Handbook of Civil Engineering Calculations

FIGURE 6

Figure 6b shows a cross section of the oil film, the shaded portion being an elemen-
tal surface. Let m = thickness of film; R = radius of plates; co = angular velocity of one
plate relative to the other; dA = area of elemental surface; dF = shearing force on ele-
mental surface; dT = torque of dF with respect to the axis through the center of the
plate.
Applying Eq. 12, develop these equations: dF = 27To^r^2 dr dO/m; dT= r dF = 27TO)IJLr^2
dr dSlm.

2. Integrate the foregoing equation to obtain the resulting torque;
   solve for \L
   Thus,

Tm

»

=

^

(13)

T= 0.25 ft-lb (0.339 N-m); m = 0.08 in (2.032 mm); w = 4 r/s; R = 4.5 in (114.3 mm); /x, =

0.25(0.08)(12)^3 /[7T^2 (4)(4,5)^4 ] = 0.00214 lb-s/ft^2 (0.1025 N-s/m^2 ).

APPLICATION OF BERNOULLI'S THEOREM

A steel pipe is discharging 10 ft^3 /s (283.1 L/s) of water. At section 1, the pipe diameter is

12 in (304.8 mm), the pressure is 18 lb/in^2 (124.11 kPa), and the elevation is 140 ft (42.67

m). At section 2, farther downstream, the pipe diameter is 8 in (203.2 mm), and the eleva-

tion is 106 ft (32.31 m). If there is a head loss of 9 ft (2.74 m) between these sections due

to pipe friction, what is the pressure at section 2?

Calculation Procedure:

1. Tabulate the given data
   Thus D 1 = 12 in (304.8 mm); D 2 = 8 in (203.2 mm); P 1 = 18 lb/in^2 (124.11 kPa); p 2 = ?;
   Z 1 - 140 ft (42.67 m); Z 2 = 106 ft (32.31 m).