OBLICZENIA:

PRZEWĘŻENIE ZWĘŻKI:

Β = $\frac{\mathbf{d}}{\mathbf{D}}$

B =$\ \frac{\ 24}{70}$ = 0,4

LEPKOŚĆ GAZU:

μ = μn$\mathbf{\ }\frac{\mathbf{1}\mathbf{+}\frac{\mathbf{\text{Cs}}}{\mathbf{273}\mathbf{,}\mathbf{15}}}{\mathbf{1}\mathbf{+}\frac{\mathbf{\text{Cs}}}{\mathbf{T}\mathbf{1}}}\mathbf{\ }\sqrt{\frac{\mathbf{T}\mathbf{1}}{\mathbf{273}\mathbf{,}\mathbf{15}}}$

μ1=0,0000178 $\frac{1 + \frac{113}{273,15}}{1 + \frac{113}{298,1}}\ \sqrt{\frac{298,1}{273,15}}$ = 1,90621e-05[Pa x s]

μ2= 1,9067e-05

μ3= 1,90843e-05

μ4= 1,90967e-05

μ5= 1,90992e-05

μ6= 1,91041e-05

μ7= 1,90992e-05

μ8= 1,90868e-05

ZWĘŻKA VENTURIEGO:

GĘSTOŚĆ:

K1=$\frac{\mathbf{1}}{\mathbf{\text{Zn}}}$

K1=$\frac{1}{0,99941} =$1,00059

Przed zwężką:

ρ₁ = $\mathbf{\rho}_{\mathbf{n}}\frac{\mathbf{\rho}_{\mathbf{1}}\mathbf{\ }\mathbf{T}_{\mathbf{n}}}{\mathbf{p}_{\mathbf{n}}\mathbf{\ }\mathbf{T}_{\mathbf{1}}\mathbf{\ }\mathbf{K}_{\mathbf{1}}}$

ρ₁= 1,292923 $\frac{224x237,15}{101,325x298,1\ \text{x\ }1,00059} =$2,62$\left\lbrack \frac{\text{kg}}{m^{3}} \right\rbrack$

ρ₂=2,6

ρ₃=2,63

ρ₄=2,59

ρ₅=2,59

ρ₆=2,59

ρ₇=2,61

ρ₈=2,61

Za zwężką:

ρ₁=1,47$\left\lbrack \frac{\text{kg}}{m^{3}} \right\rbrack$

ρ₂=1,47

ρ₃=1,48

ρ₄=1,52

ρ₅=1,6

ρ₆=1,83

ρ₇=2,34

ρ₈=2,54

Współczynnik izentropy przyjmuje κ=1,6

LICZBA EKSPANSJI Ɛ:

Ɛ1${\mathbf{=}\left\lbrack \left( \frac{\mathbf{\kappa}\mathbf{\tau}^{\mathbf{2}\mathbf{/}\mathbf{\kappa}}}{\mathbf{\kappa}\mathbf{-}\mathbf{1}} \right)\left( \frac{\mathbf{1}\mathbf{-}\mathbf{\beta}^{\mathbf{4}}}{\mathbf{1}\mathbf{-}\mathbf{\beta}^{\mathbf{4}\mathbf{\ }}\mathbf{\tau}^{\mathbf{2}\mathbf{/}\mathbf{\kappa}}} \right)\mathbf{x}\left( \frac{\mathbf{1}\mathbf{-}\mathbf{\tau}^{\frac{\mathbf{\kappa}\mathbf{-}\mathbf{1}}{\mathbf{\kappa}}}}{\mathbf{1}\mathbf{-}\mathbf{\tau}} \right) \right\rbrack}^{\mathbf{1}\mathbf{/}\mathbf{2}}$

Ɛ1= $\left\lbrack \left( \frac{1,6x{0,560267857}^{2/1,6}}{1,6 - 1} \right)\left( \frac{1 - {0,4}^{4}}{1 - {0,4}^{4}{0,560267857}^{\frac{2}{1,6}}} \right)\left( \frac{1 - {0,560267857}^{\frac{1,6 - 1}{1,6}}}{1 - 0,560267857} \right) \right\rbrack^{1/2}$

