Politecznika Wrosławska |
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03.11.2009 |
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Wydział Mechaniczno - Energetyczny |
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Energetyka |
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Rok 3 semestr 5 |
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Temat: Wyznaczenie linii nasycenia wody |
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Paweł Kois |
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Robert Paliszkiewicz |
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Dawid Noga |
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1. Cel ćwiczenia |
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Wyznaczenie linii nasycenia wody. |
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2. Rysunek stanowiska |
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3.Czynności wykonane |
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a) Sprawdzenie stanowiska pomiarowego. |
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b) Sprawdzenie waruknów otoczenia panujących w laboratorium. |
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c) Uruchomienie stanowiska pomiarowego, po zgłoszeniu gotowości prowadzącemu. |
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d) Podgrzewanie zbiorniczka z wodą. |
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e) Doczytywanie wyników na termometrze i wakuometrze. |
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f) Przeliczenie odczytanych wartości na jednostki niezbędne do obliczeń. |
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g) Opracowanie wyników i sporządzenie wykresów. |
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h) Sporządzenie wniosków do otrzymanych wyników. |
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4. Wyniki pomiarów |
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Warunki otoczenia |
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100200 |
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Temperatura otoczenia t0 |
18o C |
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Ciśnienie p0 |
1002 hPa |
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Wilgotność powietrza φ0 |
65% |
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Wyniki uzyskane podczas pomiarów |
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LP |
Temperatura |
Temperatura |
Wskazanie wakuometru |
Przeliczona różnica ciśnień |
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tn |
Tn |
∆pn1 |
∆pn2 |
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oC |
K |
atm techniczna |
Pa |
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1 |
58 |
331,15 |
0,86 |
-84366 |
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84366,00 |
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63 |
336,15 |
0,82 |
-80442 |
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80442,00 |
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67 |
340,15 |
0,78 |
-76518 |
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76518,00 |
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70 |
343,15 |
0,74 |
-72594 |
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72594,00 |
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73 |
346,15 |
0,70 |
-68670 |
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68670,00 |
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76 |
349,15 |
0,66 |
-64746 |
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64746,00 |
7 |
79 |
352,15 |
0,62 |
-60822 |
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60822,00 |
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81 |
354,15 |
0,58 |
-56898 |
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56898,00 |
9 |
83 |
356,15 |
0,54 |
-52974 |
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52974,00 |
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85 |
358,15 |
0,50 |
-49050 |
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49050,00 |
11 |
87 |
360,15 |
0,46 |
-45126 |
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45126,00 |
12 |
89 |
362,15 |
0,42 |
-41202 |
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41202,00 |
13 |
90 |
363,15 |
0,38 |
-37278 |
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37278,00 |
14 |
92 |
365,15 |
0,34 |
-33354 |
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33354,00 |
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93 |
366,15 |
0,30 |
-29430 |
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29430,00 |
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94 |
367,15 |
0,26 |
-25506 |
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25506,00 |
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95 |
368,15 |
0,22 |
-21582 |
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21582,00 |
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97 |
370,15 |
0,18 |
-17658 |
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17658,00 |
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98 |
371,15 |
0,14 |
-13734 |
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13734,00 |
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99 |
372,15 |
0,10 |
-9810 |
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9810,00 |
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Do obliczeń niezbędne śa przekształenia jednostek. |
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Mianowicie temperature w oC zmieniamy na Kelwiny, oraz atmosfery techniczne na Pa. |
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Następnie obliczamy różnicę ciśnień pomiędzy wskazaniem wakuometru a ciśnieniem atmosferycznym. |
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Lp |
Temperatura |
Ciśnienie wewnątrz pojemnika |
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Tn |
pn |
r z tablic parowych |
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K |
Pa |
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1 |
331,15 |
15834,00 |
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2 |
336,15 |
19758,00 |
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3 |
340,15 |
23682,00 |
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4 |
343,15 |
27606,00 |
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5 |
346,15 |
31530,00 |
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6 |
349,15 |
35454,00 |
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7 |
352,15 |
39378,00 |
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8 |
354,15 |
43302,00 |
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9 |
356,15 |
47226,00 |
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10 |
358,15 |
51150,00 |
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11 |
360,15 |
55074,00 |
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12 |
362,15 |
58998,00 |
