VE12. Rõhkude vahe torukonstruktsioonis

Ülesanne. Seinale on kinnitatud selline veetoru, nagu on joonisel:

Seinal olev toru

Seinal olev toru

Suurema osa ristlõikepindala (S1) ja väiksema osa ristlõikepindala (S2) suhe on 4:1. Kõrgused: H=5 m, h=1 m. Vee voolukiirus laias osas on 36 km/h. Leida…

a) … vee voolukiirus kitsas osas.
b) … rõhkude vahe laias ja kitsas osas.

Lahendus.

Paneme kirja andmed:
H equals 5 space m h equals 1 space m S subscript 1 over S subscript 2 equals 4 colon 1 v subscript 1 equals 36 fraction numerator k m over denominator h end fraction equals 10 m over s minus negative negative negative negative negative negative negative negative v subscript 2 minus ? left parenthesis p subscript 1 minus p subscript 2 right parenthesis minus ?

a) Voolukiiruse saame vooluhulkade võrdsuse tingimusest:
v subscript 1 S subscript 1 equals v subscript 2 S subscript 2 v subscript 2 equals fraction numerator v subscript 1 S subscript 1 over denominator S subscript 2 end fraction equals 10 m over s times 4 equals 40 m over s

b) Rõhkude vahe saame Bernoulli võrrandist:
p subscript 1 plus 1 half rho v subscript 1 squared plus rho g h subscript 1 equals p subscript 2 plus 1 half rho v subscript 2 squared plus rho g h subscript 2 p subscript 1 minus p subscript 2 equals 1 half rho left parenthesis v subscript 2 squared minus v subscript 1 squared right parenthesis plus rho g left parenthesis h subscript 2 minus h subscript 1 right parenthesis p subscript 1 minus p subscript 2 equals 1 half 1000 fraction numerator k g over denominator m cubed end fraction left parenthesis left parenthesis 40 m over s right parenthesis squared minus left parenthesis 10 m over s right parenthesis squared right parenthesis plus 1000 fraction numerator k g over denominator m cubed end fraction 9 comma 8 m over s squared left parenthesis 1 space m minus 5 space m right parenthesis p subscript 1 minus p subscript 2 equals 710000 space P a

VASTUS: a) 40 m/s, b) 710000 Pa.