ENERGY ANALYSIS OF CLOSED SYSTEMSOmil C.Chatib, M.Si

1. Moving Boundary Work

1. Moving Boundary Work

1. Moving Boundary Work (Isokhoric)

1. Moving Boundary Work (Isobaric)

1. Moving Boundary Work (Isobaric)

1. Moving Boundary Work (Isothermal)

1. Moving Boundary Work (Polytropic)

1. Moving Boundary Work (Polytropic)

1. Moving Boundary Work (Polytropic)

1. Moving Boundary Work (Polytropic)

1. Moving Boundary Work (Polytropic)

Contoh Soal :

1. Suatu fluida pada tekanan 3 bar dengan volume spesifik 0,18 m3/kg,diisikan dalam suatu silinder yang berpiston (torak), berekspansisecara reversibel mencapai tekanan 0,6 bar yang mengikutipersamaan p = c/v2 , dengan c adalah konstanta. Hitung kerja yangdilakukan oleh fluida pada piston !

1. Moving Boundary Work (Polytropic)

Contoh Soal :

2. Sebanyak 1 kg fluida diisikan dalam sebuah silinder pada tekanan awal 20bar. Dilanjutkan ekspansi secara reversibel dibelakang suatu piston denganmengikuti hukum pv2 = konstan sampai volumenya mencapai 2 kali. Fluidakemudian didinginkan secara reversibel pada tekanan konstan sampai pistonkembali ke posisi awalnya. Panas kemudian diberikan secara reversibeldengan piston dikunci tertutup dalam suatu posisi sampai tekanannya naikmencapai nilai awal 20 bar. Hitung kerja bersih yang dilakukan oleh fluida !

Untuk volume awal 0,05 m3.

1. Moving Boundary Work (Polytropic)

Contoh Soal :

3. Fluida tertentu pada 10 bar diisikan pada silinder yangberdampingan dengan suatu piston, volume awalnya 0,05 m3. Hitungkerja yang dilakukan bila fluida tersebut mengembang secarareversibel …

a. Pada tekanan konstan sampai volume akhirnya 0,2 m3.

b. Menurut persamaan linier sampai volume akhirnya 0,2 m3 dantekanan akhirnya 2 bar.

c. Menurut persamaan p.V = c sampai volume akhirnya 0,1 m3.

d. Menurut persamaan p.V3 = c sampai volume akhirnya 0,06 m3.

e. Menurut persamaan p = (A/V2) – (B/V) sampai volume akhirnya0,1 m3 dan tekanan akhirnya 1 bar.

2. Energy Balance for Closed Systems

For constant rates, the total quantities during a time interval ∆t are related to the quantities per unit time as ...

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

2. Energy Balance for Closed Systems

2. Energy Balance for Closed Systems

2. Energy Balance for Closed Systems

2. Energy Balance for Closed Systems

2. Energy Balance for Closed Systems

2. Energy Balance for Closed Systems

2. Energy Balance for Closed Systems

3. 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 “

In general, this energy depends on how the process isexecuted.

In thermodynamics, we are interested in two kinds ofspecific heats: specific heat at constant volume cv andspecific heat at constant pressure cp.

3. Specific HeatsThe specific heat at constant pressure cp is always greater thancv because at constant pressure the system is allowed to expandand the energy for this expansion work must also be supplied tothe system.

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

4. Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

5. Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids

Internal Energy Changes

Enthalpy Changes

5. Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids

Enthalpy Changes

5. Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids

5. Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids

5. Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids

MASS AND ENERGY ANALYSIS OF CONTROL VOLUMESOmil C.Chatib, M.Si

Flow Work and The Energy of a Flowing FluidUnlike closed systems, control volumes involve mass flow across their boundaries, and some work is required to push the mass into or out of the control volume. This work is known as the flow work, or flow energy, and is necessary for maintaining a continuous flow through a control volume.

Flow Work and The Energy of a Flowing Fluid

To push the entire fluid element into the control volume, this force must act through a distance L. Thus, the work done in pushing the fluid element across the boundary (i.e., the flow work) is ...

Flow Work and The Energy of a Flowing FluidTotal Energy of a Flowing Fluid

Flow Work and The Energy of a Flowing FluidEnergy Transport by Mass

Flow Work and The Energy of a Flowing FluidEnergy Transport by Mass

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of the exit opening is 8 mm2. Determine ...

(a) the mass flow rate of the steam and the exit velocity,

(b) the total and flow energies of the steam per unit mass, and

(c) the rate at which energy leaves the cooker by steam.

Flow Work and The Energy of a Flowing FluidEnergy Transport by Mass(a) Saturation conditions exist in a pressure cooker at all times after the steady operating

conditions are established. Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure. The amount of liquid that has evaporated, the mass flow rate of the exiting steam, and the exit velocity are ...

Flow Work and The Energy of a Flowing FluidEnergy Transport by Mass(b) Noting that h = u + Pv and that the kinetic and potential energies are disregarded, the

flow and total energies of the exiting steam are...

(c) The rate at which energy is leaving the cooker by mass is simply the product of the mass flow rate and the total energy of the exiting steam per unit mass,

Energy Analysis of Steady-Flow Systems

Energy Analysis of Steady-Flow Systems

Energy Analysis of Steady-Flow Systems

Some Stedy-Flow Engineering Devices

1. Nozzels and Diffusers

Some Stedy-Flow Engineering Devices

1. Nozzels and Diffusers

Some Stedy-Flow Engineering Devices

1. Nozzels and Diffusers

Some Stedy-Flow Engineering Devices

2. Turbines and Compressors

Some Stedy-Flow Engineering Devices

2. Turbines and Compressors

Some Stedy-Flow Engineering Devices

2. Turbines and Compressors

Some Stedy-Flow Engineering Devices

2. Turbines and Compressors

Some Stedy-Flow Engineering Devices3. Throttling Valves

Some Stedy-Flow Engineering Devices3. Throttling Valves

Some Stedy-Flow Engineering Devices4. Mixing Chambers and Heat Exchangers

Some Stedy-Flow Engineering Devices4. Mixing Chambers and Heat Exchangers

Some Stedy-Flow Engineering Devices4. Mixing Chambers and Heat Exchangers

Some Stedy-Flow Engineering Devices4. Mixing Chambers and Heat Exchangers

Some Stedy-Flow Engineering Devices4. Mixing Chambers and Heat Exchangers

Some Stedy-Flow Engineering Devices4. Mixing Chambers and Heat Exchangers

Some Stedy-Flow Engineering Devices5. Pipe and Duct Flow

Some Stedy-Flow Engineering Devices5. Pipe and Duct Flow

Energy Analysis of Unstedy-Flow Processes

Energy Analysis of Unstedy-Flow Processes

Energy Analysis of Unstedy-Flow Processes

Energy Analysis of Unstedy-Flow Processes

Energy Analysis of Unstedy-Flow Processes

Energy Analysis of Unstedy-Flow Processes