= 0,64

Ɛ2= 0,64

Ɛ3= 0,64

Ɛ4= 0,67

Ɛ5= 0,70

Ɛ6= 0,80

Ɛ7= 1,01

Ɛ8= 1,1

Współczynnik przepływu C dla zwężki Venturiego przyjmujemy C=0,995

Qm=$\frac{\mathbf{C}}{\sqrt{\mathbf{1}\mathbf{-}\mathbf{\beta}^{\mathbf{4}}}}\mathbf{x}\mathbf{\ }\mathbf{\epsilon}\mathbf{\ }\mathbf{x}\mathbf{\ }\mathbf{1\ }\mathbf{x}\mathbf{\ }\frac{\mathbf{\pi}}{\mathbf{4}}{\mathbf{\ }\mathbf{d}}^{\mathbf{2}}\sqrt{\mathbf{2}\mathbf{}\mathbf{\text{pxρ}}\mathbf{1}}$

Qm =$\ \frac{0,995}{\sqrt{1 - {0,4}^{4}}}$ x 0,636954 x$\ \frac{3,14}{4}x\ 28^{2\ }x\ \sqrt{2\ x\ 197\ x\ 2,617502}\ $= 8972,74[kg/s]

Qv = $\frac{\mathbf{\text{Qm}}}{\mathbf{\rho}\mathbf{1}}$

Qv = $\frac{8972,738}{2,617502}\ $= 3427,98[m^3/s]

KRYZA POMIAROWA:

Przed zwężką:

Ρ1 = ρn $\frac{\mathbf{\rho}\mathbf{1\ }\mathbf{\text{Tn}}}{\mathbf{\text{pn}}\mathbf{\ }\mathbf{T}\mathbf{1\ }\mathbf{K}\mathbf{1}}$

P1=1,292923 x $\frac{182,5x237,15}{101,325x298,1x1,00059} =$2,13$\left\lbrack \frac{\text{kg}}{m^{3}} \right\rbrack$

P2=2,14

P3=2,14

P4=2,15

P5=2,18

P6=2,29

P7=2,5

P8=2,57

Za zwężką:

P1=-0,46$\left\lbrack \frac{\text{kg}}{m^{3}} \right\rbrack$

P2=-0,45

P3=-0,43

P4=-0,3

P5=-0,029

P6=0,59

P7=1,96

P8=2,44

LICZBA EKSPANSJI Ɛ:

E₁= 1-$\left( \mathbf{0}\mathbf{,}\mathbf{351}\mathbf{+}\mathbf{0}\mathbf{,}\mathbf{256}\mathbf{x}\mathbf{\beta}^{\mathbf{4}}\mathbf{+}\mathbf{0}\mathbf{,}\mathbf{93}\mathbf{x}\mathbf{\beta}^{\mathbf{8}} \right)\left\lbrack \mathbf{1}\mathbf{-}\left( \frac{\mathbf{p}_{\mathbf{2}}}{\mathbf{p}_{\mathbf{1}}} \right)^{\mathbf{1}\mathbf{/}\mathbf{\kappa}} \right\rbrack$

E₁= 1-(0,351+0,256x0, 4⁴+0,93x0, 4⁸)[1−(−0216438356)^1/1, 6]

= 0,78

Ɛ2= 0,78

Ɛ3=0,77

Ɛ4=0,75

Ɛ5=0,67

Ɛ6=0,80

Ɛ7=0,95

Ɛ8=0,99

WSPÓŁCZYNNIK PRZEPŁYWU C OKREŚLA WZÓR READER-HARRIS/GALLAGHERA:

C = 0,5961 + 0,0261 x β² - 0,216 x$\mathbf{\ }\mathbf{\beta}^{\mathbf{8}}\mathbf{+}\mathbf{0}\mathbf{,}\mathbf{000521}\left\lbrack \frac{\mathbf{10}^{\mathbf{6}}\mathbf{\beta}}{\mathbf{\text{ReD}}} \right\rbrack^{\mathbf{0}\mathbf{,}\mathbf{7}}$+ (0,0188+0,0063A)${\mathbf{\ }\mathbf{\beta}}^{\mathbf{3}\mathbf{,}\mathbf{5}}\left\lbrack \frac{\mathbf{10}^{\mathbf{6}}}{\mathbf{\text{ReD}}} \right\rbrack^{\mathbf{0}\mathbf{,}\mathbf{3}}$+

