Complete This Table For H2o

Complete this table for H2O:

T,°C	P, kPa	h, kJ/kg	x	Phase description
	200		0.7
140		1800
	950		0.0
lxxx	500
	800	3162.2

Answer:

b) Given T = 140 ^?C

h = 1800 kJ/kg

Only temperature and enthalpy are given. We do not know which table to use to determine the missing backdrop because we have no inkling equally to whether we have saturated mixture, compressed liquid or superheated vapor, which can exist determined from the enthalpy of saturated liquid (h_f) and enthalpy of saturated vapor (h_g) from the saturated water – temperature table (Table A-four).

At T= 140 ^?C

h_f = 589.16 kJ/kg

h_fg = 2144.iii kJ/kg

h_g = 2733.5 kJ/kg

Thus, the given value h = 1800 kJ/kg is in between the values of h_f and h_g. Therefore, we take saturated liquid-vapor mixture and the pressure is the saturation pressure at given temperature (Tabular array A-iv)

Hence, P = P_saturday @ 140 ^?C = 361.53 kPa.

The quality is adamant from

= 0.565

c) Given Force per unit area, P = 950 kPa

quality, x = 0.0

The quality, x = 0.0 implies that 100 percent of the mass is in the liquid state. Therefore, we have saturated liquid at force per unit area of 950 kPa. From the saturated water – pressure table (Table A-5)

T = T_saturday @ 950 kPa = 177.66 ^oC.

and h = h_f = 752.74 kJ/kg

d) Given T = eighty ^oC

P = 500 kPa

In this example, the temperature and force per unit area are given, only again nosotros do not know whether we have saturated mixture, compressed liquid or superheated vapor. To determine the region we are in, we go the saturation h2o – pressure table (Tabular array A-5) and determine the saturation temperature at the given pressure. Thus, for P = 500 kPa, we have T_sat = 151.83 ^oC. We and so compare the given T value to this T_sat value, keeping in mind that

if T < T_{sat @ given P} we have compressed liquid

if T = T_{sabbatum @ given P} we have saturated mixture

if T > T_{sat @ given P} we have superheated vapor.

In our case, the given T value is lxxx ^oC, which is less than the T_sat value at the given pressure. Therefore, we have compressed liquid and commonly we would make up one's mind the enthalpy value from the compressed liquid table (Table A-vii). Simply in this example the given pressure is lower than the lowest pressure value in the compressed liquid table (which is v MPa), and therefore we are justified to treat the compressed liquid as saturated liquid at the given temperature (non pressure) and obtain the value of enthalpy from Tabular array A-iv.

h = h_f @ fourscore ^oC = 335.02 kJ/kg.

Every bit we have compressed liquid, at that place is no meaning for quality in the compressed liquid region.

e) Given P = 800 kPa

h = 3162.2 kJ/kg

Merely pressure and enthalpy are given. We practise not know which tabular array to use to determine the missing properties because nosotros have no clue as to whether nosotros have saturated mixture, compressed liquid or superheated vapor, which can be determined from the enthalpy of saturated liquid (h_f) and enthalpy of saturated vapor (h_g) from the saturated water – pressure tabular array (Table A-five).

At P= 800 kPa

h_f = 720.87 kJ/kg

h_g = 2768.3 kJ/kg

Thus, the given value h = 3162.2 kJ/kg is greater than the value of h_g. Therefore, we have superheated vapor and the temperature is adamant from the superheated vapor table (Tabular array A-6) at force per unit area, P=0.eight MPa and h= 3162.2 kJ/kg

Hence, T = 350 ^oC.

The quality has no meaning equally we have superheated vapor.