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13 |
363,15 |
62922,00 |
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14 |
365,15 |
66846,00 |
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-80442 |
15 |
366,15 |
70770,00 |
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-76518 |
16 |
367,15 |
74694,00 |
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-72594 |
17 |
368,15 |
78618,00 |
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-68670 |
18 |
370,15 |
82542,00 |
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-64746 |
19 |
371,15 |
86466,00 |
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-60822 |
20 |
372,15 |
90390,00 |
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-56898 |
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-52974 |
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-49050 |
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-45126 |
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-41202 |
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-37278 |
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-33354 |
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-29430 |
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-25506 |
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-21582 |
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-17658 |
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-13734 |
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-9810 |
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-100200 |
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Sporządzamy wykres ciśnienia w funkcji temperatury |
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Lp |
X |
Y |
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1 |
0,0000000 |
0,0000000 |
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2 |
0,0000449 |
0,2213989 |
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3 |
0,0000799 |
0,4025557 |
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4 |
0,0001056 |
0,5558736 |
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5 |
0,0001309 |
0,6887799 |
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6 |
0,0001557 |
0,8060766 |
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7 |
0,0001801 |
0,9110478 |
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8 |
0,0001961 |
1,0060393 |
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9 |
0,0002120 |
1,0927851 |
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10 |
0,0002277 |
1,1726030 |
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11 |
0,0002432 |
1,2465182 |
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12 |
0,0002585 |
1,3153440 |
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13 |
0,0002661 |
1,3797363 |
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14 |
0,0002812 |
1,4402319 |
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15 |
0,0002887 |
1,4972757 |
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16 |
0,0002961 |
1,5512402 |
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17 |
0,0003035 |
1,6024412 |
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18 |
0,0003182 |
1,6511477 |
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19 |
0,0003255 |
1,6975917 |
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20 |
0,0003327 |
1,7419741 |
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|
Wartość współczynnika a= |
0,000191296687051 |
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5. Użyte wzory |
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|
R=8314,7/18≈461,9 J/kgK |
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|
Teoretyczna wartość współczynnika a= 5125 (wedłu tablic parowych). |
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|
Wartość doświadczalną współczynnik a obliczyliśmy za pomocą prograamu EXEL. |
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|
Tn=tn+273,15 [K] |
|
#DIV/0! |
0 |
|
Przeszkształcenie atmosfer technicznych na Pascale odbywa się z własności: 1Atmosfera techniczna = 98100 Pa (według tabeli zamieszczonej w laboratoeium) |
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|
0 |
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|
#VALUE! |
0 |
|
∆pn2=∆pn1x98100 [Pa] |
|
461,9 |
0 |
|
pni=pb-|∆pn2i| |
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|
0 |
|
Yi=ln(pn,i/pn,1) |
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|
0 |
|
Xi=1/Tn,i-1-Tn,1 |
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|
0 |
|
a =5224 |
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|
0 |
|
R=461,9 J/kgK |
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|
#REF! |
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0 |
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0 |
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6. Przykładowe obliczenia |
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0 |
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0 |
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0 |
|
58+273,15=331,15 |
T1 |
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0 |
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98100x0,86=84366 |
∆p2 |
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100200 - |-84366||=15834 |
p1 |
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0 |
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0 |
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0 |
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0 |
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#DIV/0! |
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7. Wnioski |
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Przeprowadzając badania poznaliśmy jeden ze sposobów wyznaczenia linii nasycenia dla wody. |
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Badanie uznajemy za wykonane poprawnie ponieważ założenie, że linia nasycenia po przekształceniach matematycznych zawartych w instrukcji ma być zbliżona do lini prostej zostało spełnione. |
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Niewielkie odchylenia od prostej mogą być wynikami brak precyzji w odczytywaniu wyników. |
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Otrzymany współczynnik adośw jest bardzo zbliżony do teoretycznego współczynnika. |
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Wynika z tego że jesteśmy w stanie wyznaczyć linie nasycenia dla wody z dość dużą dokładnością. |
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8. Załączniki |
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a) protokół pomiarowy sporządzony: 24.10.2009 Wrocław |
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Sprawozdanie wykonał / Data |
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Sprawozdanie Sprawdził / data |
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Paweł Kois / 04.11.2009 |
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