$\left( \mathbf{0}\mathbf{,}\mathbf{043}\mathbf{+}\mathbf{0}\mathbf{,}\mathbf{080}\mathbf{e}^{\mathbf{-}\mathbf{10}\mathbf{L}\mathbf{1}}\mathbf{+}\mathbf{0}\mathbf{,}\mathbf{123}\mathbf{e}^{\mathbf{-}\mathbf{7}\mathbf{L}\mathbf{1}} \right)\mathbf{\ }\left( \mathbf{1}\mathbf{-}\mathbf{0}\mathbf{,}\mathbf{11}\mathbf{A} \right)\mathbf{\ }\frac{\mathbf{\beta}^{\mathbf{4}}}{\mathbf{1}\mathbf{-}\mathbf{\beta}^{\mathbf{4}}}\mathbf{\ }$+

0,031 (M′₂-0,8M′₂^1,1) β^1,3 + 0,11 (0,75-β)$\mathbf{\ }\left( \mathbf{2}\mathbf{,}\mathbf{8}\mathbf{-}\frac{\mathbf{D}}{\mathbf{25}\mathbf{,}\mathbf{4}} \right)$

C= 0,5961+0,0261x0, 4²-0,216x${0,4}^{8} + 0,000521\left\lbrack \frac{10000000x0,4}{10000000} \right\rbrack^{0,7}$+

(0,0188+0,0063x0,003196325)${\ 0,4}^{3,5}\left\lbrack \frac{10000000}{10000000} \right\rbrack^{0,3}$+

$\left( 0,043 + 0,080e^{- 10x1} + 0,123e^{- 7x1} \right)\ \left( 1 - \ 0,11\ x\ 0,003196325 \right)\frac{{0,4}^{4}}{1 - {0,4}^{4}}\ $+

0,031 ($1,566666667 - 0,8x1,566666667\hat{}1,1$) 0, 4^1, 3 + 0,11 (0,75-0,4)$\ \left( 2,8 - \frac{70}{25,4} \right)$ = 0,6

W celu obliczenia współczynnika C korzystałem ze wzorów:

${\mathbf{M}\mathbf{'}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}{\mathbf{L}\mathbf{'}}_{\mathbf{2}}}{\mathbf{1}\mathbf{-}\mathbf{\beta}}$

M′₂= $\frac{2\ x\ 0,47}{1 - 0,4} = \ $1,57

A = $\left( \frac{\mathbf{19000}\mathbf{\ }\mathbf{\beta}}{\mathbf{\text{Re}}_{\mathbf{D}}} \right)^{\mathbf{0}\mathbf{,}\mathbf{8}}$

A = $\left( \frac{19000\ x\ 0,4}{10000000} \right)^{0,8}$ = 0,0032

Qm=$\frac{\mathbf{C}}{\sqrt{\mathbf{1}\mathbf{-}\mathbf{\beta}^{\mathbf{4}}}}\mathbf{\epsilon}\mathbf{1}\mathbf{x}\frac{\mathbf{\pi}}{\mathbf{4}}\mathbf{d}^{\mathbf{2}}\sqrt{\mathbf{2}\mathbf{}\mathbf{\text{pxρ}}\mathbf{1}}$

Qm = $\frac{0,600060238}{\sqrt{1 - {0,4}^{4}}}$ x 0,78

x$\ \frac{3,14}{4}x\ 28^{2\ }x\ \sqrt{2\ x\ 444\ x\ 2,132562839}$ = 8973,12[kg/s]

Qv = $\frac{\mathbf{\text{Qm}}}{\mathbf{\rho}\mathbf{1}}$

Qv= $\frac{8972,738}{2,132562839}$= 4207,67[m^3/s]