Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

136 | Thermodynamics

saturation table (Table A–4) and determine the ufand ugvalues at the given

temperature. At 125°C, we read uf
524.83 kJ/kg and ug
2534.3 kJ/kg.

Next we compare the given uvalue to these ufand ugvalues, keeping in

mind that

In our case the given uvalue is 1600, which falls between the ufand ugval-

ues at 125°C. Therefore, we have saturated liquid–vapor mixture. Then the

pressure must be the saturation pressure at the given temperature:

The quality is determined from

The criteria above for determining whether we have compressed liquid,

saturated mixture, or superheated vapor can also be used when enthalpy hor

specific volume vis given instead of internal energy u, or when pressure is

given instead of temperature.

(c) This is similar to case (b), except pressure is given instead of tempera-

ture. Following the argument given above, we read the ufand ugvalues at the

specified pressure. At 1 MPa, we have uf
761.39 kJ/kg and ug
2582.8

kJ/kg. The specified uvalue is 2950 kJ/kg, which is greater than the ugvalue

at 1 MPa. Therefore, we have superheated vapor, and the temperature at this

state is determined from the superheated vapor table by interpolation to be

We would leave the quality column blank in this case since quality has no

meaning for a superheated vapor.

(d) In this case the temperature and pressure are given, but again we cannot

tell which table to use to determine the missing properties because we do

not know whether we have saturated mixture, compressed liquid, or super-

heated vapor. To determine the region we are in, we go to the saturation

table (Table A–5) and determine the saturation temperature value at the

given pressure. At 500 kPa, we have Tsat
151.83°C. We then compare the

given Tvalue to this Tsatvalue, keeping in mind that

In our case, the given Tvalue is 75°C, which is less than the Tsatvalue at

the specified pressure. Therefore, we have compressed liquid (Fig. 3–44),

and normally we would determine the internal energy value from the com-

pressed liquid table. But in this case the given pressure is much lower than

the lowest pressure value in the compressed liquid table (which is 5 MPa),

and therefore we are justified to treat the compressed liquid as saturated liq-

uid at the given temperature (notpressure):

u
uf @ 75°C
313.99 kJ>kg¬¬ 1 Table A–4 2

if¬¬T 7 Tsat @ given (^) P ¬¬we have superheated vapor
if¬¬T
Tsat @ given (^) P ¬¬we have saturated mixture
if¬¬T 6 Tsat @ given (^) P ¬¬we have compressed liquid
T
395.2°C¬¬ 1 Table A–6 2
x

uuf
ufg


1600 524.83
2009.5

0.535
P
Psat @ 125°C
232.23 kPa¬¬ 1 Table A–4 2
if¬¬u 7 ug¬¬ we have superheated vapor
if¬¬ufuug¬¬ we have saturated mixture
if¬¬u 6 uf¬¬ we have compressed liquid
u
T, °C
P = 500 kPa
75
151.83
u ~ = uf @ 75°C
FIGURE 3–44
At a given Pand T, a pure substance
will exist as a compressed liquid if
TTsat @ P.