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Properties-of-Working-Substance

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THERMODYNAMICS
Thermodynamics – is that branch of the physical sciences that
treats of various phenomena of energy
and the related properties of matter,
especially of the laws of transformation of
heat into other forms of energy and vice
versa.
a. Pure substance - a working substance whose chemical
composition remains the same even if
there is a change in phase. (most
common example is water)
b. Ideal gas - a working substance which remains in gaseous
state during its operating cycle and whose
equation of state is PV = mRT. (most common
example is air)
Newton's law states that 'the acceleration of
a particular body is directly proportional to the
resultant force acting on it and inversely
proportional to its mass.”
a=
𝑘𝐹
,
𝑚
F=
𝑚𝑎
,
𝑘
k is a proportionality constant
k=
𝑚𝑎
𝐹
System of units where k is unity but not dimensionless:
cgs system: 1 dyne force accelerates 1 gram at 1 cm/s2
mks system: 1 newton force accelerates 1 kg mass at 1 m/s2
fps system: 1 lb force accelerates 1 slug mass at 1 ft/s2
System of units where k is not unity:
If the same word is used for both mass and force in a
given system, k is either unity or dimensionless:
1 lb force accelerates a 1 lb mass at 32.174 ft/s2
1 g force accelerates a 1 g mass at 980.66 cm/s2
1 kg force accelerates a 1 kg mass at 9.8066 m/s2
The mass of body is absolute quantity of matter in it.
The weight of the body means the force of gravity Fg on
the body.
𝑚 𝐹 𝐹𝑔
= =
𝑘 𝑎
𝑔
where:
g = acceleration produced by force Fg
a = acceleration produced by another force F
At or near the surface of the earth, k and g are
numerically equal, so are m and Fg.
Sample Problems
1. What is the weight of a 66-kgm man at standard condition?
2. The weight of an object is 50 lb. What is its mass at
standard condition?
Properties of Working Substance:
Formulas:
1. Pressure - is the ratio of force per unit area.
Gage pressure – is the pressure reading from the gage
pressure instrument which is higher or
lower than the atmospheric pressure.
Vacuum pressure – is a pressure that is less than
atmospheric pressure.
a. Pressure =
𝑭𝒐𝒓𝒄𝒆
, KN/m2, lb/in2
𝑨𝒓𝒆𝒂
𝑭
𝑷=
𝑨
Absolute pressure(Pabs) = Gauge pressure + Atmospheric Pressure
Pabs = Pg + Patm
b. 1 atm = 0 Kpag
= 101.325 Kpa
= 14.7 psi
= 29.92 in Hg
= 760 mm Hg
= 1.033 kg/ cm2
c. Hydrostatic Pressure, P:
P
h
𝑷 = 𝒘𝒉
where:
𝑤 = density of fluid
= (𝑆đē đ‘Ĩ 𝑤𝑤)
ℎ = height of fluid
𝑤𝑤 = density of water
= 1000 kg/m3 = 9.81 KN/m3
= 62.4 lb/ft3 = 1 kg/li
2. Temperature - is the degree of hotness or coldness of a substance or
body.
Thermometer - is an instrument used to measure the temperature
of a body or a substance.
Pyrometer - is an instrument used to measure high temperature
gases.
a. Relation between ī‚°C and ī‚°F scales
𝟓
ī‚°C = (oF-32)
𝟗
𝟗
ī‚°F = đ‘Ē + 𝟑𝟐
𝟓
b. Absolute temperatures K and R:
K = ī‚°C + 273
R = ī‚°F + 460
c. Temperature Difference:
ī„ī‚°C = 5/9 ī„ī‚°F
ī„ī‚°F = 9/5 ī„ī‚°C
ī„ī‚°C = ī„K
ī„ī‚°F = ī„R
3. Density, Specific Volume and Specific Gravity:
𝒎 𝟏
a. 𝒘 =
=
đ‘Ŋ 𝒗
𝐕
𝟏
=
b. 𝒗 =
đĻ 𝐰
𝐃𝐞𝐧đŦđĸ𝐭𝐲 𝐨𝐟 𝐚𝐧𝐲 𝐟đĨ𝐮đĸ𝐝
c. đ‘ē𝑮 =
𝐃𝐞𝐧đŦđĸ𝐭𝐲 𝐨𝐟 𝐰𝐚𝐭𝐞đĢ
D. If two or more fluids are mixed together:
𝜮𝒎 𝒎𝟏 + 𝒎𝟐 + ⋯
𝑾𝒎 =
=
𝜮đ‘Ŋ
đ‘Ŋ𝟏 + đ‘Ŋ𝟐 + ⋯
𝒎𝟏 𝒎𝟐
+
+⋯
𝟏
𝜮đ‘Ŋ
đ‘Ŋ𝟏 + đ‘Ŋ 𝟐 + ⋯
𝑾𝟏 𝑾𝟐
𝒗𝒎 =
=
=
=
𝑾𝒎 𝜮𝒎 𝒎𝟏 + 𝒎𝟐 + ⋯ 𝒎𝟏 + 𝒎𝟐 + ⋯
𝑾𝒎
đ‘ē𝑮𝒎 =
𝑾𝒘
Where:
𝑾𝒎 = 𝑑𝑒𝑛𝑠𝑖𝑡đ‘Ļ 𝑜𝑓 𝑚𝑖đ‘Ĩ𝑡đ‘ĸ𝑟𝑒
𝒗𝒎 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 đ‘Ŗ𝑜𝑙đ‘ĸ𝑚𝑒 𝑜𝑓 𝑚𝑖đ‘Ĩ𝑡đ‘ĸ𝑟𝑒
đ‘ē𝑮𝒎 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎đ‘Ŗ𝑖𝑡đ‘Ļ 𝑜𝑓 𝑚𝑖đ‘Ĩ𝑡đ‘ĸ𝑟𝑒
Sample Problems
1. Find the pressure at the 100 fathom depth of water in kpag.
2. Convert 750°R to °K.
3. An iron block weighs 5 N and has a volume of 200 cm3.
What is the density of the block?
