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1.1 What is the value of  gc and what are its units in a system in which the second, the foot,and the pound mass are defined as in Sec. 1.2, and the unit of force is the poundal, defined as theforce required to give l(lbm) an acceleration of l(ft)(s)-2?

 Answer

= 1  

1 () = 1 (1 )(1 ) 

= 1 ∙

∙ 

This system of units is the English-system equivalent to the SI system.

1.5  Pressures up to 3000 bar are measured with a dead-weight gauge. The piston diameteris 4 mm. What is the approximate mass in kg of the weights required?

 Answer

The pressure in a dead weight gauge is given by the formula:

=   

In an ideal dead-weight gauge, the force (F) is exerted only by the weights (the product of massand the local gravity) and A is the transversal area, replacing both F and A in the last equationgives.

=

4 

4 =   4 = (.1) 

Beforehand, a conversion of units is necessary

3000 = 10 1 =310 = 310(

) 

4 = 110−1 =410− 

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Replacing the given values in Eq. 1

310

∙(410−)

4∙9.807 = 384.41  

1.6 Pressures up to 3000 atm are measured with a dead-weight gauge. The piston diameteris 0.17 (in). What is the approximate mass in (lbm) of the weights required?

 Answer

This problem is similar to exercise 1.6; the same equation can be applied.

4 =  

Beforehand, a conversion of units is necessary

3000 = 101325 1 =303975000  

0.17 =2.54

1 1

100 = 0.004318  

Replacing the given values

303975000∙ (0.004318)4∙9.807 = 453.89  In Lbm

453.89 ∙ 1 0,45359 =1000,6757  

1.7 The reading on a mercury manometer at 298.15 K (25°C) (open to the atmosphere atone end) is 56.38 cm. The local acceleration of gravity is 9.832 m/s2. Atmospheric pressure is101.78 kPa. What is the absolute pressure in kPa being measured? The density of mercury at298.15 K (25°C) is 13.534 g / cm3

 Answer

The gauge pressure measured by the mercury manometer is given by the equation:

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= ℎ = 13.534 ∙1

1000 ∙100

1 ∙9.832 ∙ 56.38 ∙

1100 = 75022,77  

The absolute pressure being measured is the addition of gauge and atmospheric pressure

= + = 75,02 + 101,78 = 176,8  

1.10  the first accurate measurements of the properties of high-pressure gases were made by

E.H. Amagat in france between 1869 and 1893. Before developing the dead-weight

gauge. He worked in mine shaft and used a mercury manometer for measurements of

 pressure to more than 400 bar. Estimate the height of manometer requires?

Answer

ρ = 13.5 gr/cm3

g = 9.8 m/s2

  p = 400 bar

Ditanya : h?

h =P

ρ . g  =

13.5 . 9.8= 302.3 m

1.14 A 70 W outdoor security light burns, on average, 10 hours a day. A new bulb costs $5.00,and the lifetime is about 1000 hours. If electricity costs $0.10 per kWh, what is the yearly price

of "security," per light?

 Answer

Assuming there are 365 days per year and that the security light will work every day, the totalworking hours are 3650. The yearly cost per light is the sum of the cost of consumed electricity(variable) and the cost of buying new bulbs as necessary (fixed).

The fixed cost is:

36501000 ≅ 4 → 4 ∙ $5.00 = $20.00*The number of total bulbs must be rounded up as you can’t buy a fraction of a bulb. 

The variable cost is:

0.1 $ ∙ ℎ ∙ 3650ℎ ∙ 0.07 = $25,55 The total cost is:

=$20.00+$25.55= $45.55  

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1.15 A gas is confined in n 1.25 (ft)- diameter cylinder by a piston, on which rests a

weight. The mass of the piston and weight together in 250 (lb m). The local

acceleration of gravity is 32.169 (ft)(s)-2. The atmospheric pressure is 30.12 (in Hg).

(a) what is the force in (lbf ) exerted on the gas by atmosphere, the piston and the

weight, assuming no friction between the piston and cylinder?

(b) what is the pressure of the gas in (psia)

(c) if the gas in the cylinder is heated, it expands, pushing the piston and weight

upward. If the piston and weight are raised 1.7 (ft), what is the work done by

gas in (ft)? what is the change in potential energy of the piston and weight?  

D = 1.25 ft g = 32.169 ft/s2 

Patm  = 30.12 in Hg A = π . DMass = 250 lbm A = .  . (1.25)2 = 1.227 ft2 Jawab

(a) F = Patm . A + mass . g

= 30.12 . 1.227 + 250 . 32.169

= 2.857 x 103 lbf

(b) Pabs =FA 

=.

= 16.166 psia(c) Δl = 1.7 ft

Work = F . Δl 

= 2.857 x 103 . 1.7

= 4.857 x 103 

ΔPE = mass . g . Δl 

= 250 . 32.169 . 4.857 x 103 

= 424.933 ft.lbf