3b. 1st law of thermodynamics closed systems · first law of thermodynamics in closed systems...

First Law of ThermodynamicsClosed Systems

Content

• The First Law of Thermodynamics• Energy Balance• Energy Change of a System• Mechanisms of Energy Transfer• First Law of Thermodynamics in Closed Systems• Moving Boundary Work

The First Law of Thermodynamics

• The first law of thermodynamics, also known as the conservation of energy principle, provides a sound basis for studying the relationships among the various forms of energy and energy interactions.

• Based on experimental observations, the first law of thermodynamics states that energy can be neither created nor destroyed during a process; it can only change forms

• Therefore, every bit of energy should be accounted for during a process.

The First Law of Thermodynamics

Energy Balance

• The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process. That is,

or∆

• This relation is often referred to as the energy balance and is applicable to any kind of system undergoing any kind of process.

Energy Change of a System

• The determination of the energy change of a system during a process involves the evaluation of the energy of the system at the beginning and at the end of the process, and taking their difference. That is,

Energy change = Energy at final state ‐ Energy at initial stateor

Esystem = Efinal ‐ Einitial = E2 ‐ E1• Note that energy is a property, and the value of a property does not change unless the state of the system changes.

• Therefore, the energy change of a system is zero if the state of the system does not change during the process.


• Energy can exist in numerous forms such as internal (sensible, latent, chemical, and nuclear), kinetic, potential, electric, and magnetic, and their sum constitutes the total energy E of a system.

• In the absence of electric, magnetic, and surface tension effects (i.e., for simple compressible systems), the change in the total energy of a system during a process is the sum of the changes in its internal, kinetic, and potential energies and can be expressed as

∆ ∆ ∆ ∆where

∆∆∆


• Most systems encountered in practice are stationary, that is, they do not involve any changes in their velocity or elevation during a process.

• Thus, for stationary systems, the changes in kinetic and potential energies are zero, that is,

KE = PE = 0 and the total energy change relation it reduces to

E = U for such systems.

• Also, the energy of a system during a process will change even if only one form of its energy changes while the other forms of energy remain unchanged.

Mechanisms of Energy Transfer

• Energy can be transferred to or from a system in three forms:• heat,• work, and• mass flow

• Energy interactions are recognized at the system boundary as they cross it, and they represent the energy gained or lost by a system during a process.

• The only two forms of energy interactions associated with a fixed mass or closed system are heat transfer and work.


Heat Transfer, Q• Heat transfer to a system (heat gain) increases the energy of the molecules and thus the internal energy of the system, and heat transfer from a system (heat loss) decreases it since the energy transferred out as heat comes from the energy of the molecules of the system.

• Heat is transferred by three mechanisms:o conduction,o convection, ando radiation


Work Transfer, W• An energy interaction that is not caused by a temperature difference between a system and its surroundings is work.

• A rising piston, a rotating shaft, and an electrical wire crossing the system boundaries are all associated with work interactions.

• Work transfer to a system (i.e., work done on a system) increases the energy of the system, and work transfer from a system (i.e., work done by the system) decreases it since the energy transferred out as work comes from the energy contained in the system.

• Car engines and hydraulic, steam, or gas turbines produce work while compressors, pumps, and mixers consume work.


Mass Flow, m• Mass flow in and out of the system serves as an additional mechanism of energy transfer.

• When mass enters a system, the energy of the system increases because mass carries energy with it (in fact, mass is energy).

• Likewise, when some mass leaves the system, the energy contained within the system decreases because the leaving mass takes out some energy with it.

• For example, when some hot water is taken out of a water heater and is replaced by the same amount of cold water, the energy content of the hot‐water tank (the control volume) decreases as a result of this mass interaction


• Noting that energy can be transferred in the forms of heat, work, and mass, and that the net transfer of a quantity is equal to the difference between the amounts transferred in and out, the energy balance can be written more explicitly as

, , ∆

where the subscripts “in” and “out” denote quantities that enter and leave the system, respectively.• All six quantities on the right side of the equation represent “amounts,” and thus they are positive quantities.

