Солнечная электростанция 30кВт - бизнес под ключ за 27000$

15.08.2018 Солнце в сеть




Производство оборудования и технологии
Рубрики

Specific Weight and Density

Specific Weight and Density

Throughout this book a distinction is made between force (F) and mass (m). and specific weight (7) and density (p) because such distinctions are essential for arriving at the correct results. Weight is not the same as mass. A body of mass m is accelerated towards the center of the earth by gravity g with a force of magnitude mg. This force mg is defined as the weight of the body having mass m.

Specific weight 7 is defined as the weight per unit volume. Density p is defined as the mass per unit volume. Thus, the relation between the density and specific weight is:

7 = P9

EXAMPLE 1:

Given the situation in Fig. D. l, determine the resultant force on the blade.

A = 0.1 sq ft V,= 130 ft/s

I

SPECIFIC WEIGHT OF MUD (MUD BALANCE READING) = 70 Ib/cu ft

Fig. D. l: Diagram of mud jet striking blade. Example 1.

Weight rate of flow = area x velocity x specific weight = constant ^l^l7l = ^2^272

For most purposes, liquids are considered to be incompressible and, therefore, the specific weight 7 is a constant.

The rate of discharge of the mud flow in Fig. D. l is:

Q = volume rate of flow.

= A1V1 = 0.1 x 130

= 13 ft3/s

TOC o "1-5" h z The blade in the path of the jet deflects only that portion of the total discharge which overtakes it. If Q is the discharge in Fig. D. l. the discharge Q’ overtaking the blade is:

a, a /Vi-v) /130-30

Q — Q x —=— = 13 x

Vx j V 130

= 10 ft3/s

If blades are in a row, then Q = Q’. The fluid velocity relative to the blade at the entrance is (Vj — v). The exit value of the fluid is the vector sum of (Vj — v) and v. Thus, the magnitudes of the force components of the blade on the fluid are:

= pQ{y2x-vXl)

= pQ [(^u — «) cost? — (Vjx — r)]

70

= 13 x — x [(130 — 30) cos в — (130 — 30)] = -378.62 lbf

i. e., summation of forces in the x-direction is equal to the volumetric rate of flow (Q in ft3/sec) times the density of fluid (p) times the change in velocity (V in ft/sec) in the x-direction (x-component of the final velocity, V2x, minus the x-component of the initial velocity, Vx). In this problem,

70 Ib/cuft о it 1 / r

P = 32.2 ft/sec/sec = 2Л7 sluSs/cuft

= +Fy

= pQ (y2y — Ky)

= [(Й., — v) sin# — o]

70

= 13 x ^ x K130 — 30)sin 0 — °)1

ij z.. z

= 1413.04 lbf

i. e., summation of forces in the (/-direction is equal to the volumetric rate of flow (Q in ft3/sec) times the density of fluid (p) times the change in velocity (V in ft/sec) in the (/-direction (//-component of the final velocity. V2y — minus the //-component of the initial velocity, Viy).

Thus the resultant force is:

fr =

= ^/(—378.62)2 + (1413.04)2 = 1,463 lbf

Clearly the distinction between specific weight and density is important in order to obtain the correct result.

In many flow problems it is necessary to know the specific weight of a fluid. If the specific gravity (SG) of the fluid is known, the specific weight (-,/) in lb/ft3 can be calculated:

7/ = SG x 62.4 (D. l)

(Specific weight of water (7Ш) at 60°F is equal to 62.4 lb/ft3.)

EXAMPLE 2:

Turpentine has a specific gravity of 0.87 © 60 °F. If the weight of one cubic foot of pure water is assumed to be 62.4 lbf, then how much would one cubic foot of turpentine weigh?

Solution:

By simple substitution into Eq. D. l:

[1] Y„

In the elastic range, a formula similar to Eq. 2.136 has been employed. The constant in the numerator has been set equal to a higher value on the basis of the results of 147 tests. The minimum for the collapse resistance includes 99.5 ‘/c of the measured data and amounts to 75.6 c/c of the average test results, or 72.7 Tt of the theoretical values:

[2] J final

[3] For the buildup section, from 7.000 ft to 6.463 ft (bottom part of buildup section):

[4] Load calculation criteria are not mentioned. The entire design may not meet the design demands or, on the contrary, may exceed the loading require­ments and result in expensive, over-designed, combination casing strings.

• Limited in use to a given casing diameter.

• Limited in use to a given mud weight.

• Limited to vertical wells.

• If many manufacturers are considered, problems will arise when non-API

casing is selected.

Оставить комментарий