Constant Pressure Process MCQ Quiz in मल्याळम - Objective Question with Answer for Constant Pressure Process - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 10, 2025
Latest Constant Pressure Process MCQ Objective Questions
Top Constant Pressure Process MCQ Objective Questions
Constant Pressure Process Question 1:
In a reversible, constant-pressure, non-flow process, heat input is given by:
Answer (Detailed Solution Below)
Constant Pressure Process Question 1 Detailed Solution
Explanation:
In a reversible constant pressure process, the heat input is the change in enthalpy.
According to the first law of thermodynamics:
δQ = δW + ΔU
For isobaric Process: δW = PdV = P(V2 – V1)
Heat, δQ = δW + ΔU = mcpΔT = dh
So, heat added at constant pressure is equal to change in Enthalpy and it not only increases the temperature (i.e. internal energy) but also does the work.
Non-Flow Processes
- These are compression and expansion processes on gases in a cylinder with complete leak proof. In these, there is only energy transfer with zero mass transfer.
- These nonflow processes can be the followings:
- constant pressure process
- constant volume process
- constant temperature process
- reversible adiabatic process
- poly-tropic process
- constant internal energy process
Important Points
Isochoric process means volume is constant while all other variables change.
As volume is kept constant therefore no work is done on or by the gas.
Heat absorbed by the gas is completely used to change its internal energy and its temperature.
From First law of Thermodynamics
δQ= ΔU + δW ⇒ δQ = ΔU
so, heat added at constant volume is equal to change in internal energy.
Constant Pressure Process Question 2:
The ratio of isentropic work to Euler work is called-
Answer (Detailed Solution Below)
Constant Pressure Process Question 2 Detailed Solution
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 cθ1 and cθ2.
- The value of the work given by wT = u1cθ1 - u2cθ2 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
Constant Pressure Process Question 3:
Air (an ideal gas) is stored in a rigid closed tank of volume V at absolute temperature of T0 and pressure P. Ignoring the effect of motion and gravity, what will be the specific exergy (availability per unit mass) of air? [Given: Environmental absolute temperature is T0; Environmental pressure is P0; Gas constant is R.]
Answer (Detailed Solution Below)
Constant Pressure Process Question 3 Detailed Solution
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:
\( e = R T_0 \left[ \frac{P_0}{P} - 1 + \ln \frac{P}{P_0} \right] \)
Constant Pressure Process Question 4:
Which of the following is the property of a system?
Answer (Detailed Solution Below)
Constant Pressure Process Question 4 Detailed Solution
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.