• The direction of any energy transfer is described by the subscripts “in” and “out.”• The heat transfer Q is zero for adiabatic systems, the work transfer W is zero for systems that involve no work interactions, and the energy transport with mass Emass is zero for systems that involve no mass flow across their boundaries (i.e., closed systems).


• For a closed system undergoing a cycle, the initial and final states are identical, and thus

Esystem = E2 ‐ E1 = 0• Then the energy balance for a cycle simplifies to

Ein ‐ Eout = 0 or Ein = Eout• Noting that a closed system does not involve any mass flow across its boundaries, the energy balance for a cycle can be expressed in terms of heat and work interactions as

, , , ,

• That is, the net work output during a cycle is equal to net heat input.

First Law of Thermodynamics in Closed Systems

• Steady statemeans that all the thermodynamic properties of the system remain unchanged over time.

• The relations of the energy balance (or 1st Law) given above are intuitive in nature and are easy to use when the magnitudes of the heat transfer and work are known.

• However, when carrying out a general analytical study or want to solve a problem involving an unknown interaction of work or heat, we need to assume a direction for the interactions of heat or work.

• In such cases, it is common practice to use convection signs of classical thermodynamics and assume that the heat will be transferred into the system (heat input) in an amount Q, and the work will be done by the system (work output) in an amount W, and then solve the problem.


• The ratio of the energy balance for this case can be expressed as:

Qnet,in ‐ Wnet,out = Esystem or Q ‐ W = E

where Q = Qnet,in = Qin ‐ Qout is the net heat input, and W = Wnet,out = Wout ‐ Win is the net work output . If we obtain a negative amount of Q or W simply means that you took the wrong direction and we should invert it.


• Sometimes it is convenient to consider the term work in two parts: Wothers and Wb, where Wothers represents all forms of work except the boundary work. Then the first law takes the following form:

Q ‐ Wothers ‐ Wb = Esystem


• The energy balance for a closed system can be expressed as:

Q ‐ W = ΔEsystem = U + KE + PE

KE = PE = 0 for closed systems

Then,

Q ‐ W = U or Q ‐ Wother ‐ Wb = U


• For a constant pressure process, the boundary work is given byWb = P0(V2 ‐ V1)

• Substituting this formula in the above relation we got:Q ‐ Wother ‐ P0(V2 ‐ V1) = U

However,P0 = P1 = P2 → Q ‐ Wother = (U2 + P2V2) ‐ (U1 + P1V1)

in addition, H = U + PV, therefore,Q ‐ Wother = H2 ‐ H1 for constant pressure process

Moving Boundary Work

• One form of mechanical work frequently encountered in practice is associated with the expansion or compression of a gas in a piston–cylinder device.

• During this process, part of the boundary (the inner face of the piston) moves back and forth.

• Therefore, the expansion and compression work is often called moving boundary work, or simply boundary work.

• Some call it the P dV work for reasons explained later.


• Consider the gas enclosed in the piston–cylinder device.• The initial pressure of the gas is P, the total volume is V, and the crosssectional 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

• That is, the boundary work in the differential form is equal to the product of the absolute pressure P and the differential change in the volume dV of the system.

• This expression also explains why the moving boundary work is sometimes called the P dV work.


• 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:

• The quasi‐equilibrium expansion process described is shown on a P‐V diagram. On this diagram, the differential area dA is equal to P dV, which is the differential work. The total area A under the process curve 1–2 is obtained by adding these differential areas:


Constant Pressure

PV = mRT = Constant ln

PVn = Constant (polytropic Process)

Since and , then

1 (kJ)

Examples (textbook)

EXAMPLE 4–1 Boundary Work for a Constant‐Volume ProcessA rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively. Determine the boundary work done during this process.

