College Of Engineering

| College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 390Fluid Mechanics |
| Spring 2008 Number: 11971 Instructor: Larry Caretto |

Solutions to Exercise Nine – Flow in Pipes
1 A 70-ft-long, 0.5-in-diameter hose with a roughness of  = 0.00009 ft is fastened to a water faucet where the pressure is p1. Determine p1 if there isno nozzle attached and the average velocity in the hose is 6 ft/s. Neglect minor losses and elevation changes.
We start by computing the Reynolds number for this flow to determine if the flow is laminar or turbulent. We use the properties for water from Table 1.6:  = 1.21x10-5 ft2/s and  = 1.94 slugs/ft3.

Therefore the flow is turbulent and we can find the friction factor,f, from the Moody diagram on page 434. The relative roughness of the pipe, /D = (0.00009 ft) / (0.5/12 ft) = 0.0216. This value of /D and Re = 2.07x104 gives f = 0.03. Checking this with the Colebrook equation gives the following value.

The pressure drop in the 70-ft length of hose can now be found.

We assume that the hose flow exits to the atmosphere where the (gage) pressure iszero so the pressure drop computed above is the same as the pressure, p1, at the water faucet. Dividing the result by 144 to get psi gives p1 = 12.2 psi.
2 Blood (assume  = 4.5x10-5 lbfs/ft2 and SG = 1.0) flows through an artery in the neck of a giraffe from its heart to its head at a rate of 2.5x10-4 ft3/s. Assume that the length of the artery is 10 ft and that its diameter is0.20 in. If the pressure at the beginning of the artery (outlet of the heart) is equivalent to 0.70 ft Hg, determine the pressure at the end of the artery when the head is (a) 8 ft above the heart or (b) 6 ft below the heart. Assume steady flow. How much of this pressure difference is due to elevation effects and how much is due to friction?
We can write the head loss in the energyequation as p/ = p/g =[ f (l/D) V2/2 ] / (g) = f(l/D)V2/2g so that the energy can be written to give the following relationship between the end position in the giraffe’s head (2) and the inlet at the heart (1).

Since the area is assumed constant and the blood is assumed to have constant density, the velocity at all points in the artery is the same. Thus we can cancel theV12 and V22 terms and write the following equation for the pressure difference between the head (p2) and the heart (p1).

To compute the friction factor we first compute the Reynolds number to determine if the flow is laminar or turbulent. Note that a specific gravity of 1 gives a density  = 1.94 slug/ft3 and a specific weight of 62.4 lbf/ft3 for the blood. The diameter of 0.20 in= 0.01667 ft. giving a cross sectional area of D2/4 = (0.01667 ft)2/4 = 0.000218 ft2. We can compute the velocity from the flow rate: V = Q/A = (2.5x10-4 ft3/s) / (0.000218 ft2) = 1.146 ft/s.

Thus, the flow is laminar so f = 64/Re = 64/823 = 0.0777. We can now compute the head loss term.

The pressure at the start of the artery is given as equivalentto 0.70 ft of Hg. Multiplying this height by the specific weight of mercury, Hg = 847 lbf/ft3, from Table 1.5 gives p1 = 593 lbf/ft2. Substituting this value for p1 and the value of 0.9517 ft found for the head loss into the energy equation gives.

When the head is 8 ft above the heart, z2 – z1 = 8 ft and the value of p2 is
=
When the head is6 ft below the heart, z1 – z2 = 6 ft and the value of p2 is
=
In both cases, the head loss (due to frictional effects) is (0.9517 ft) = (0.9517 ft)(62.4 lbf/ft3) = 59.4 lbf/ft2. When the head is above the heart the initial pressure of 593 lbf /ft2 is reduced by 59.4 lbf/ft2 by frictional effects and an additional (8 ft)(62.4lbf/ft2) = 499 lbf/ft2 by the elevation...