Constant Pressure Process MCQ Quiz in मराठी - Objective Question with Answer for Constant Pressure Process - मोफत PDF डाउनलोड करा

Explanation:

Euler Work

• The work done by the fluid during an isentropic flow through a turbine stage with perfect guidance by its blades is the maximum work that can be expected in the stage.
• In such a case the deflection of the fluid through the rotor provides ideal values of the tangential velocity components c_θ1 and c_θ2.
• The value of the work given by w_T = u₁c_θ1 - u₂c_θ2 is the ideal work with perfect guidance by the blades. this is known as Euler's work

Isentropic work

• The deflection of the fluid in an actual turbine stage with the isentropic flow of fluid is not the same as dictated by its blades.
• This is due to the imperfect guidance given by the blades to the fluid.
• The inability of the blade passages to provide sufficient constraint on the flow leads to a lesser pressure drop and work in the stage.
• This will happen regardless of the fact whether the flow is reversible or irreversible.
• The ratio of isentropic work to Euler work is known as the Pressure coefficient

Concept:

Apply the first law of thermodynamics:

\( \Delta U = Q - W \)

Given:

• Initial volume: \( V_1 = 0.3~m^3 \)
• Final volume: \( V_2 = 0.15~m^3 \)
• Pressure: \( P = 0.105~MPa = 105~kPa \)
• Heat rejected: \( Q = -37.6~kJ \)

Step 1: Work done

\( W = P (V_2 - V_1) = 105 \times (-0.15) = -15.75~kJ \)

Step 2: Change in internal energy

\( \Delta U = -37.6 - (-15.75) = -21.85~kJ \)

Concept:

Exergy is the maximum useful work possible when a system is brought to equilibrium with its surroundings.

For an ideal gas, the specific exergy (availability per unit mass) is given by:

\( e = (h - h_0) - T_0 (s - s_0) \)

For an ideal gas with constant specific heats, the entropy change is:

\( s - s_0 = C_p \ln \frac{T}{T_0} - R \ln \frac{P}{P_0} \)

For a rigid closed tank, volume remains constant, and thus:

\( h - h_0 = 0 \) (since no change in enthalpy in a rigid tank)

Therefore, specific exergy simplifies to:

\( e = - T_0 \left( C_p \ln \frac{T}{T_0} - R \ln \frac{P}{P_0} \right) \)

Using the relation for an ideal gas at constant volume:

\( \frac{T}{T_0} = \frac{P}{P_0} \)

Substituting this into the equation:

\( e = R T_0 \left[ \frac{P_0}{P} - 1 + \ln \frac{P}{P_0} \right] \)

Explanation:

Properties

All measurable characteristics of a system are known as properties.

Eg. Pressure, volume, temperature, internal energy, density etc.

There are two types of properties:

Extensive property

• Those properties which depend on mass are known as extensive properties.
• Examples are volume, energy, enthalpy, etc.

Intensive property

• Those properties which don't depend on mass are known as intensive properties.
• Examples are pressure, temperature, density, viscosity.

Important Points

• The ratio of two extensive properties is an intensive property.
• Specific properties are intensive properties. For example specific volume, specific energy etc.
• If a property divides with space then it is extensive property otherwise the property will be intensive.
• Property is a point function.
• Property is independent of past history.
• In a cycle change in property is equal to zero.
• Properties are exact differential.

Note: Heat, work, and entropy generation are path functions, they are not properties.