New answers tagged

1

Why? The expression given by: $$ \int_{S} \ dS \ \hat{\mathbf{n}} \cdot \left( \nabla \times \mathbf{A} \right) = \oint_{C} \ \mathbf{A} \cdot d\mathbf{l} $$ is just a vector calculus rule called Stokes' theorem and is valid regardless of the type of flow. Because the $S$ is not simply connected? The $S$ in the expression represents the closed ...


1

Exactly because the region is not simply connected. The stokes (or Green in 2d) theorem no longer holds. Consider for example the two dimensional vector field $$\vec F(x,y)=\frac{-y\hat i+x\hat j}{x^2+y^2},$$ which has vanishing curl and circulation $2\pi$ around a unit circle centerd at the origin. If this vector field is meant to be a flow velocity field ...


2

Compute the gradient of this velocity field: $$\operatorname{grad}{\bf u}=\nabla_i u_j= \begin{bmatrix} \partial_x v_{x}& \partial_x v_{y}&\partial_x v_{z}\\ \partial_y v_{x}& \partial_y v_{y}&\partial_y v_{z}\\ \partial_z v_{x}& \partial_z v_{y}&\partial_z v_{z} \end{bmatrix}$$ This matrix can be represented as a sum of an ...


0

(Subscripts $w,i,m$ correspond to water, insecticide, and mixture respectively.) First, $Q_m=Q_w+Q_i$ only if $\rho_w=\rho_i=\rho_m$. Otherwise you must equate mass flow rates, $\dot{M}_m=\dot{M}_w+\dot{M}_i$, to find $v_m$, assuming that mixture density is uniform over the cross-section at point 3. Second, the form of Bernoulli equation you have written ...


0

In principle you can't apply Bernoulli to what is in effect a (simple) network of pipes but in some cases approximations will do. Let's apply Bernoulli's equation to the left and middle sections of the pipe: $$P_1+\frac12 \rho v_1^2=P_2+\frac12 \rho v_2^2$$ As liquids are incompressible ($A$ is the cross-section of the pipe): $$A_1v_1=A_2v_2$$ So with ...


0

You can either measure $F$ and $v$ then calculate $C_D$ from the eqn, or you can look up $C_D$ in a table for the corresponding Reynolds number [1], and use that to calculate $F$. You cannot calculate both quantities using one equation. [1] http://www.thermopedia.com/content/707/


3

I've never been in a jacuzzi, but I'll try to imagine one. I read elsewhere that the bubbles are added using venturi-effect, or using a blower. Main thing is, they aren't pumped together with the water (the pump wouldn't like that anyway). The Venturi effect creates low pressure (suction) by increasing the velocity of the water in a narrow section of the ...



Top 50 recent answers are included