Holooly Plus Logo
Holooly Plus Logo

Question 8.4: Water flows steadily up a short, vertical, 2.54-cm-diameter ...

Water flows steadily up a short, vertical, 2.54-cm-diameter pipe and discharges to atmospheric pressure (Fig. 8.8). If a pressure of 16 kPa drives the fluid at a volumetric flow rate Q of 5 L/min, what height does the fluid reach?FIGURE 8.8 Flow from a vertical tube/pump that discharges to atmosphere. Because of the influence of gravity, fluid particles will rise to a particular height and then fall.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Given:

z_{1}=0 A_{1}=\pi (1.27)^{2}=5.07 cm^{2}
z_{2}=hm Q=5\frac{L}{min}
g=9.81\frac{m} {s^{2}} p_{1}=16KPa
\rho =1000\frac{kg}{m^{3}}. p_{2}=0(gauge)

Assume:
1. Incompressible
2. Inviscid
3. Steady flow (given)
4. Along a streamline (given)
5. Constant gravitational forces

Moreover, let us assume that the velocity \nu _{2}=0 at the maximum height of the fluid column. Hence, from mass balance

                                 Q=\nu _{1}A_{1}\rightarrow \nu _{1}=\frac{Q}{A_{1}}=\frac{5L/\min }{5.07  cm^{2}}\left(\frac{1000  cm^{3}}{1 L} \right)=986\frac{cm}{\min }

or \nu _{1}=0.164  m/s. Hence, from Bernoulli, Fundamental Balance Relations

                            \frac{p_{1}}{\rho }+gz_{1}+\frac{\nu ^{2}_{1} }{2}=\frac{p_{2}}{\rho }+gz_{2}+\frac{\nu ^{2}_{2} }{2}\rightarrow \frac{p_{1}}{\rho }+\frac{1}{2}\nu ^{2}_{1}=gh,
or
        h=\frac{p_{1}}{\rho g}+\frac{1}{2g}\nu ^{2}_{1}=\frac{16000N/m^{2}}{(1000  kg/m^{3})(9.81m/s^{2})}+\frac{(0.164m/s^{2})}{2(9.81m/s^{2})}=1.64m.

Note that \rho g is sometimes called the specific weight and denoted by \gamma, not to be confused with the specific gravity      SG=\rho /\rho H_{2O}    at   4  ^{\circ }C. Given that 1 kPa=7.5 mmHg, what might this suggest with regard to how far blood might travel if an open needle (having a different diameter) were placed in the heart?

Related Answered Questions

Evaluate the pressure difference between points A and B in Fig. 8.11. Assume aortic values such that v at A is 0.15 m/s, that the diameter at A is 0.03 m, and that the diameter at B is 0.01 m. Assume that ρ=1,060 kg/m^3.

Verified Answer:
Although Bernoulli should not be used across a sud...

Note that Bernoulli and mass balance provide two equations:p1/ρ+gz1+1/2ν2^1=p2/ρ+gz2+1/2ν2^2, ν1A1=ν2A2, which can be used to solve for the two velocities, v1 and v2, along a straight horizontal streamline s in a steady, converging, ideal flow, with A1 and A2 known. To do so, however, we must

Verified Answer:
Because we do not know \nu as a fun...

Determine the time tf it takes for a cylindrical container with a small central hole to drain. Fundamental Balance Relations FIGURE 8.13 Fluid draining from a reservoir through a centrally located bottom hole.

Verified Answer:
Referring to Fig. 8.13, let us consider a streamli...

Under what conditions can you compute the pressure in the system in Fig. 8.15? Recall that streamlines must be parallel and straight, where the radius of curvature is infinity, in order for ∂p/∂n=0. Consider multiple possibilities.

Verified Answer:
For flow along a streamline between points 2 and 4...

A forced vortex flow is given by v=rωoeθ, where ωo is constant (Fig. 8.16). Determine if Bernoulli’s equation can be used to determine the pressure differ-ence between two radial locations. Ignore gravity.

Verified Answer:
For \nu _{r}\hat{e}_{r}+\nu _{\theta }\hat{...

Is the following velocity field a possible incompressible flow? ν(x,y,z,t)=ρg sin θ/μ(yh-y^2/2)i where ρ is the mass density of the fluid, g (=9.81 m/s^2) is the gravitational constant, μ is the viscosity of the fluid, θ is some fixed angle (number) relative to a horizontal datum, and h is some depth of a fluid film.

Verified Answer:
\triangledown \cdot \nu =\frac{\partial}{\p...

Consider a container of fluid at rest with a Cartesian coordinate defined as positive downward and the origin located at the surface of the fluid (Fig. 8.3). Find the hydrostatic pressure p at the depth h.

Verified Answer:
From Eqs. (8.36), (8.40), and (8.42), with ...

It can be shown experimentally that Bernoulli’s equation can be used in computations for flows through constrictions but not for flows through expansions. The reason for this is that in the latter case, adverse pressure gradients can disturb the flow such that there is a reversal and thus

Verified Answer:
Assuming that we know the cross-sectional area wit...