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Problem 1
Determine the pressure, both absolute and gauge, exerted at the bottom of the column of liquid 1 meter high, with density of 1200 kg / m3
Solution
Pgauge = (density of liquid) X ( acceleration due to gravity) X (height of liquid column)
Pgauge = ( 1200 kg/m3 ) X ( 9.7 m/s2 ) X ( 1 N/kg/m/s2)
Pgauge = 11.760 kN/m2
Pgauge = 11.760 kPa
Pabsolute = Pgauge + Patmospheric
Pabsolute = 11.760 kPa + 101.3 kPa
Pabsolute = 113.06 kPa
EQUATION
OF STATE
The relationship between the three state variables may be written as:
For a perfect gas,
where,
The specific gas constant for dry air, R d = 0.287 Joules / gm /oK,
and for water vapor, R v = 0.461 Joules / gm /oK.
The specific gas constant for wet air will not be constant quantity but
will depend on mixing ratio.
Problem 2
Calculate the density of a gas with a molecular weight of 29 @ 1 atm(absolute) and 80 oF. Gas constant, R = 0.7302 ft3atm/lb-moleoR
Solution
Absolute Temperature = 80 oF + 460 = 540 oR
Density = P ( molecular weight) / RT
Density = ( 1atm. ) . (29 lb/lb mole) / ( 0.7302 ft3atm/lb-moleoR). (540 oR)
Density = 0.073546 lb/ ft3
Solve this problem online using applet DENS 1.0.
First
Law of Thermodynamics
The first law of thermodynamics is based on the law of conservation of total energy of a thermodynamic system which undergoes a change of state. Mathematically one can write it as:
Heat added per unit mass = (Change in internal energy per unit mass)
+ (work done by a unit mass)
Second
Law of thermodynamics
There are several different statements for the second law of thermodynamics in the physics books. In summary, the law can be stated as "no cyclic process exists having the transference of heat from a colder to hotter body as its sole effect".
Specific
Heats
The specific heat is defined as the amount of heat needed to change the temperature of unit mass by 1oK. The specific heat can be computed at the constant volume or at constant pressure. The definitions are:
Specific Heat at Constant Volume
Specific Heat at Constant Pressure
The Carnot's Law provides a relationship between Cp and Cv. For a perfect gas,
C p - C v = R
For a perfect diatomic gas ( approximation to dry air), Cp and Cv can be obtained from,
C p = (7/2) . R, and
C v = (5/2) . R
For dry air, the value of
C pd = 1.003 Joules / gm / oK, and
C vd = 0.717 Joules / gm / oK
The ratio of Cp, and Cv is given by
=
Cp / Cv = 1.4 (for dry air). The value of
for dry air is 1.4. The value of
will be different for different gases.