fundamentals of mechanical engineering 6.pdfduring a constant volume process, the specific internal...
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Fundamentals of mechanical engineering
Mohammed abass ali
polytropic Process
• In practice, it is found that many processes
approximate to a reversible law in the form
• pVn = constant
• where n is a constant. Vapours and perfect gases obey this type of law closely in many non-flow processes. Such processes are internally reversible. If a piston in a cylinder is cooled perfectly and a compression or expansion is carried out slowly, the process will be isothermal where n = 1.
• If a compression is carried out rapidly and again the piston and cylinder are perfectly insulated, the process will be adiabatic and in this case the constant n = γ.
• From Figure above depicts the polytropic
process.
• Work Transfer
• Referring to Figure above, there is an increase in
volume during the process and as the fluid
expands, the expansion work is given by
• Heat Transfer
• The energy balance is applied to this case as
• EXAMPLE
• Following compression, the combustion gases in a
petrol engine are at 35 bar and 900°C. The gases
then expand through a volume ratio (V2/V1) of
8.5/1 and occupy 0.51 × 10−3 m3 after expansion.
The polytropic expansion index n = 1.15 when the
engine is air cooled. Calculate the temperature and
pressure of the gas after expansion and establish
what the work output will be?
• Solution
• From the question, p1 = 30 bar, t1 = 900°C (T1 =
900 + 273 = 1173 K), V2 = 0.51 × 10−3 m3, n = 1.15 and V2/V1 = 8.5.
• Treating the air as a perfect gas, for a polytropic
process the property relationship is given by
• Equation
constant Volume Process
• In certain chemical processes, fluids are held in
fixed volume rigid-walled vessels whilst the fluid
or gas is either heated or cooled as shown in
Figure below. In this case, the process is
considered a constant volume process as the
vessel has a fixed volume. The general property
relation between the initial and final states of a
perfect gas is applied as
• V1=V2
• Work Transfer
• Work transfer (p dV) will be zero as the change in
volume (dV) during the process will also be zero.
• Heat Transfer
• Applying the non-flow energy equation from
• This result is important and shows that the net
amount of heat energy supplied to or taken from
the fluid during a constant volume process is
equal to the change in the internal energy of the
fluid.
• EXAMPLE
• During a constant volume process, the specific internal energy of a fluid is increased from 120 kJ/ kg to 180 kJ/kg. Calculate the amount of heat energy supplied to 2 kg of fluid to increase the internal energy.
• Solution
• From the non-flow energy equation:
• Q – W = U2 – U1
• For a constant volume process:
• W = 0
• Therefore, the equation becomes
• Q = U2 – U1
• = (180 – 120) kJ/kg = 60 kJ/kg
• For a mass of 2 kg of fluid:
• Q = 60 × 2 = 120 kJ
Constant Pressure Process
• Intensifiers are used in hydraulic and gas supply systems to maintain a constant pressure within the system, even though the flow may be varying due to usage. These consist of having a cylinder fitted with a piston that has a constant load applied to it as depicted in Figure below. This is an example of a constant pressure process.
• The general property relation between the initial and final states of a perfect gas is applied as
• On a p–V diagram, the area under the process
line represents the amount of work transfer.
From Figure above
• W = Area of the shaded rectangle
• = Height × width
• = p(V2 – V1)
• Heat Transfer
• Again applying the non-flow energy equation
• Q – W = U2 – U1
• or
• Q = (U2 – U1) + W
• Part of the heat supplied is converted into work energy and the remainder is utilised in increasing the internal energy of the system.
• Q = (U2 – U1) + p(V2 – V1)
• = U2 – U1 + p2V2 – p1V1 (since p2 = p1)
• = (U2 + p2V2) – (U1 + p1V1)
• Now
• H = U + pV
• Q = H2 – H1
• EXAMPLE
• A cylinder contains a fluid with a volume of 0.1 m3 at a constant pressure of 7 bar and having a specific enthalpy of 210 kJ/kg. The volume expands to 0.2 m3 following the application of heat energy to the fluid and the specific enthalpy increases to 280 kJ/kg, m = 2.25 kg.
• Determine:
• 1. The quantity of heat energy supplied to the fluid.
• 2. The change in the internal energy of the fluid.
• Solution
• From the question, p = 7.0 bar, V1 = 0.1 m3 and
V2 = 0.2 m3.
• 1. Heat energy supplied = change in enthalpy of
the fluid: