Fundamentals of mechanical engineering

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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: