ENGR 2213 Thermodynamics

ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics

F. C. Lai

School of Aerospace and Mechanical

Engineering

University of Oklahoma

Thermodynamic Properties of WaterThermodynamic Properties of Water

Phase Change of Water

T

v


Phase Change of Water

T

v

●Critical point

Saturated liquid lineSaturated vapor line

Sub

cool

ed li

quid

Saturated mixture

Sup

erhe

ated

vap

or


Saturation StateA state at which a phase change begins or ends

Critical PointIntersection of saturated liquid and saturated vapor

Single-Phase RegionsSubcooled liquid (compressed liquid), superheated vapor

Two-Phase RegionLiquid-vapor mixture

States of WaterStates of Water

Compressed liquid (Subcooled liquid)● For a given T, P > Psat

● For a given P, T < Tsat

● For a given v, v < vf

compressed liquid

subcooled liquid

Superheated vapor● For a given P, T > Tsat

superheated vapor

● For a given T, P < Psat

● For a given v, v > vg

States of WaterStates of Water

Saturated mixture● For a given T, P = Psat

● For a given P, T = Tsat

● For a given v, vf < v < vg

Qualityg

f g

mx

m m

0 ≤ x ≤ 1

Saturated Mixture of WaterSaturated Mixture of Water

gfVV V

vm m m

g gf fm vm v

m m

f g(1 x)v xv

V = Vf + Vg

f g fv x(v v )


P-v-T Relations of Water

water most substances


P-T Diagram of Water

●

●

Triple point

Critical point

Temperature

Pre

ssur

e

solid

vapor

liquid

Pre

ssur

e

Temperature

●

●Triple point

Critical point

liquid

solid

vapor

Thermodynamic Property Tables of WaterThermodynamic Property Tables of Water

Simple SystemA system for which there is only one way the system energy can be altered by work as the system undergoes a quasi-equilibrium process.

Simple Compressible SystemThe only mode of energy transfer by work isassociated with the volume change (expansion orcompression).


State Principle (State Postulate)The number of independent properties required tospecify the state of a system is one plus the numberof relevant work interactions.

Number of independent properties

= 1 + Number of work interactions The number of independent properties required tospecify the state of a simple compressible system is two.


Single-Phase Region (Compressed Liquid,Superheated Vapor)● Temperature and Pressure● Temperature and Specific Volume● Pressure and Specific Volume

Two-Phase Region (Saturated Mixture)● Temperature and Specific Volume● Temperature and Quality● Pressure and Specific Volume● Pressure and Quality


● Saturated Water – Temperature Table (A-4)● Saturated Water – Pressure Table (A-5)● Superheated Vapor (A-6)● Compressed Liquid (A-7)

Linear interpolation is required when the states encountered in problems do not fall exactly on the values provided by the tables.

Example 1Example 1

A rigid tank contains 50 kg of saturated liquid water at 100 ºC. Determine the pressure in the tank and the volume of the tank.

Table A-4, Psat = 101.35 kPa, P = Psat = 101.35 kPa

Table A-4, vf = 0.001044 m3/kg, V = mvf = 50 (0.001044) = 0.052 m3

A rigid tank contains 50 kg of saturated liquid water at 100 ºC. Determine the pressure in the tank and the volume of the tank.

Example 2Example 2

Table A-5, Tsat = 99.63 ºC, T = Tsat = 99.63 ºC

Table A-5, vg = 1.694 m3/kg, m = V/vg = 2/(1.694) = 1.18 kg

A piston-cylinder device contains 2 m3 of saturated water vapor at 100 kPa pressure. Determine the temperature of the vapor and the mass of the vapor Inside the cylinder.

A piston-cylinder device contains 2 m3 of saturated water vapor at 100 kPa pressure. Determine the temperature of the vapor and the mass of the vapor inside the cylinder.

Example 3Example 3

Table A-4, Psat = 70.14 kPa, P = Psat = 70.14 kPa

Table A-4, vf = 0.001036 m3/kg, vg = 2.361 m3/kg

V = Vf + Vg = mfvf + mgvg = 8(0.001036) + 2(2.361) = 4.73 m3

A rigid tank contains 10 kg of water at 90 ºC. If 8 kg of the water is in the liquid form and the rest is in the vapor form. Determine the pressure in the tank and the volume of the tank .

A rigid tank contains 10 kg of water at 90 ºC. If 8 kg of the water is in the liquid form and the rest is in the vapor form. Determine the pressure in the tank and the volume of the tank .

v = vf + x(vg - vf) = 0.001036 + 0.2(2.361 – 0.001036) = 0.473 m3/kg

V = mv = 10(0.473) = 4.73 m3

ENGR 2213 Thermodynamics

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