Polytropic Process
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Calculate total work done in a polytropic process.
In a polytropic process both heat and work transfers take place. It follows the equation
\[Pv^{n} = c\]
where n is called the index of process or polytropic index.
Work done, \[W = \dfrac{p_1 V_1 - p_2 V_2}{n-1} = \dfrac{mR(T_1- T_2)}{n-1}\] Change in internal energy, \[\Delta U = mC_v (T_2- T_1)\] \[Q = \left(\dfrac{n - \gamma}{n - 1}\right) mC_V (T_2 - T_1)\] Depending upon various values of n, there will be following cases:
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n = 0, constant pressure (Isobaric) process
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n = 1, constant temperature (Isothermal) process
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n = \(\gamma\) , reversible adiabatic process
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n = \(\infty\) ,constant volume (Isochoric) process
Solved Example:
72-6-01
If the value of n = 0 in the equation pv$^n$ = C, then the process is called: (AFCAT EKT Mech Paper I- Set 02/2016)
Solved Example:
72-6-02
The value of n = 1 in the polytropic process indicates it to be: (UPSSSC JE Mech 2016 Paper 2)
Solved Example:
72-6-03
If value of n is infinitely large in a polytropic process pV$^n$ = C, then the process is known as constant: (UPPSC AE Mech 2019)
Solved Example:
72-6-04
During polytropic process:
Solved Example:
72-6-05
A fluid at 0.7 bar occupying 0.09 m$^3$ is compressed reversibly to a pressure of 3.5 bar according to pv$^n$ = constant. The fluid is then heated reversibly at constant volume until the pressure is 4 bar; the specific volume is then 0.5 m$^3$/kg. A reversible expansion according to the law pv$^{1.6}$ = constant restores the fluid to its initial state. Calculate the net work of the cycle.
Solved Example:
72-6-06
A mass of gas at initial pressure of 30 bar, and with an internal energy of 1500kJ, is contained in a well-insulated cylinder of volume 0.1m$^3$. The gas is allowed to expand behind a piston until its internal energy is 1300 kJ, the law of expansion is pv$^2$ = constant. Calculate the final pressure.
Solved Example:
72-6-07
A fluid at a pressure of 3 bar and with specific volume of 0.18m$^3$/kg, contained in a cylinder behind a piston expands reversibly to a pressure of 0.6 bar according to the law p = $\dfrac{C}{V^2}$, where C is a constant. The work done by the fluid on the piston is:
Solved Example:
72-6-08
Properties of substances like pressure, temperature and density, in thermodynamic coordinates are:
Solved Example:
72-6-09
Which of the following quantities is not the property of the system: (SJVNL JE Mech 2018)
Solved Example:
72-6-10
According to Avogadro's law, for a given pressure and temperature, each molecule of a gas:
Solved Example:
72-6-11
Which of the following is not the intensive property? (SSC JE ME Paper 11- March 2021 Morning)
Solved Example:
72-6-12
Which of the following items is not a path function: (BPSC AE NE 2012- Part 5)
Solved Example:
72-6-13
Heat and work are: (GATE ME 2011)
Solved Example:
72-6-14
Which of the following parameters is constant for a mole for most of the gases at a given temperature and pressure:
Solved Example:
72-6-15
For an ideal gas with constant values of specific heats, for calculation of the specific enthalpy: (GATE ME 2015- Shift I)
Solved Example:
72-6-16
The area under the temperature-entropy curve (T-s curve) of any thermodynamic process represents: (Based on JKSSB JE ME 2015)
Solved Example:
72-6-17
A mixture of gas expands from 0.03 m$^3$ to 0.06 m$^3$ at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is: (ISRO RAC 2018)
Solved Example:
72-6-18
Heating wet steam at constant temperature is heating it at constant:
Solved Example:
72-6-19
Intensive property of a system is one whose value:
Solved Example:
72-6-20
The index of compression n tends to reach ratio of specific heats $\gamma$ when:
Solved Example:
72-6-21
Change in enthalpy of a system is the heat supplied at: (UPSSSC JE Mech 2016- Paper 2)
Solved Example:
72-6-22
The term N.T.P. stands for:
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72-6-23
In an isothermal process, the internal energy of gas molecules: (ISRO VSSC Tech Asst. Mech Feb 2015)
Solved Example:
72-6-24
Work done is zero for the following process:
Solved Example:
72-6-25
In a non-flow reversible process for which p = (- 3V+ 15) $\times$ 10$^5$ N/m$^2$,V changes from 1 to 2 m$^3$. The work done will be about: (ESE Mech 2013- Paper 1)
Solved Example:
72-6-26
In a free expansion process:
Solved Example:
72-6-27
If a fluid expands suddenly into vacuum through an orifice of large dimension, then such a process is called:
Solved Example:
72-6-28
Which of the following processes is irreversible process: (VIZAG MT Mech 2017)
Solved Example:
72-6-29
For a thermodynamic process to be reversible, the temperature difference between hot body and working substance should be:
Solved Example:
72-6-30
Minimum work in compressor is possible when the value of adiabatic index n is equal to:
Solved Example:
72-6-31
A gas is compressed in a cylinder by a movable piston to a volume one-half its original volume. During the process 300 kJ heat left the gas and internal energy remained same. The work done on gas in Nm will be: (TANGEDCO AE EC 2018)
Solved Example:
72-6-32
For reversible adiabatic process, change in entropy is: (BPSC AE ME 2019- Part 5)
Solved Example:
72-6-33
Change in enthalpy in a closed system is equal to heat transferred if the reversible process takes place at constant: (UPPSC AE Mech 2019- Paper II)