chapter 4 : first law of thermodynamicsportal.unimap.edu.my/portal/page/portal30/lecture...

CHAPTER 4 : FIRST LAW OF THERMODYNAMICS

Do you still remember what 1ST LAW OF THERMO tell us???

What about CLOSED and

OPEN system??


Energy CANNOT

be created or destroyed


TYPES OF SYSTEM

ISOLATED CLOSED OPEN

neither mass nor energy can cross the selected boundary

Only energy can cross the selected boundary

Both mass and energy can cross the selected boundary


Known as fixed mass

No mass can across its boundary

Energy in the form of work /heat can cross the boundary

Work : moving boundary work

Heat : specific heat, internal energy, enthalpy

Volume does not have to be fixed

Example : piston-cylinder

CLOSED SYSTEM


CLOSED SYSTEM (WORK)

Moving boundary work : 𝑾𝒃 = 𝑷 𝒅𝑽 = 𝒅𝑨𝟐

𝟏

𝟐

𝟏

Wb is positive for expansion (amount of energy transferred from system)

Wb is negative for compression (amount of energy transferred to system)

Figure 1 : piston-cylinder system



Moving boundary work done will be dependence to the chosen path

Energy balance of closed system :

𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚


EXAMPLE 1

A frictionless piston-cylinder device contain 5kg of steam at 400kPa and 200oC. Heat is now transferred to the steam until the temperature reach 250oC. If the system is not attached to a shaft and its mass is constant, calculate the work done during this process.

Steam in piston-cylinder is heated at constant pressure ( isobaric)

From : 𝑾𝒃 = 𝑷 𝒅𝑽

𝟐

𝟏= 𝑷 𝒅𝑽

𝟐

𝟏= 𝐏 𝑽𝟐 − 𝑽𝟏 = 𝒎𝑷(𝒗𝟐 − 𝒗𝟏)

𝑾𝒃 = 𝟓 𝟒𝟎𝟎𝒌𝑷𝒂 𝟎. 𝟓𝟗𝟓𝟐𝟎 − 𝟎. 𝟓𝟑𝟒𝟑𝟒 = 𝟏𝟐𝟐 𝒌𝑱


m = 5kg

P1 = 400 kPa

T1= 200oC

m = 5kg

P2 = 400 kPa

T2= 250oC

Superheated water (Table A-6)

P=0.40MPa

T(oC) v(m3/kg) u(kJ/kg) h(kJ/kg) s(kJ/kg.K)

200 0.53434 2647.2 2860.9 7.1723

250 0.59520 2726.4 2964.5 7.3804


EXAMPLE 2

A piston-cylinder device initially contains 0.4m3 of air at 100kPa and 80oC. The air is now compressed to 0.1m3 but the temperature inside the cylinder is remain constant. Calculate the work done during this process.

Air

P1 = 100 kPa

V1 = 0.4 m3

T1 = 80OC

V2 = 0.4 m3

T2 = 80OC

Air in piston-cylinder is compressed isothermally

By considering air as an ideal gas: 𝑃𝑉 = 𝑚𝑅𝑇 = 𝐶 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 P = C/V

From : 𝑾𝒃 = 𝑷 𝒅𝑽

𝟐

𝟏=

𝑪

𝑽 𝒅𝑽

𝟐

𝟏= 𝐂 𝐥𝐧

𝑽𝟐

𝑽𝟏= 𝑷𝟏𝑽𝟏 𝐥𝐧

𝑽𝟐

𝑽𝟏

𝑾𝒃 = 𝟏𝟎𝟎𝒌𝑷𝒂 𝟎. 𝟒 𝒍𝒏 𝟎. 𝟏

𝟎. 𝟒= −𝟓𝟓. 𝟓𝒌𝑱

*Wb is negative energy of 55.5kJ are transferred to the system



EXAMPLE 3

A piston-cylinder device initially contains 0.05m3 of a gas initially at 200kPa. At this state, a linear spring with k = 150 kN/m is touching the piston but exerting no force on it. Heat is transferred to the gas causing the piston to rise and compress the spring until the volume inside the cylinder doubles. If the cross sectional area of the piston is 0.25m2, determine :

i. Final pressure inside the cylinder ii. Total work done by the gas

Air in piston-cylinder is expanding

The enclosed volume at final state: 𝑉2 = 2𝑉1 = 0.1𝑚3

Displacement of piston :

𝑥 = ∆𝑉

𝐴=(0.1 − 0.05)𝑚3

0.25𝑚2= 0.2𝑚

Additional pressure applied by spring:

𝑃 =𝐹

𝐴=

𝑘𝑥

𝐴=

(150 k𝑁/𝑚)(0.2𝑚)

0.25𝑚2= 120 k𝑃𝑎

Therefore, final pressure : 200 kPa + 120 kPa = 320 kPa

Work , W = area = 1

2200 + 320 0.05 = 13 k𝐽


A = 0.25m2

P1 = 200 kPa

V1 = 0.05 m3

P2 = ??

V1 = 0.1 m3


CLOSED SYSTEM (HEAT)

When we talk about energy transfer by heat, few keywords need to be remember..

