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Author: Admin | 2025-04-28
Follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. Let the subscripts 1 and 2 refer to any two points along the path that the bit of fluid follows; Bernoulli’s equation becomes[latex]{P_1\:+}[/latex] [latex]{\frac{1}{2}}[/latex] [latex]{\rho{v}_1^2+\rho{g}h_1=P_2\:+}[/latex] [latex]{\frac{1}{2}}[/latex] [latex]{\rho{v}_2^2+\rho{g}h_2.}[/latex]Bernoulli’s equation is a form of the conservation of energy principle. Note that the second and third terms are the kinetic and potential energy with [latex]{m}[/latex] replaced by [latex]{\rho}.[/latex] In fact, each term in the equation has units of energy per unit volume. We can prove this for the second term by substituting [latex]{\rho=m/V}[/latex] into it and gathering terms:[latex]{\frac{1}{2}}[/latex] [latex]{\rho{v}^2\:=}[/latex] [latex]{\frac{\frac{1}{2}mv^2}{V}}[/latex] [latex]{=}[/latex] [latex]{\frac{\text{KE}}{V}}.[/latex]So [latex]{\frac{1}{2}\rho{v}^2}[/latex] is the kinetic energy per unit volume. Making the same substitution into the third term in the equation, we find[latex]{\rho{gh}\:=}[/latex] [latex]{\frac{mgh}{V}}[/latex] [latex]{=}[/latex] [latex]{\frac{\text{PE}_{\text{g}}}{V}},[/latex]so [latex]{\rho{gh}}[/latex] is the gravitational potential energy per unit volume. Note that pressure [latex]{P}[/latex] has units of energy per unit volume, too. Since [latex]{P=F/A},[/latex] its units are [latex]{\text{N/m}^2}.[/latex] If we multiply these by m/m, we obtain [latex]{\text{N}\cdotp\text{m/m}^3=\text{J/m}^3},[/latex] or energy per unit volume. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction.MAKING CONNECTIONS: CONSERVATION OF ENERGYConservation of energy applied to fluid flow produces Bernoulli’s equation. The net work done by the fluid’s pressure results in changes in the fluid’s [latex]\text{KE}[/latex] and [latex]{\text{PE}_{\text{g}}}[/latex] per unit volume. If other forms of energy are involved in fluid flow, Bernoulli’s equation can be modified to take these forms into account. Such forms of energy include thermal energy dissipated because of fluid viscosity.The general form of Bernoulli’s equation has three terms in it, and it is broadly applicable. To understand it better, we will look at a number of specific situations that simplify and illustrate its use and meaning.Let us first consider the very simple situation where the fluid is static—that is, [latex]{v_1=v_2=0}.[/latex] Bernoulli’s equation in that case is[latex]{P_1+\rho{gh}_1=P_2+\rho{gh}_2}.[/latex]We can further simplify the equation by taking [latex]{h_2=0}[/latex] (we can always choose some height to be zero, just as we often have done for other situations involving the gravitational force, and take all other heights to be relative to this). In that case, we get[latex]{P_2=P_1+\rho{gh}_1}.[/latex]This equation tells us that, in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by [latex]{h_1},[/latex] and consequently, [latex]{P_2}[/latex] is greater than [latex]{P_1}[/latex] by an amount [latex]{\rho{gh}_1}.[/latex] In the very simplest case, [latex]{P_1}[/latex] is zero at the top of the fluid, and we get the familiar relationship [latex]{P=\rho{gh}}.[/latex] (Recall that [latex]{P=\rho{gh}}[/latex] and [latex]{\rho\text{PE}_{\text{g}}=mgh.}[/latex] ) Bernoulli’s equation includes the fact that the pressure due to the weight of a fluid is [latex]{\rho{gh}}.[/latex] Although we introduce Bernoulli’s equation for fluid flow, it includes much of what we studied for static fluids in the preceding chapter.Another important situation is one in which the fluid moves but its depth is constant—that is, [latex]{h_1=h_2}.[/latex] Under that condition, Bernoulli’s equation becomes[latex]{P_1\:+}[/latex] [latex]{\frac{1}{2}}[/latex] [latex]{\rho{v}_1^2=P_2\:+}[/latex] [latex]{\frac{1}{2}}[/latex] [latex]{\rho{v}_2^2}.[/latex]Situations in which fluid flows at
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