1. Schemat pomiarowy

2. Wykorzystane wzory obliczeniowe:

∆p= ρ_m ∙ ∆h ∙ q q_v= A∙V_sr

V = $\sqrt{\frac{2p}{\rho}}$ ρ₀= $\frac{1}{R_{s}}\frac{1 + 0,622 \bullet \varphi \bullet \frac{p_{s}}{p - p_{s \bullet \varphi}}}{1 + \varphi \bullet \frac{p_{s}}{p - p_{s \bullet \varphi}}}$ ∙ $\frac{P}{T}$

A= $\frac{\pi({2r}^{\begin{matrix} 2 \\ ) \\ \end{matrix}}}{4}$ $\frac{V_{t}}{V_{\max}}$= ${(1 - \frac{r}{R}\ )\ }^{\frac{1}{2,1logRe - 1,9}}$

Re= $\frac{\rho_{0 \bullet V_{sr} \bullet (d \bullet 2)}}{\mu}$ p₀ = 9,8065 ∙ 10⁵ ∙ $\frac{e^{0,01028 \bullet T - \ \frac{7821,541}{T} + 82,86568}}{T^{11,48776}}$

T = 273,16 + t

	y	r	∆h
	mm	mm	mm
1	66	39	12
2	65	38	15
3	64	37	20
4	63	36	22
5	61,5	34,5	25
6	60	33	28
7	58,5	31,5	28
8	57	30	29
9	55	28	30
10	53	26	30
11	51	24	29
12	49	22	30
13	47	20	31
14	45	18	30
15	43	16	31
16	41	14	31
17	38	11	31
18	35	8	32
19	32	5	31
20	29	2	32
21	27	0	32

Stałe:

q_m=1000kg/m³

t_o=20°C

p_o=998hPa

φ=49%

D=80mm

μ₀=17,08·10^-6

R_s=281,1 J/(kgK)

C=112

3.Przykłądowe obliczenia:

p_s = 9,8065 ∙ 10⁵ ∙ $\frac{e^{0,01028 \bullet 293,16 - \ \frac{7821,541}{T} + 82,86568}}{{293,16}^{11,48776}}$ ≈ 1877 Pa

ρ₀= $\frac{1}{R_{s}}\frac{1 + 0,622 \bullet 0,49 \bullet \frac{1877}{99800 - 1877_{\bullet 0,49}}}{1 + 0,49 \bullet \frac{1877}{99800 - 1877_{\bullet 0,49}}}$ ∙ $\frac{99800}{293,16}$ ≈ 1,18

∆p = 1000 ∙ 0,012 ∙ 9,81 ≈ 118Pa

V = $\sqrt{\frac{2 \bullet 275}{2,75}}$ ≈ 14,13$\frac{m}{s}$ $\frac{V}{V_{\max}}$ = $\frac{21,58}{23,07}$ ≈ 0,94

$\frac{r}{R}\ $= $\frac{33}{40}\ $= 0,825 V_sr = $\frac{14,13 + 22,33 + 23,07}{4}$ ≈ 20,56

A= $\frac{3,14({2 \bullet 0,04}^{\begin{matrix} 2 \\ ) \\ \end{matrix}}}{4}$ ≈ 50,24 ∙ 10⁻⁴ q_v = 50,24 ∙ 10⁻⁴ ∙ 20,56 ≈ 10,33∙10⁻² $\frac{m}{s^{2}}$

𝝁 = 17,08 ∙ 10⁻⁶ ∙ $\frac{273 + 112}{293,16 + 112}$ ($\frac{{293,16}^{3/2}}{273})$ ≈ 0,00002

Re = $\frac{1,18\ \bullet 20,56\ \bullet 0,08}{0,00002\ }$ ≈ 107463

$\frac{V_{t}}{V_{\max}}$= ${(1 - 0,825)\ }^{\frac{1}{2,1log(107463) - 1,9}\ }$ ≈ 0,82