Question

STATE BERNOULLIES THEOREM ?

Answer 1

$<font color="violet"><b> <i >HELLO !!!$

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BERNOULLIES THEOREM :

✏✏ It state that in case of regular flow of
non viscous incompressible liquid the sum of kinetic energy per unit mass , potential energy per unit mass , pressure energy remains constant ...

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HOPE IT HELPS UH ✌✌

Answer 2

Bernoulli's Theorem:

➟ According to Bernoulli's during the flow of ideal liquid through pipe of different cross section area its kinetic energy, potential energy & pressure energy remains constant at both the ends.

➟ P.E + K.E + Pot. E = const.              ..........(1)

Proof:

Workdone at end 'P':

⇒ W₁ = F.S

⇒ W₁ = P₁ A₁ V₁ Δt

Workdone at end 'Q':

⇒ W₂ = F.S

⇒ W₂ = P₂ A₂ V₂ Δt

Net Workdone, W = W₁ - W₂

⇒ W = P₁ A₁ V₁ Δt - P₂ A₂ V₂ Δt

Now, from equation of continuity, i.e. A₁ V₁ = A₂ V₂ = const.

⇒ W = (P₁ - P₂) A₁ V₁ Δt

$\rule{150}{1}$

$\implies \sf K.E.\;at\;'P'\;end\;(K.E.)_{P}=\dfrac{1}{2}mv_{1}^{2}\\ \\ \\ \implies \sf K.E.\;at\;'Q'\;end\;(K.E.)_{Q}=\dfrac{1}{2}mv_{2}^{2}\\ \\ \\ \implies \sf Net\;K.E=(K.E)_{Q}-(K.E.)_{P}\\ \\ \\ \implies \sf (K.E.)_{Net}=\dfrac{1}{2}m(v_{2}^{2}-v_{1}^{2})\\ \\ \\ \rule{150}{1}$

$\implies \sf P.E.\;at\;end\;'P'\;(P.E.)_{P}=mgh_{1}\\ \\ \\ \implies \sf P.E.\;at\;end\;'Q'\;(P.E.)_{Q}=mgh_{2}\\ \\ \\ \implies \sf Net\;P.E.=(P.E.)_{Q}-(P.E.)_{P}\\ \\ \\ \implies \sf (P.E.)_{Net}=mg(h_{2}-h_{1})\\ \\ \\ \rule{150}{1}$

$\sf Now,\;from\;energy\;consevation,\\ \\ \\ \implies \sf Workdone=K.E+P.E\\ \\ \\ \implies (P_{1}-P_{2})A_{1} V_{1}\Delta t=\dfrac{1}{2}m(v_{2}^{2}-v_{1}^{2})+mg(h_{2}-h_{1})\;\;\;\;.........(2)\\ \\ \\ \sf Now,\;divide\;equation\;(2)\;by\;'v',\\ \\ \\ \implies \sf (P_{1}-P_{2})\;\dfrac{A_{1}V_{1}\Delta t}{v}=\dfrac{1}{2} \dfrac{m}{v}(v_{2}^{2}-v_{1}^{2})+\dfrac{mg}{v}(h_{2}-h_{1})$

$\implies \sf P_{1}-P_{2}=\dfrac{1}{2}\rho (v_{2}^{2}-v_{1}^{2})+\rho g(h_{2}-h_{1})\\ \\ \\ \implies \sf P_{1}+\dfrac{1}{2}\rho v_{1}^{2}+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^{2}+\rho gh_{2}\\ \\ \\ \implies \sf P+\dfrac{1}{2}\rho v^{2}+\rho gh = const.\;\;\;\;..........(3)\\ \\ \\ \sf Now,\;divide\;equation\;(3)\;by\;'\rho gh',$

$\implies \sf \dfrac{P}{\rho gh}+\dfrac{\dfrac{1}{2}\rho v^{2}}{\rho gh}+\dfrac{\rho gh}{\rho gh}=\dfrac{const.}{\rho gh}\\ \\ \\ \implies \sf \dfrac{P}{\rho gh}+\dfrac{v^{2}}{2gh}+1=\dfrac{const.}{\rho gh}\\ \\ \\ \implies \sf \dfrac{P}{\rho gh}= Pressure\;head\\ \\ \\ \implies \sf \dfrac{v^{2}}{2gh}= Velocity\;head\\ \\ \\ \implies \sf 1 = P.E.\;head\\ \\ \\ \implies P.E+K.E+Pot.\;E=const.\\ \\ \\ \bf Hence\;Proved!!$