TITLE.PM5

BASIC CONCEPTS OF THERMODYNAMICS 43

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M-therm/th2-2.pm5

Example 2.4. A vacuum recorded in the condenser of a steam power plant is 740 mm of

Hg. Find the absolute pressure in the condenser in Pa. The barometric reading is 760 mm of Hg.

Solution. Vacuum recorded in the condenser = 740 mm of Hg

Barometric reading = 760 mm of Hg

We know that,

Absolute pressure in the condenser

= Barometric reading – vacuum in the condenser

= 760 – 740 = 20 mm of Hg

= 20 × 133.4 N/m^2 (Q 1 mm of Hg 133.4 N / m )=^2

= 2668 N/m^2 = 2668 Pa.(Ans.)

Example 2.5. A vessel of cylindrical shape is 50 cm in diameter and 75 cm high. It con-

tains 4 kg of a gas. The pressure measured with manometer indicates 620 mm of Hg above

atmosphere when barometer reads 760 mm of Hg. Determine :

(i)The absolute pressure of the gas in the vessel in bar.

(ii)Specific volume and density of the gas.

Solution. Diameter of the vessel, d = 50 cm ( = 0.5 m)

Height of the vessel, h = 75 cm ( = 0.75 m)

Mass of gas in the vessel, m = 4 kg

Manometer reading = 620 mm of Hg above atmosphere

Barometer reading = 760 mm of Hg

Now, volume of the vessel = π 4 d^2 × h = π 4 × (0.5)^2 × (0.75) = 0.147 m^3.

(i)Total pressure in the vessel

= 760 + 620 = 1380 mm of Hg

= 1380 × 133.4 N/m^2 [1mmofHQ g=133.4 N / m ]^2

= 1.841 × 10^5 N/m^2 = 1.841 bar.(Ans.) [Q 1 bar = 10^5 N/m^2 ]

(ii) Specific volume =

0.147

4

= 0.03675 m^3 /kg.(Ans.)

Density =^4

0.147

= 27.21 kg/m^3 .(Ans.)

Example 2.6. In a pipe line the pressure of gas is measured with a mercury manometer

having one limb open to the atmosphere (Fig. 2.25). If the difference in the height of mercury in

the two limbs is 550 mm, calculate the gas pressure.

Given : Barometric reading = 761 mm of Hg

Acceleration due to gravity = 9.79 m/s^2

Density of mercury = 13640 kg/m^3.

Solution. At the plane LM, we have

p = p 0 + ρgh

Now p 0 = ρgh 0

where h 0 = barometric height ; ρ = density of mercury ; p 0 = atmospheric pressure

Therefore, p = ρgh 0 + ρgh = ρg (h 0 + h)