Heat transfer, Q

Specific heat, C

Enthalpy, h Internal energy, u

Entropy, s

Conduction (direct contact) Convection (conduction + fluid motion) Radiation (emission of EM waves)

Energy required to raise temp. of

substances by 1oC Specific heat at constant volume

𝑐𝑣 = 𝜕𝑢

𝜕𝑇 𝑣

(changes in internal energy) Specific heat at constant pressure

𝑐𝑝 = 𝜕ℎ

𝜕𝑇𝑝

(changes in enthalpy)

h = u + PV = u + RT

∆ℎ = ℎ2 − ℎ1 = 𝑐𝑝 𝑇 𝑑𝑇2

1

microscopic energy U = latent + sensible + chemical +

nuclear

∆𝑢 = 𝑢2 − 𝑢1 = 𝑐𝑣 𝑇 𝑑𝑇2

1

Relates to 2nd law of

thermodynamics movement of particle/atom


FIRST LAW RELATION FOR CLOSED SYSTEM

General 𝑸−𝑾 = ∆𝑬

Stationary systems 𝑸−𝑾 = ∆𝑼

Per unit mass 𝒒 − 𝒘 = ∆𝒆

Differential form 𝜹𝒒 − 𝜹𝒘 = 𝒅𝒆


OPENED SYSTEM

Also known as control volume

Mass and energy can across its boundary

Involve mass flow

Conservation of mass: Mass cannot be created/destroyed during a process.

Example : compressor, turbine, nozzle


OPENED SYSTEM

Conservation of Energy : 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚

Conservation of Mass : 𝑚𝑖𝑛 −𝑚𝑜𝑢𝑡 = ∆𝑚𝑐𝑣

Relation between mass (𝑚 ) and volume flow rate (𝑉 )

𝑚 = 𝜌𝑉

Total mass within control volume (CV) :

𝑚𝑐𝑣 = 𝜌 𝑑𝑉𝑐𝑣


OPENED SYSTEM

MASS BALANCE FOR STEADY FLOW PROCESSES

Steady flow process : The amount of mass in CV remain the same Total amount of mass entering CV = total amount leaving CV

Steady flow: 𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡 (kg/s)


OPENED SYSTEM

MASS BALANCE FOR STEADY FLOW PROCESSES

Single stream steady flow : involve one inlet and one outlet i.e. nozzles, turbines, diffusers, turbines, compressor.

Steady flow (single stream) : 𝑚 1 = 𝑚 2 𝜌1𝑉1𝐴1 = 𝜌2𝑉2𝐴2

Volume doesn’t need to be conserved.


Special Case: Incompressible Flow

Steady,

incompressible

Steady,

incompressible

flow (single stream)

There is no such thing as a “conservation of volume” principle.

For steady flow of liquids, the volume flow

rates, as well as the mass flow rates, remain constant since liquids are essentially incompressible substances.


OPENED SYSTEM EXAMPLE 1

A garden hose attached with a nozzle is used to fill a 40L bucket. The inner diameter of the hose is 2cm and it reduces to 0.8cm at the nozzle exit. If it takes 50s to fill the bucket with water, determine (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit. SOLUTION Assume the density of water 1000 kg/m3 = 1 kg/L a) Noting that 40L of water are discharged in 50s, the mass and flow rates of

water. The volume and mass flow rates are ?? b) The cross-sectional area of the nozzle exit is ?? The volume flow rate through the hose and nozzle is constant. The average velocity of water at nozzle exit is ??


FLOW WORK AND THE ENERGY OF A FLOWING FLUID

Flow work, or flow energy: The work (or energy) required to push the mass into or out of the control volume. This work is necessary for maintaining a continuous flow through a control volume.


FLOW WORK AND THE ENERGY OF A FLOWING FLUID

The flow work relation is the same whether the fluid is pushed into or out of the control volume


TOTAL ENERGY OF A FLOWING FLUID

The total energy consists of three parts for a non flowing fluid and four parts for a flowing fluid.

Where enthalpy, h = u + Pv


Energy Transport by Mass

When the kinetic and potential energies of a fluid stream are negligible

When the properties of the mass at

each inlet or exit change with time

as well as over the cross section


ENERGY ANALYSIS OF STEADY-FLOW SYSTEMS


Mass and Energy balances for a steady-flow process

Mass balance

Energy balance


Energy balance (OTHER EQUATIONS)

A water heater in

steady operation.


Energy balance relations with sign conventions (i.e., heat input and

work output are positive)

where


Under steady operation, shaft work and electrical work are the only forms of work a simple compressible system may involve.

• Q is the rate of heat transfer between

the control volume and its surroundings

• when the CV is losing heat : Q is

negative

• if the CV is insulated (adiabatic): Q =0

• W = power

• For steady flow devices, the CV

constant, no boundary work, W = 0

NOTE: Enthalpy, h in kJ/kg Kinetic energy ,ke in J/kg To add (h + ke), ke must be in kJ/kg by dividing it by 1000 ALSO NOTE THAT:


SOME STEADY-FLOW ENGINEERING DEVICES

Many engineering devices operate essentially under the same conditions for long periods of time. The components of a steam power plant (turbines, compressors, heat exchangers, and pumps), for example, operate nonstop for months before the system is shut down for maintenance. Therefore, these devices can be conveniently analyzed as steady-flow devices.

A modern land-based gas turbine used for electric power production. This is a General Electric LM5000 turbine. It has a length of 6.2 m, it weighs 12.5 tons, and produces 55.2 MW at 3600 rpm with steam injection.


Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even garden hoses.

A nozzle is a device that increases the velocity of a fluid at the expense of pressure.

A diffuser is a device that increases the pressure of a fluid by slowing it down.

The cross-sectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flows. The reverse is true for diffusers.

Energy

balance for

a nozzle or

diffuser:


Turbines and Compressors Turbine drives the electric generator in steam, gas, or hydroelectric power plants.

As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work.

Compressors, as well as pumps and fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft. Therefore, it involves work input.

A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas.

A compressor is capable of compressing the gas to very high pressures.

Pumps work very much like compressors except that they handle liquids instead of gases.

Energy balance for the

compressor in this figure:

chapter 4 : first law of thermodynamicsportal.unimap.edu.my/portal/page/portal30/lecture...

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