],x=x[1]) h[2] = h[1] T[2] = temperature(R134a,P=P[2],h=h[2]) "Data for plot of x_1 versus T_2" h_2 = enthalpy(R134a,T=T_2,P=P[2]) h_1 = h_2 x_1 = quality(R134a,T=T[1],h=h_1)
![Page 8: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/8.jpg)
![Page 9: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/9.jpg)
(SP 3)
Given:
4L of water inside rigid body cooker from State 1 to State 2
State 1: 2L of saturated liquid water @ ๐1 = 200 ๐๐๐
2L of saturated water vapor @ ๐1 = 200 ๐๐๐
State 2: 4L of saturated water vapor @ ๐2 = 200 ๐๐๐
๐๐ ๐ก๐๐ฃ๐ = 1000 โ, ๐0 = 20โ, ๏ฟฝ๏ฟฝ๐๐๐ ๐ = 300 ๐, ๏ฟฝ๏ฟฝ๐๐ = ๐๐๐๐ ๐ก๐๐๐ก, โ๐ก = 1 โ๐๐ข๐
Find:
(a) The total mass of the steam (kg) entering the environment.
(b) The rate of heat transfer (kW) from the source.
(c) The total entropy production (kJ/K) in one hour.
(d) The total exergy destroyed (kJ) in one hour.
System: open system with saturated water inside 4L pressure cooker
Assumptions:
1. Steady flow (No steady state!)
2. No change in PE and KE
3. No work
4. Temperature and pressure are uniform throughout the cooker
State1 2L saturated liquid water + 2L saturated water vapor at P= 200 kPa
State 2
4L saturated water vapor
at P= 200 kPa
![Page 10: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/10.jpg)
5. Constant properties
6. A half of the volume is saturated liquid water and a half of the volume is saturated water
vapor inside the cooker at state 1
7. Water vapor inside the cooker at state 2
8. Water vapor leaves the cooker at state 2
Basic Equations:
๐๐๐๐ฃ
๐๐ก= โ ๏ฟฝ๏ฟฝ๐๐ โ โ ๏ฟฝ๏ฟฝ๐๐ข๐ก
๐๐ธ๐๐ฃ
๐๐ก= ๐๐๐ฃ
โ ๐๐๐ฃ + โ ๏ฟฝ๏ฟฝ๐๐(โ + ๐๐ธ + ๐พ๐ธ)๐๐ โ โ ๏ฟฝ๏ฟฝ๐๐ข๐ก(โ + ๐๐ธ + ๐พ๐ธ)๐๐ข๐ก
๐๐๐๐ฃ
๐๐ก= โ
๏ฟฝ๏ฟฝ๐๐ฃ
๐๐+ โ ๏ฟฝ๏ฟฝ๐๐๐ ๐๐ โ โ ๏ฟฝ๏ฟฝ๐๐ข๐ก๐ ๐๐ข๐ก + ๏ฟฝ๏ฟฝ๐ถ๐
๐ธ๏ฟฝ๏ฟฝ = ๐0๐๐๐ฃ
Solution:
(a)
Because pressure 200 kPa is given for the saturated water, we can find properties of
saturated water from table A-3.
Tsat = 120.2โ
vf = 0.0010605 m3/kg vg = 0.8857 m3/kg
uf = 504.49 kJ/kg ug = 2529.5 kJ/kg
hf = 504.7 kJ/kg hg = 2706.7 kJ/kg
sf = 1.5301 kJ/kgK sg = 7.1271 kJ/kgK
First, mass balance equation for open system is
๐๐๐๐ฃ
๐๐ก= โ ๏ฟฝ๏ฟฝ๐๐ โ โ ๏ฟฝ๏ฟฝ๐๐ข๐ก
Integrate the equation w.r.t. time and then the equation can be written as
โ๐ = ๐2 โ ๐1 = โ ๐๐๐ข๐ก
The mass at state 1 is
๐1 = ๐๐ + ๐๐ = ๐๐
๐ฃ๐โ +
๐๐๐ฃ๐
โ = 2๐ฟ0.0010605 ๐3/๐๐โ + 2๐ฟ
0.8857 ๐3/๐๐โ
โต no inlet
![Page 11: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/11.jpg)
= 0.002๐3
0.0010605๐3
