STEM Problem Detail

Problem Statement

Prompt

Please give me step wise solution for Example 2.2-2.

Solution Rendering...

We designate the manometric liquid as the system to which we apply the macroscopic balances. In that case, there are no planes 1 and 2 through which liquid enters or exits. The free liquid surfaces are capable of performing work on the surroundings, $$W_m$$, and hence play the role of the moving mechanical parts in $$\S 7.4$$. We apply the mechanical energy balance of Eq. 7.4-2, with $$E_t$$ set equal to zero (since the manometer liquid is regarded as incompressible). Because of the choice of the system, both $$w_1$$ and $$w_2$$ are zero, so that the only terms on the right side are $$-W_m$$ and $$-E_v$$.

To evaluate $$dK_{tot}/dt$$ and $$E_v$$ it is necessary to make some assumption about the velocity profile. Here we take the velocity profile to be parabolic:

$$ v(r, t) = 2\langle v \rangle \left[ 1 - \left( \frac{r}{R} \right)^2 \right] $$ (7.7-14)

in which $$\langle v \rangle = dh/dt$$ is a function of time, defined to be positive when the flow is from left to right.

The kinetic energy term may then be evaluated as follows:

$$ \frac{dK_{tot}}{dt} = \frac{d}{dt} \int_0^L \int_0^{2\pi} \int_0^R \frac{1}{2} \rho v^2 r dr d\theta dl $$

$$ = 2\pi L \left( \frac{1}{2} \rho \right) \int_0^R v^2 r dr $$

$$ = 2\pi L R^2 \left( \frac{1}{2} \rho \right) \frac{d}{dt} \int_0^1 (2\langle v \rangle (1 - \xi^2))^2 \xi d\xi $$

$$ = \frac{4}{3} \rho LS \langle v \rangle \frac{d \langle v \rangle}{dt} $$ (7.7-15)

Here $$L$$ is a coordinate running along the axis of the manometer tube, and $$l$$ is the distance along this axis from one manometer interface to the other—that is, the total length of the manometer fluid. The dimensionless coordinate $$\xi$$ is $$r/R$$, and $$S$$ is the cross-sectional area of the tube.

The change of potential energy with time is given by

$$ \frac{d\Phi_{tot}}{dt} = \frac{d}{dt} \int_0^L \int_0^{2\pi} \int_0^R (\rho g z) r dr d\theta dl $$

$$ = \frac{d}{dt} \left[ \int_0^L (\text{integral over portion below } z=0, \text{ which is constant}) + \rho g S \int_0^H zdz + \rho g S \int_H^{H+h} zdz \right] $$

$$ = 2\rho g S h \frac{dh}{dt} $$ (7.7-16)

The viscous loss term can also be evaluated as follows:

$$ E_v = \int_0^L \int_0^{2\pi} \int_0^R \mu (\nabla v \cdot \nabla v) r dr d\theta dl $$

$$ = 2\pi L \mu \int_0^R \left( \frac{dv}{dr} \right)^2 r dr $$

$$ = 8\pi L \mu \langle v \rangle^2 \int_0^1 (-2\xi)^2 \xi d\xi $$

$$ = 8LS \mu \frac{\langle v \rangle^2}{R^2} $$ (7.7-17)

Furthermore, the net work done by the surroundings on the system is

$$ W_m = (p_a - p_b) S \langle v \rangle $$ (7.7-18)

Substitution of the above terms into the mechanical energy balance and letting $$\langle v \rangle = dh/dt$$ then gives the differential equation for $$h(t)$$ as

$$ \frac{d^2h}{dt^2} + \left( \frac{6\mu}{\rho R^2} \right) \frac{dh}{dt} + \left( \frac{3g}{2L} \right) h = \frac{3}{4} \frac{(p_a - p_b)}{\rho L} $$ (7.7-19)

which is to be solved with the initial conditions that $$h=0$$ and $$dh/dt = 0$$ at $$t=0$$. This second-order, linear, nonhomogeneous equation can be rendered homogeneous by introducing a new variable $$k$$ defined by

$$ k = 2h - \frac{p_a - p_b}{\rho L} $$ (7.7-20)

Then the equation for the motion of the manometer liquid is

$$ \frac{d^2k}{dt^2} + \left( \frac{6\mu}{\rho R^2} \right) \frac{dk}{dt} + \left( \frac{3g}{2L} \right) k = 0 $$ (7.7-21)

This equation also arises in describing the motion of a mass connected to a spring and dashpot as well as the current in an RLC circuit (see Eq. C.1-7).

We now try a solution of the form $$k = e^{mt}$$. Substituting this trial function into Eq. 7.7-21 shows that there are two admissible values for $$m$$:

$$ m_{\pm} = \frac{1}{2} \left( -\frac{6\mu}{\rho R^2} \right) \pm \sqrt{\left( \frac{6\mu}{\rho R^2} \right)^2 - \left( \frac{6g}{L} \right)} $$ (7.7-22)

and the solution is

$$ k = C_1 e^{m_+ t} + C_2 e^{m_- t} \quad \text{when } m_+ \neq m_- $$ (7.7-23)

$$ k = C_1 e^{m t} + C_2 t e^{m t} \quad \text{when } m_+ = m_- = m $$ (7.7-24)

with the constants being determined by the initial conditions.

The type of motion that the manometer liquid exhibits depends on the value of the discriminant in Eq. 7.7-22:

(a) If $$(6\mu / \rho R^2)^2 > (6g/L)$$, the system is overdamped, and the liquid moves slowly to its final position.

(b) If $$(6\mu / \rho R^2)^2 < (6g/L)$$, the system is underdamped, and the liquid oscillates about its final position, the oscillations becoming smaller and smaller.

(c) If $$(6\mu / \rho R^2)^2 = (6g/L)$$, the system is critically damped, and the liquid moves to its final position in the most rapid monotone fashion.

The tube radius for critical damping is then

$$ R_{cr} = \left( \frac{6\mu^2 L}{ \rho^2 g} \right)^{1/4} $$ (7.7-25)

If the tube radius $$R$$ is greater than $$R_{cr}$$ an oscillatory motion occurs.

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Problem #20