Question 10.4: Uniform air flow with speed U = 1.5 m/s approaches a cylinde...
Uniform air flow with speed U = 1.5 m/s approaches a cylinder as shown in Figure 10.13. The velocity distribution at the location shown downstream in the wake of the cylinder may be approximated by
u(y)=1.25+ \frac{y^2}{4} −1<y<1 ,
where u(y) is in m/s and y is in meters. Determine (a) the mass flux across the surface AB per meter of depth (into the page) and (b) the drag force per meter of length acting on the cylinder.
Given: Flow over cylinder; upstream and downstream velocity profiles.
Find: Mass flux across surface AB, drag force on cylinder.
Assume: Air has constant, uniform density 1.23 kg/m³. Flow is symmetrical and steady. Pressure differences, gravity, and viscous effects may be neglected.
We first select a control volume. It is generally wisest to choose control volumes on whose surfaces we have information about the flow. It is also useful to take advantage of symmetry to simplify our calculations.
Our choice of control volume is shown in Figure 10.14.
At its left surface, the normal velocity is U = 1.5 m/s, into the CV. Its top surface is a plane of symmetry for the flow, so there is no mass flux across it. At the right, the wake velocity profile is given by u(y) above, and outside the wake, the velocity is 1.5 m/s, out of the CV. We have a steady flow, so the conservation of mass is written as
0= \int\limits_{CS}{\rho \underline{V}\cdot \underline{n}dA}0= \int\limits_{CS}{\rho \underline{V}\cdot \underline{n}dA} + \int\limits_{AB}{\rho \underline{V}\cdot \underline{n}dA} + \int\limits_{\underset{CS}{right}}{\rho \underline{V}\cdot \underline{n}dA}
= – \rho UH+\dot{m}_{AB}+\int\limits_{0}^{H}{\rho u(y)dy} .
Note that since we are not given the length of the cylinder into the page, we must find the mass flux across AB per meter of cylinder length. We account for this by assuming a unit cylinder length. The areas of our control surfaces are thus ldy, with a unit length l. Hence, the area of the left control surface is simply H (m²).
H, however, is not known. To complete the solution we must investigate the flow field further. Outside the wake region, which is 1 m wide at the control surface, the flow out of the right CS has speed 1.5 m/s. The left control surface has a uniform inflow of 1.5 m/s. Hence, more than 1 m away from the cylinder axis, the flow is unaffected by the cylinder and simply proceeds with constant speed 1.5 m/s. We can therefore assess the amount of mass flux forced across AB by integrating only from 0 to 1, instead of 0 to H:
\dot{m}_{AB} = \rho U(1) – \int\limits_{0}^{1}{\rho (1.25 + \frac{y^2}{4})dy}
\dot{m}_{AB} = (1.23 \frac{\textrm{kg}}{\textrm{m}^3})\left\{1.5\frac{\textrm{m}^2}{s}-\left[1.25y+\frac{y^3}{12} \right]^{1}_{0}\right\}
\dot{m}_{AB} = 0.205 \frac{\textrm{kg}}{s} per meter of cylinder length .
To address part (b) of this problem, we conserve linear momentum in the x direction. We may either continue with the same control volume as in part (a), multiplying the fluxes by 2 to obtain the force on the whole cylinder, or we may now use a CV that consists of all the fluid between –H and +H, or equivalently –1 and 1. The drag force on the cylinder is in the +x direction; hence, there is an equal and opposite force on the fluid in the –x direction.
Conserving x momentum, we have
Under the assumptions of steady flow, with negligible contributions from pressure gradients, gravity, and viscous effects, this becomes
-F_x= \int\limits_{CS}{ρ u \underline{V}\cdot \underline{n} dA}We evaluate the flux at all three control surfaces of the initial CV and multiply each by 2 due to symmetry:
= -2 \int\limits_{0}^{1}{\rho U \cdot U 1 \cdot dy+2} \int\limits_{AB}{ρ u \underline{V}\cdot \underline{n} dA +2 } \int\limits_{0}^{1}{ \rho u(y) \cdot u(y) 1 \cdot dy} .
We have again assumed a unit cylinder length into the page so that the area of both right and left control surfaces is 1dy. We next recognize that the second integral contains the mass flux we just solved for,
\dot{m}_{AB}=\int\limits_{AB}{\rho \underline{V}\cdot \underline{n}dA } ,
and differs from this only by the value of u, the x component of velocity at the surface AB. The surface AB is at a distance of H from the cylinder axis, where, as we have discussed, the cylinder does not influence the x directional flow. The velocity u is therefore U = 1.5 m/s on AB. We thus get something even simpler:
-F_x = -2 ρ U^2 (1)^2 + 2U \dot{m}_{AB} + 2 \int\limits_{0}^{1}{ρ \left[1.25 + \frac{y^2}{4} \right]^2} 1\cdot dy-F_x = -2 \rho U^2 +2U\dot{m}_{AB}+2\rho \left[1.25^2 y +\frac{2.5}{12}y^3+\frac{y^4}{16}\right]^{1}_{0}
-F_x = -2(1.23 ^\textrm{kg}/_{\textrm{m}^3})(1.5 ^\textrm{m}/_{s})^2+2(1.5 ^\textrm{m}/_{s})(0.205 ^\textrm{kg}/_{s})+2(1.23 ^\textrm{kg}/_{\textrm{m}^3})(1.833 ^{\textrm{m}^2}/_{s^2})
F_x = 4.07 N per meter of cylinder length.