0

Examples (textbook)

EXAMPLE 4–2 Boundary Work for a Constant‐Pressure ProcessA frictionless piston–cylinder device contains 10 lbm of steam at 60 psia and 320°F. Heat is now transferred to the steam until the temperature reaches 400°F. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this process.

Examples (textbook)

Data:= 10 lbm = 4.5 kg= 60 psia = 414 kPa→ kPa= 320°F = 160°C → °C= 400°F = 204.4°C → °C= ?

@ , 0.47088m3

kg TableA6

@ , 0.53434m3

kg TableA6

4.5kg 400kPa 0.53434 0.47088m3

kg1kJ

1kPam3

.

.

Examples (textbook)

EXAMPLE 4–3 Isothermal Compression of an Ideal GasA piston–cylinder device initially contains 0.4 m3 of air at 100 kPa and 80°C. 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.

⇒

ln ln ln ln ln

Examples (textbook)

Data:= 0.4 m3, = 0.1 m3

= 100 kPa= 80°C = constant= ?

ln

100kPa 0.4m3 ln0.1m3

0.4m3

1kJ1kPam3

.

Examples (textbook)

EXAMPLE 4–4 Expansion of a Gas against a SpringA piston–cylinder 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.

Examples (textbook)

Data:= 0.05 m3, = 2 = 2(0.05 m3) = 0.1 m3

= 200 kPa= 150 kN/m= 0.25 m2

a) = ?∆

0.1 0.05 m3

0.25m2 0.2m 150kN/m 0.2m 30kN

30kN0.25m2 120 kPa 200 kPa 120 kPa

Examples (textbook)b) = ?

areaI12 areaII

areaI 200kPa 0.1 0.05 m31kJ

1kPam3 ⇒ 10kJ

areaII12 0.1 0.05 m3 320 200 kPa

1kJ1kPam3 ⇒ 3kJ

10kJ 3kJ

Examples (textbook)c) = ?

areaII

12

12 150kN/m 0.2m 0

1kJ1kNm

Examples (textbook)

EXAMPLE 4–5 Electric Heating of a Gas at Constant PressureA piston–cylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120‐V source. At the same time, a heat loss of 3.7 kJ occurs. Determine the final temperature of the steam.@ constant ⇒ ∆

Examples (textbook)

Data:= 25 g (saturated water vapor)= = 300 kPa= 0.2 A, ∆ = 5 min = 300 s, = 120 V= 3.7 kJ

⇒

∆ 120V 0.2A 300s1kJ/s1000VA

7.2kJ

@ 2724.9kJ/kg TablaA53.7kJ 7.2kJ

0.025kg 2724.9 kJ/kg 2864.9 kJ/kg @ , °Table A6

Examples (textbook)

EXAMPLE 4–6 Unrestrained Expansion of WaterA rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25°C, 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 25°C. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process.

Examples (textbook)

Data:= 5 kg (water)= 200 kPa= = 25°C

a) = ?

2 ≅ @ 0.001003m3/kg TableA4

2 5kg 0.001003m3/kg

.

Examples (textbook)b) = ?

0.01m3

50.002m3/kg

@ 25°C TableA40.001003 3/43.340m3/kg

@ . TableA4

Examples (textbook)c) = = ?

∆ 0 ⇒

≅ @ ° 104.83kJ/kg TablaA4

0.002 0.00100343.34 0.001003 2.3 10

@ 25°C104.83kJ/kg2304.3kJ/kg

Table A4

104.83kJ/kg 2.3 10 2304.3kJ/kg 104.88kJ/kg

5kg 104.88 104.83kJkg

.

Homework 3b

Problems from the textbook (Thermodynamics, Yunus, 8th ed.):• Answer the following conceptual problems:

• Chapter 4, problem: 1

• Choose 5 problems and answer them (those who you consider to provide better understanding to the subject seen in this section)• Chapter 4, problems: 2‐42

First Law of ThermodynamicsClosed Systems

3b. 1st law of thermodynamics closed systems · first law of thermodynamics in closed systems...

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