Advanced Thermodynamics: Lecture 3 Shivasubramanian Gopalakrishnan - - PowerPoint PPT Presentation

Advanced Thermodynamics: Lecture 3

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Moving Boundary Work: Closed systems

Moving boundary work is the primary form of work involved in automobile engines. During their expansion, the combustion gases force the piston to move, which in turn forces the crankshaft to rotate. Also called PdV work.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Moving Boundary Work: Closed systems

We analyze the moving boundary work for a quasi- equilibrium process,

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition

The initial pressure of the gas is P, the total volume is V, and the cross- sectional area of the piston is A. If the piston is allowed to move a distance ds in a quasi-equilibrium manner, the differential work done during this process is δWb = Fds = PAds = PdV

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Moving Boundary Work: Closed systems

The total boundary work done during the entire process as the piston moves is obtained by adding all the differential works from the initial state to the final state

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition

Wb = Area under curve = A = Z 2

1

dA = Z 2

1

PdV

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Work: Path Function

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Work: Cycle

The net work done during a cycle is the difference between the work done by the system and the work done on the system.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Boundary Work for constant volume process

A rigid tank contains air at 500 kPa and 150oC. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65oC and 400 kPa, respectively. Determine the boundary work done during this process.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Boundary Work for constant pressure process

A frictionless pistoncylinder device contains 10 lbm of steam at 60 psia and 320oF. Heat is now transferred to the steam until the temperature reaches 400oF. If the piston is not attached to a shaft and its mass is con- stant, determine the work done by the steam during this process.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Polytropic process

For a polytropic process the pressure and volume are related by the relation PV n = C Where n and C are constants.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Work done during a Polytropic process

Wb = Z 2

1

PdV = Z 2

1

CV −ndV = C V −n+1

2

V −n+1

1

n + 1 For and ideal gas PV = mRT, this equation becomes Wb = mR(T2 T1) 1 n for n 6= 1 For special case of n = 1, boundary work is Wb = Z 2

1

PdV = Z 2

1

CV −1dV = Cln ✓V2 V1 ◆

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Isothermal compression of an Ideal gas

A pistoncylinder device initially contains 0.4 m3 of air at 100 kPa and 80oC. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Expansion of a Gas against a Spring

A pistoncylinder device contains 0.05 m3 of a gas initially at 200

• kPa. At this state, a linear spring that has a spring constant of 150

kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, determine (a) the final pressure inside the cylinder, (b) the total work done by the gas, and (c) the fraction of this work done against the spring to compress it.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Unrestrained Expansion of Water

A rigid tank is divided into two equal parts by a partition. Initially,

• ne side of the tank contains 5 kg of water at 200 kPa and 25OC,

and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25OC. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Specific heats

The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. It depends on how the process is executed.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Specific heats

The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. It depends on how the process is executed. Two kinds of specific heats: specific heat at constant volume cv and specific heat at constant pressure cp. cv can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. The energy required to do the same as the pressure is maintained constant is the specific heat at constant pressure cp.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Consider a fixed mass in a stationary closed system undergoing a constant–volume process. The conservation of energy principle for this process can be expressed in the differential form as δein δeout = du The left–hand side of this equation represents the net amount of energy transferred to the system. From the definition of cv, this energy must be equal to cvdT, where dT is the differential change in temperature. cvdT = du at constant volume cv = ✓ ∂u ∂T ◆

v

cv is defined as the change in the internal energy of a substance per unit change in temperature at constant volume.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Similarly for a constant pressure process, the energy balance can be written as δein δeout = dh From the definition of cp, this energy must be equal to cpdT, where dT is the differential change in temperature. cpdT = dh at constant pressure cp = ✓ ∂h ∂T ◆

p

cp is defined as the change in the internal energy of a substance per unit change in temperature at constant pressure.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Joule’s experiment

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition

u = u(T) For an ideal gas the internal energy is a function of temperature alone.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Using the definition of enthalpy and the equation of state of an ideal gas, we have h = u + Pv Pv = RT h = u + RT ∴ h = h(T) Since u and h depend only on temperature for an ideal gas, the specific heats cv and cp also depend, at most, on temperature

• nly. Then the differential changes in the internal energy and

enthalpy of an ideal gas can be expressed as du = cv(T)dT dh = cp(T)dT

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by ∆u = u2 u1 = Z 2

1

cv(T)dT ∆h = h2 h1 = Z 2

1

cp(T)dT

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by ∆u = u2 u1 = Z 2

1

cv(T)dT ∆h = h2 h1 = Z 2

1

cp(T)dT If variation of specific heats are smooth and approximately linear, the integrations can be replaced by u2 u1 = cv,avg(T2 T1) h2 h1 = cp,avg(T2 T1)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Specific Heat Relations of Ideal Gases

Differentiating the relation, h = u + RT yields dh = du + RdT Replacing dh by cpdT and du by cvdT and dividing the resulting expression by dT, we obtain cp = cv + R ✓ kJ kg · K ◆ in Molar basis ¯ cp = ¯ cv + Ru ✓ kJ kmol · K ◆

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

We introduce another ideal gas property called specific heat ratio k defined as k = cp cv For monatomic gases, its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Evaluation of the ∆u of an Ideal Gas

Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy of air per unit mass, using the functional form of the specific heat ¯ cP(T) = a + bT + cT 2dT 3 where a = 28.11, b = 0.1967 ⇥ 10−2,c = 0.4802 ⇥ 10−5 and d = 1.966 ⇥ 10−9

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Heating of a Gas in a Tank by Stirring

An insulated rigid tank initially contains 1.5 lbm of helium at 80F and 50 psia. A paddle wheel with a power rating of 0.02 hp is

• perated within the tank for 30 min. Determine (a) the final

temperature and (b) the final pressure of the helium gas.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Heating of a Gas by a resistance heater

A piston–cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27C. An electric heater within the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120-V

• source. Nitrogen expands at constant pressure, and a heat loss of

2800 J occurs during the process. Determine the final temperature

• f nitrogen.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Heating of a Gas at Constant Pressure

A pistoncylinder device initially contains air at 150 kPa and 27C. At this state, the piston is resting on a pair of stops and the enclosed volume is 400 L. The mass of the piston is such that a 350-kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat transferred to the air.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Specific heats for solids and liquids

The specific heats of incompressible substances depend on temperature only. Enthalpy change for a solid or an incompressible liquid. h = u + Pv dh = du + vdP +

• *0

Pdv ∆h = ∆u + v∆P ⇡ cavg∆T + v∆P Constant–pressure processes, as in heaters (∆P = 0) ∆h = ∆u = cavg∆T Constant–temperature processes, as in pump (∆T = 0) ∆h = v∆P

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661