4. A storage tank contains oil with specific gravity of 0.88
and depth of 20 m. What is the hydrostatic pressure at the
bottom of the tank in kg/cm2.
4. Internal Energy(U) - heat energy due to the movement of
molecules within the system that is
brought about its temperature.
5. Flow Work,Wf = Pressure x Specific Volume
𝑲𝑱
𝑾𝒇 = 𝑷 𝒗𝟐 − 𝒗𝟏 ,
𝑲𝒈
𝑾𝒇 = 𝑷 đ‘Ŋ𝟐 − đ‘Ŋ𝟏 , 𝑲𝑱
where: P = pressure, KPA
đ‘Ŋ = volume, m3
𝒗 = specific volume, m3/Kg
6. Enthalpy, h:
𝒉= Internal Energy + Flow Work
𝒉 = đ‘ŧ + 𝑷đ‘Ŋ
𝒉 = đ‘ŧ + (𝑷₂đ‘Ŋ₂ − 𝑷₁đ‘Ŋ₁)
7. Stagnation enthalpy, hs:
đ‘Ŋ𝒐 𝟐 𝑲𝑱
𝒉𝒔 = 𝒉𝒐 +
,
𝟐𝟎𝟎𝟎 𝒌𝒈
where:
𝒗𝒐 = initial velocity, m/s
𝒉𝒐 = ordinary enthalpy, KJ/kg
Sample Problems
1. 100 g of water are mixed with150 g of alcohol ( W= 790
kg/ m3). What is the specific volume of the resulting mixture,
assuming the fluids mixed completely?
2. Steam flows through a nozzle at 400 oC and 1 Mpa ( h= 3263.9
KJ/kg) with velocity of 300 m/s. Find the stagnation enthalpy.
3. Given the steam pressure of 900 lb/ft2 , the temperature of 300
oF, specific volume of 5.8 ft3/lb. If the specific enthalpy is 9500 ftlb/lb, what is the internal energy per lb of the system?
8. Entropy, s:
𝒅𝑸
āļą 𝒅𝒔 = āļą
đ‘ģ
𝑸
𝒔=
đ‘ģ
General Flow Equations:
1. Volume flow rate, (𝑸) 𝒎𝟑/𝒔𝒆𝒄:
𝑸 = 𝑨𝒓𝒆𝒂 𝒙 đ‘Ŋ𝒆𝒍𝒐𝒄𝒊𝒕𝒚
= 𝑨𝒙đ‘Ŋ
where:
𝝅
𝟒
A= đ‘Ģ𝟐 (𝑭𝒐𝒓 𝒄𝒊𝒓𝒄𝒖𝒍𝒂𝒓 𝒄𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏)
𝑨 = đ‘ŗ 𝒙 𝒘 (𝒇𝒐𝒓 𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒖𝒍𝒂𝒓)
2. Mass flow rate (m), kg/sec:
𝐐
𝒎=
𝒗
where:
𝒗 = specific volume
đ‘Ŋ= volume
3. If Q is constant:
𝑸𝟏 = 𝑸 𝟐
𝑨𝟏 đ‘Ŋ 𝟏 = 𝑨𝟐 đ‘Ŋ 𝟐
𝝅
𝝅
𝟐
đ‘Ģ𝟏 đ‘Ŋ𝟏 =
đ‘Ģ𝟐 𝟐 đ‘Ŋ𝟐
𝟒
𝟒
đ‘Ģ𝟏 𝟐 đ‘Ŋ𝟏 = đ‘Ģ𝟐 𝟐 đ‘Ŋ𝟐
Where:
𝑨𝟏
𝑨𝟐
đ‘Ŋ𝟏
đ‘Ŋ𝟐
= 𝑖𝑛𝑙𝑒𝑡 𝑎𝑟𝑒𝑎
= 𝑒đ‘Ĩ𝑖𝑡 𝑎𝑟𝑒𝑎
= 𝑖𝑛𝑙𝑒𝑡 đ‘Ŗ𝑒𝑙𝑜𝑐𝑖𝑡đ‘Ļ
= 𝑒đ‘Ĩ𝑖𝑡 đ‘Ŗ𝑒𝑙𝑜𝑐𝑖𝑡đ‘Ļ
Sample Problems
1. Air enters a nozzle steadily at 2.21 kg/m3 and 30 m/s.
What is the mass flow rate through the nozzle if the inlet
area of the nozzle is 80 cm2?
2. A pipe has a diameter of 4” at section AA, and a
diameter of 2” at section BB. For an ideal fluid flow, the
velocity is given as 1 ft/sec at section AA. What is the
flow velocity at section BB?
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