Question

Water, in a 150 in^3 rigid tank, initially has a temperature of 70°F and an enthalpy of 723.5 Btu/lbm. Heat is added until the water becomes a saturated vapor. Calculate: a) The initial quality of the water b) The total mass of the water c) The temperature of the water at the final state.

Answer 1

a)initial quality of water 0.724.

b)mass of water =2.45 kg

c)T=70°F

Step-by-step explanation:

h=732.5 Btu/lbm

h=763.33 KJ/kg (1 Btu=1.05 kj)

T=70°F⇒T=21.1°C

`V=150 in^3=0.002458m^3`

(a)From steam table

Properties of saturated steam at 21.1°C

`h_f= 88.56(KJ)/(Kg) ,h_g= 2539.49(KJ)/(Kg)`

To find dryness fraction

`h=h_f+x(h_g-h_f)(KJ)/(Kg)`

`763.33=88.56+x(2539.49-88.56)(KJ)/(Kg)`

x=0.27

So initial quality of water 0.724.

(b)

`v=v_f+x(v_g-v_f)(m^3)/(Kg)`

where v is specific volume

From steam table at 21.1°C

`v_f= 0.001(m^3)/(Kg) ,v_g= 54.13(m^3)/(Kg)`

V=
`m_f* v_f`

0.002458=
`m_f* 0.001`

So mass of water =2.45 kg

(c)

Actually water will take latent heat ,it means that heating of will take place at constant temperature and constant pressure.So we can say that final temperature of water will remain same (T=70°F).