๐๐โ
+ 0.002๐3
0.8857๐3
๐๐โ
= 1.886 ๐๐ + 0.002 ๐๐
= 1.888 ๐๐
The mass at state 2 is
๐2 = ๐๐ฃ๐
โ = 4๐ฟ0.8857 ๐3/๐๐โ
= 0.004๐3
0.8857๐3
๐๐โ
= 0.005๐๐
Thus,
โ๐ = ๐2 โ ๐1 = 0.005 ๐๐ โ 1.888 ๐๐ = โ1.883 ๐๐
โ ๐๐๐ข๐ก = โ1.883 ๐๐
๐๐๐ข๐ก = 1.883 ๐๐
Therefore, 1.883 kg of steam enters into the environment
(b)
Now, the energy balance equation for open system is
๐๐ธ๐๐ฃ
๐๐ก= ๐๐๐ฃ
โ ๐๐๐ฃ + โ ๏ฟฝ๏ฟฝ๐๐(โ + ๐๐ธ + ๐พ๐ธ)๐๐ โ โ ๏ฟฝ๏ฟฝ๐๐ข๐ก(โ + ๐๐ธ + ๐พ๐ธ)๐๐ข๐ก
Apply no work, no KE, no PE, and one outflow, then, the equation above can be
๐๐ธ๐๐ฃ
๐๐ก= ๐๐๐ฃ
โ ๐๐๐ฃ + โ ๏ฟฝ๏ฟฝ๐๐(โ + ๐๐ธ + ๐พ๐ธ)๐๐ โ โ ๏ฟฝ๏ฟฝ๐๐ข๐ก(โ + ๐๐ธ + ๐พ๐ธ)๐๐ข๐ก
Rewrite the equation,
![Page 12: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/12.jpg)
๐๐ธ๐๐ฃ
๐๐ก= ๐๐๐ฃ
โ ๏ฟฝ๏ฟฝ๐๐ข๐กโ๐๐ข๐ก
Integrate the equation w.r.t. time,
โ๐ธ๐๐ฃ = ๐๐๐ฃ โ ๐๐๐ข๐กโ๐๐ข๐ก
Because no KE and no PE, the equation above can be written as
๐2๐ข2 โ ๐1๐ข1 = ๐๐๐ โ ๐๐๐๐ ๐ โ ๐๐๐ข๐กโ๐๐ข๐ก
0.005๐๐ โ 2529.5๐๐ฝ/๐๐ โ (1.886๐๐ โ 504.49๐๐ฝ/๐๐ + 0.002๐๐ โ 2529.6๐๐ฝ/๐๐)
= ๐๐๐ โ 0.3๐๐ฝ/๐ โ 3600๐ โ 1.883๐๐ โ 2706.7๐๐ฝ/๐๐
๐๐๐ = 5232.8 kJ
The rate of heat transfer is
๏ฟฝ๏ฟฝ๐๐ =๐๐๐
โ๐กโ =5232.8 ๐๐ฝ
3600 ๐ = 1.454 kW
Therefore, ๐. ๐๐๐ ๐๐พ of the rate of heat transfer is entering into the cooker from
the stove.
(c)
The entropy balance equation for open system is
๐๐๐๐ฃ
๐๐ก= โ
๏ฟฝ๏ฟฝ๐๐ฃ
๐๐+ โ ๏ฟฝ๏ฟฝ๐๐๐ ๐๐ โ โ ๏ฟฝ๏ฟฝ๐๐ข๐ก๐ ๐๐ข๐ก + ๏ฟฝ๏ฟฝ๐ถ๐
Apply assumptions and integrate the equation w.r.t. time,
โ๐๐๐ฃ = โ๐๐๐ฃ
๐๐โ ๐๐๐ข๐ก๐ ๐๐ข๐ก + ๐๐๐ฃ
Rewrite,
๐2๐ 2 โ ๐1๐ 1 =๐๐๐
๐๐ ๐ก๐๐ฃ๐โ
๐๐๐๐ ๐
๐0โ ๐๐๐ข๐ก๐ ๐๐ข๐ก + ๐๐๐ฃ
๐คโ๐๐๐ ๐1๐ 1 = ๐๐๐ ๐ + ๐๐๐ ๐
Once you substitute all known values, you will get one unknown,
![Page 13: (SP 1) - Purdue Engineeringย ยท (SP 1) Given: In a closed rigid ... (saturated vapor)" P[1] = pressure(R134a,T=T[1],x=1.0) ... System: open system with saturated water inside 4L pressure](https://reader031.vdocuments.site/reader031/viewer/2022022519/5b187fbc7f8b9a3c258bd044/html5/thumbnails/13.jpg)
0.005 ๐๐ โ 7.1271๐๐ฝ
๐๐๐พโ (1.886 ๐๐ โ 1.5301
๐๐ฝ
๐๐๐พ+ 0.002 ๐๐ โ 7.1271
๐๐ฝ
๐๐๐พ)
=5232.8 ๐๐ฝ
1273 ๐พโ
0.3 ๐๐ โ 3600 ๐
293 ๐พโ 1.883 ๐๐ โ 7.1271
๐๐ฝ
๐๐๐พ+ ๐๐๐ฃ
๐๐๐ฃ = 0.0356 โ 2.9 โ 4.111 + 3.686 + 13.420 = 10.131 kJ/K
Therefore, ๐๐. ๐๐๐ ๐ค๐/๐ of the total entropy production is generated in one hour.
(d)
The total exergy destruction can be calculated by
๐ธ๏ฟฝ๏ฟฝ = ๐0๐๐๐ฃ = 293 ๐พ โ 10.131๐๐ฝ
๐พ= 2968.3 ๐๐ฝ
Therefore, ๐๐๐๐. ๐ ๐ค๐ of the total exergy is destroyed during one hour.