جریان در لوله

L e d = 0.058 Re d {displaystyle {frac {{L}_{e}}{d}}=0.058{{operatorname {Re} }_{d}}}

( m ˙ V x ) i n − ( m ˙ V x ) o u t + ∑ f x = 0 {displaystyle {{({dot {m}}{{V}_{x}})}_{in}-{({dot {m}}{{V}_{x}})}_{out}+sum {f_{x}}=0}}

m ˙ = ρ b ∫ u ( y ) d y {displaystyle {{dot {m}}=rho bint {u(y)dy}}}

( m ˙ V x ) i n = ( m ˙ V x ) o u t = ρ b ∫ u 2 ( y ) d y → ∑ F x = 0 {displaystyle {{({dot {m}}{{V}_{x}})}_{in}={({dot {m}}{{V}_{x}})}_{out}=rho bint {u^{2}(y)dy}rightarrow sum {F_{x}}=0}}

∑ f x = ( P x − P x + d x ) b d y + ( τ y + d y − τ y ) b d x = 0 {displaystyle {sum {f_{x}}=(P_{x}-P_{x+dx})bdy+(tau _{y+dy}-tau _{y})bdx=0}}

− d P d x + d τ d y = 0 {displaystyle {-{frac {dP}{dx}}+{frac {dtau }{dy}}=0}}

τ = μ d u d y {displaystyle {tau =mu {frac {du}{dy}}}}

d 2 u d y 2 = 1 μ d P d x = − G {displaystyle {{frac {{d}^{2}u}{{dy}^{2}}}={frac {1}{mu }}{frac {dP}{dx}}=-G}}

u ( y ) = − G 2 y 2 + c 1 y + c 2 {displaystyle u(y)=-{frac {G}{2}}{{y}^{2}}+{{c}_{1}}y+{{c}_{2}}}

y = 0 → u = 0 → c 2 = 0 y = h → u = 0 → c 1 = G 2 h u ( y ) = G 2 y ( h − y ) y = h 2 → u = u max → u max = u ( h 2 ) = G h 2 8 {displaystyle {begin{aligned}&y=0to u=0to {{c}_{2}}=0\&y=hto u=0to {{c}_{1}}={frac {G}{2}}h&u(y)={frac {G}{2}}y(h-y)\&y={frac {h}{2}}to u={{u}_{max }}to {{u}_{max }}=u({frac {h}{2}})=G{frac {{h}^{2}}{8}}\end{aligned}}}

→ G = 8 u max h 2 {displaystyle to G={frac {8{{u}_{max }}}{{h}^{2}}}}

→ u ( y ) = 4 u max ( y h ) ( 1 − y h ) {displaystyle to u(y)=4{{u}_{max }}({frac {y}{h}})(1-{frac {y}{h}})}

→ u max = − 1 8 μ d p d x h 2 = p 1 − p 2 8 μ L h 2 {displaystyle to {{u}_{max }}={frac {-1}{8mu }}{frac {dp}{dx}}{{h}^{2}}={frac {{{p}_{1}}-{{p}_{2}}}{8mu L}}{{h}^{2}}}

u _ = 1 h ∫ 0 h u ( y ) d y = 2 3 u max = p 1 − p 2 12 μ L h 2 {displaystyle {overset {_}{mathop {u}}},={frac {1}{h}}int _{0}^{h}{u(y)}dy={frac {2}{3}}{{u}_{max }}={frac {{{p}_{1}}-{{p}_{2}}}{12mu L}}{{h}^{2}}}

Q = u _ A = − G H 3 b 12 μ {displaystyle Q={overset {_}{mathop {u}}},A={frac {-G{{H}^{3}}b}{12mu }}}

d p d x = − 12 μ H 2 u _ {displaystyle {frac {dp}{dx}}={frac {-12mu }{{H}^{2}}}{overset {_}{mathop {u}}},}

Δ p = − Δ x d p d x {displaystyle {Delta p}={-Delta x}{frac {dp}{dx}}}

→ Δ p = 12 μ u _ L H 2 = f L D H × 1 2 ρ u 2 {displaystyle to {Delta p}={frac {12mu {{overset {_}{mathop {u}}},}L}{{H}^{2}}}=f{frac {L}{{D}_{H}}}times {frac {1}{2}}rho {{u}^{2}}}

D h = 4 A p {displaystyle {D}_{h}={frac {4A}{p}}}

D H = 4 b H 2 ( b + H ) = 2 H {displaystyle {D}_{H}={frac {4bH}{2(b+H)}}=2H}

f = 12 L μ u _ H 2 L 2 A 1 2 ρ u _ − 2 = 48 μ × 2 ρ u × 2 H = 96 μ ρ u D H = 96 R e ( D H ) {displaystyle f={frac {12Lmu {{overset {_}{mathop {u}}},}}{{{H}^{2}}{frac {L}{2A}}{frac {1}{2}}rho {{{overset {_}{mathop {u}}},}^{-2}}}}={frac {48mu times 2}{rho utimes 2H}}={frac {96mu }{rho u{{D}_{H}}}}={frac {96}{{Re}_{({D}_{H})}}}}