Homework #13 ME 363 - Fluid Mechanics

Homework #13 ME 363 - Fluid Mechanics

Due Friday, May 2, 2008 Spring Semester 2008

1] Water is to be withdrawn from a 3-m-high water reservoir by drilling a 1.5-cm-diameter hole in the bottom surface. Disregarding the effect of the kinetic energy correction factor (α in equation 8-29), determine the flow rate in [m3/s] of water through the hole if (a) the entrance of the hole is well-rounded and (b) the entrance is sharp-edged.

2] Water at 20 degrees C is to be pumped from reservoir A (zA = 2 m) to reservoir B at a higher elevation (zB = 9 m) through two 25-m-long smooth pipes connected in parallel. The diameters of the two pipes are 3 cm and 5 cm. Water is to be pumped by a 68-percent-efficient motor-pump unit that draws 7 kW of electrical power during operation. The minor losses and the head loss in the pipes that connect the parallel pipes to the two reservoirs are considered to be negligible. Determine the flow rates through each of the parallel pipes in [m3/s].

3] A clothes dryer discharges air at 1 atm and 120 degrees F at a rate of 1.2 ft3/s when its 5-in-diameter, well-rounded vent with negligible loss is not connected to any duct. Determine the flow rate when the vent is connected to a 15-ft-long, 5-in-diameter duct made of galvanized iron (e = 0.0005 ft) with three 90-degree flanged smooth bends (K = 0.3). Do not forget to include the exit loss. Take the friction factor f of the duct to be 0.019, and assume the fan power input to remain constant.

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Problem#1 / Problem #2 / Problem #3

Problem 1 Solution

Solution Water is to be withdrawn from a water reservoir by drilling a hole at the bottom surface. The flow rate of water through the hole is to be determined for the well-rounded and sharp-edged entrance cases.

Assumptions 1 The flow is steady and incompressible. 2 The reservoir is open to the atmosphere so that the pressure is atmospheric pressure at the free surface. 3 The effect of the kinetic energy correction factor is disregarded, and thus a = 1.

Analysis The loss coefficient is KL = 0.5 for the sharp-edged entrance, and KL = 0.03 for the well-rounded entrance. We take point 1 at the free surface of the reservoir and point 2 at the exit of the hole. We also take the reference level at the exit of the hole (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is zero (V1 = 0), the energy equation for a control volume between these two points (in terms of heads) simplifies to

where the head loss is expressed as . Substituting and solving for V2 gives

since a2 = 1. Note that in the special case of KL = 0, it reduces to the Toricelli equation , as expected. Then the volume flow rate becomes

Substituting the numerical values, the flow rate for both cases are determined to be

Well-rounded entrance:

Sharp-edged entrance:

Discussion The flow rate in the case of frictionless flow (KL = 0) is 1.36´10-3 m3/s. Note that the frictional losses cause the flow rate to decrease by 1.5% for well-rounded entrance, and 18.5% for the sharp-edged entrance.

Problem 2

Solution The pumping power input to a piping system with two parallel pipes between two reservoirs is given. The flow rates are to be determined.

Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully developed. 3 The elevations of the reservoirs remain constant. 4 The minor losses and the head loss in pipes other than the parallel pipes are said to be negligible. 5 The flows through both pipes are turbulent (to be verified).

Properties The density and dynamic viscosity of water at 20°C are r = 998 kg/m3 and m = 1.002´10-3 kg/m×s. Plastic pipes are smooth, and their roughness is zero, e = 0.

Analysis This problem cannot be solved directly since the velocities (or flow rates) in the pipes are not known. Therefore, we would normally use a trial-and-error approach here. However, nowadays the equation solvers such as EES are widely available, and thus below we will simply set up the equations to be solved by an equation solver. The head supplied by the pump to the fluid is determined from

We choose points A and B at the free surfaces of the two reservoirs. Noting that the fluid at both points is open to the atmosphere (and thus PA = PB = Patm) and that the fluid velocities at both points are zero (VA = VB =0), the energy equation for a control volume between these two points simplifies to

or

(2)

where

We designate the 3-cm diameter pipe by 1 and the 5-cm diameter pipe by 2. The average velocity, Reynolds number, friction factor, and the head loss in each pipe are expressed as

(9)

(10

This is a system of 13 equations in 13 unknowns, and their simultaneous solution by an equation solver gives

,

V1 = 5.30 m/s, V2 = 7.42 m/s, , hpump,u = 26.5 m

Re1 = 158,300, Re2 = 369,700, f1 = 0.0164, f2 = 0.0139

Note that Re > 4000 for both pipes, and thus the assumption of turbulent flow is verified.

Discussion This problem can also be solved by using an iterative approach, but it will be very time consuming. Equation solvers such as EES are invaluable for this kind of problems.

Problem 3