Introduction to Desalination. Louis TheodoreЧитать онлайн книгу.

the problem be restated or simplified?

4 If I cannot solve the problem, can I solve a related problem?

5 Can part of the problem be solved? What data are lacking to solve the rest of the problem? Did I use all of the data given?

3.8.3 Carrying Out the Plan

Once a plan for the solution has been developed, carry out the plan. Choose a basis of calculation, if applicable. During various stages in the solution of a complex problem it may be necessary to use more than one basis. Also remember that the most brilliant work is useless if it cannot be conveyed clearly to other persons by written or oral presentation.

3.8.4 Looking Back and Checking the Problem Solution

The amount of time it takes to check a problem is usually only a small fraction of the time it takes to solve it. Nearly all solutions to physical problems are amendable to verification. Thus, one should always:

1 Look for order-of-magnitude errors. Can the answer be checked by an approximate solution? Does the answer make sense?

2 Attempt to solve the problem by an alternate method.

3 Can the results be verified?

3.9 Illustrative Examples

Five illustrative examples complement the material presented in this chapter.

3.9.1 Illustrative Example 1

Convert the following temperatures.

1 20°F to °C, K, and °R

Solution. The following key equations are employed:

upper T left-parenthesis degree upper F right-parenthesis equals 1.8 upper T left-parenthesis degree upper C right-parenthesis plus 32 (3.5)

upper T left-parenthesis normal upper K right-parenthesis equals upper T left-parenthesis degree upper C right-parenthesis plus 273 (3.6)

upper T left-parenthesis degree upper R right-parenthesis equals upper T left-parenthesis degree upper F right-parenthesis plus 460 (3.7)

upper T left-parenthesis degree upper R right-parenthesis equals l .8 upper T left-parenthesis normal upper K right-parenthesis (3.8)

Thus:

Equations 3.5, 3.6, and 3.7 are used as follows:

left-parenthesis 20 degree upper F right-parenthesis equals 1 .8 upper T left-parenthesis degree upper C right-parenthesis plus 32 semicolon left-parenthesis 20 degree upper F hyphen 32 right-parenthesis slash 1 .8 equals upper T left-parenthesis degree upper C right-parenthesis equals hyphen 6 .7 degree upper C

upper T left-parenthesis normal upper K right-parenthesis equals left-parenthesis hyphen 6 .7 degree upper C right-parenthesis plus 273 equals 266 .3 upper K

upper T left-parenthesis degree upper R right-parenthesis equals left-parenthesis 20 degree upper F right-parenthesis plus 460 equals 480 degree upper R

3.9.2 Illustrative Example 2

The height of a liquid column of mercury is 2.493 ft. Assume the density of mercury is 848.7 lb/ft³ and atmospheric pressure is 2116 lb_f/ft² absolute. Calculate the gauge pressure in lb_f/ft² and the absolute pressure in lb_f/ft², psia, mm Hg, and in H₂O.

Solution. Expressed in various units, the standard atmosphere is equal to:

1.0	Atmospheres (atm)
33.91	Feet of water (ft H₂O)
14.7	Pounds force per square inch absolute (psia)
2116	Pounds force per square foot absolute (psfa)
29.92	Inches of mercury (in Hg)
760.0	Millimeters of mercury (mm Hg)
1.013 x 10⁵	Newtons per square meter (N/m²)

The density equation is describing the gauge pressure in terms of the column height and liquid density is:

upper P Subscript g Baseline equals rho g h slash g Subscript c (3.9)

where P_g = gauge pressure, ρ = liquid density, g = acceleration of gravity, h = column height, and g_c = conversion constant. Thus,

upper P Subscript g Baseline equals left-parenthesis 848.7 l b slash f t cubed right-parenthesis left-parenthesis 1 StartFraction l b Subscript f Baseline Over l b EndFraction right-parenthesis left-parenthesis 2.493 f t right-parenthesis equals 2116 StartFraction l b Subscript f Baseline Over f t squared EndFraction gauge

The pressure in lb_f/ft² absolute is:

upper P Subscript a b s o l u t e Baseline equals upper P Subscript g Baseline plus upper P Subscript a t m o s p h e r i c Baseline equals 2,116 lb Subscript normal f Baseline slash ft squared plus 2,116 lb Subscript normal f Baseline slash ft squared equals 4,232 lb Subscript normal f Baseline slash ft squared absolute

The pressure in psia is;

upper P left-parenthesis psia right-parenthesis equals left-parenthesis 4,232 lb Subscript normal f Baseline slash ft squared right-parenthesis left-parenthesis 1 ft squared slash 144 in squared right-parenthesis equals 29 .4 psia

The corresponding gauge pressure in psi is:

upper P left-parenthesis psig right-parenthesis equals 29 .4 hyphen 14 .7 equals 14 .7 psig

The pressure in mm Hg is:

upper P left-parenthesis mm Hg right-parenthesis equals left-parenthesis 29 .4 psia right-parenthesis left-parenthesis 760 mm Hg right-parenthesis slash left-parenthesis 14 .7 psia right-parenthesis equals 1,520 mm Hg

Finally, the pressure in in H₂O is:

upper P left-parenthesis in upper H Subscript 2 Baseline normal upper O right-parenthesis equals left-parenthesis StartFraction 29.4 psia Over 14.7 psia slash atm EndFraction right-parenthesis left-parenthesis StartFraction 33.91 ft upper H Subscript 2 Baseline normal upper O Over 1 atm EndFraction right-parenthesis left-parenthesis StartFraction 12 in Over 1 ft EndFraction right-parenthesis equals 813.8 in upper H Subscript 2 Baseline normal upper O

The reader should note that absolute and gauge pressures are usually expressed with units of atm, psi, or mm Hg. This statement also applies to partial pressures. One of the most common units employed to describe pressure drop is inches of H₂O, with the notation in H₂O or IWC (inches of water column).

3.9.3 Illustrative Example 3

Given the

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