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Chapter 1 - Section B - Non-Numerical Solutions

1.1   This system of units is the English-system equivalent of SI. Thus,

gc  = 1(lbm)(ft)(poundal)−1(s)−2

1.2   (a) Power is power , electrical included. Thus,

Power [=]energy

time[=]

N·m

s[=]

kg·m2

s3

(b) Electric current is by definition the time rate of transfer of electrical charge. Thus

Charge [=] (electric current)(time) [=] A·s

(c) Since power is given by the product of current and electric potential, then

Electric potential [=]power

current[=]

kg·m2

A·s3

(d ) Since (by Ohm’s Law) current is electric potential divided by resistance,

Resistance [=]electric potential

current[=]

kg·m2

A2·s3

(e) Since electric potential is electric charge divided by electric capacitance,

Capacitance [=]charge

electric potential[=]

A2·s4

kg·m2

1.3  The following are general:

ln x  = ln10 × log10 x   ( A)

P  sat/kPa =   P  sat/torr × 100750.061

kPatorr

( B)

t /◦C = T /K − 273.15   (C )

By Eqs. ( B) and ( A),

ln P  sat/kPa = ln10 × log10  P sat/torr + ln

100

750.061

The given equation for log10  P sat/torr is:

log10 P sat/torr = a −

b

t /◦C + c

Combining these last two equations with Eq. (C ) gives:

ln P  sat/kPa  =   ln10

a −

b

T /K − 273.15 + c

+ ln

100

750.061

=   2.3026

a −

b

T /K − 273.15 + c

− 2.0150

Comparing this equation with the given equation for ln  P  sat/kPa shows that:

 A = 2.3026 a − 2.0150   B  = 2.3026 b C  = c − 273.15

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1.9   Reasons result from the fact that a spherical container has the minimum surface area for a given interior

volume. Therefore:

(a) A minimum quantity of metal is required for tank construction.

(b) The tensile stress within the tank wall is everywhere uniform, with no sites of stress concentration.

Moreover, the maximum stress within the tank wall is kept to a minimum.(c) The surface area that must be insulated against heat transfer by solar radiation is minimized.

1.17   Kinetic energy as given by Eq. (1.5) has units of mass·velocity2. Its fundamental units are therefore:

 E K  [=] kg·m2·s−2 [=] N·m [=] J

Potential energy as given by Eq. (1.7) has units of mass·length·acceleration. Its fundamental units are

therefore:

 E P  [=] kg·m·m·s−2 [=] N·m [=] J

1.20  See Table A.1, p. 652, of text.

•   1(atm) ≈ 1 bar = 1/0.986923 = 1.01325 bar•   1(Btu) ≈ 1 kJ = 1/0.947831 = 1.05504 kJ

•   1(hp) ≈ 0.75 kW = 1/1.34102 = 0.745701 kW

•   1(in) ≈ 2.5 cm = 2.54 cm exactly, by definition (see p. 651 of text)

•   1(lbm) ≈ 0.5 kg = 0.45359237 kg exactly, by definition (see p. 651 of text)

•   1(mile) ≈ 1.6 km = 5280/3280.84 = 1.60934 km

•   1(quart) ≈ 1 liter = 1000/(264.172 × 4) = 0.94635 liter (1 liter ≡ 1000 cm3)

•   1(yard) ≈ 1 m = (0.0254)(36) = 0.9144 m exactly, by definition of the (in) and the (yard)

An additional item could be:

•   1(mile)(hr)−1≈ 0.5 m s−1= (5280/3.28084)(1/3600) = 0.44704 m s−1

1.21   One procedure here, which gives results that are internally consistent, though not exact, is to assume:

1 Year [=] 1 Yr [=] 364 Days

This makes 1 Year equivalent to exactly 52 7-Day Weeks. Then the  average Month contains 30 13

  Days

and 4 13

 Weeks. With this understanding,

1 Year [=] 1 Yr [=] 364 Days [=] (364)(24)(3600) = 31,449,600 Seconds

Whence,

•   1 Sc [=] 31.4496 Second 1 Second [=] 0.031797 Sc•   1 Mn [=] 314.496 Second 1 Minute [=] 60 Second [=] 0.19078 Mn

•   1 Hr [=] 3144.96 Second 1 Hour [=] 3600 Second [=] 1.14469 Hr

•   1 Dy [=] 31449.6 Second 1 Day [=] (24)(3600) Second [=] 2.74725 Dy

•   1 Wk [=] 314496. Second 1 Week [=] (7)(24)(3600) Second [=] 1.92308 Wk 

•   1 Mo [=] 3144960 Second 1 Month [=] (4 13

)(7)(24)(3600) Second[=] 0.83333 Mo

The final item is obviously also the ratio 10/12.

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