A 160L vessel contains 8 kg of refrigerant134a at a pressure

A 160-L vessel contains 8 kg of refrigerant-134a at a pressure of 160kPa. Determine (i) thetemperature, (ii) the quality, (iii) the enthalpy of the refrigerant and (iv) the volume occupied by the vaporphase.

(b) A rigid tank contains 100 kg of saturated liquid water at 90 Celsius. Determine the pressure in the tank and the volume of the tank. Imagine now that a second tank contains water at a state of P=0.5 MPa and that h=2890 kJ/kg – what is the temperature of water?

Solution

Ans-a. Given

Volume=160 L = 160/1000= 0.16 m³

Mass=8 kg

Specific Volume = Volume/mass

=0.16/8

=0.02 m³/kg

Properties of Refrigerant 134a from refrigeration tables,

At P_sat = 160KPa,

Specific Volume v_f = 0.0007437 m³/kg

Specific Volume v_g= 0.12348 m³/kg

So it is clear that v_f <v < v_g

Hence Refrigerant 134a lies in the saturated liquid vapour mixture.

a.The Temperature

Hence T=T_sat (at P_sat = 160kpa) = -15.60⁰ c

b. Quality

x=(v-v_f) / v_fg

x= (0.02-0.0007437) / (0.12348-0.0007437) (as v_fg = v_g - v_f )

x = 0.157

c. Enthalpy of Refrigerant

At 160 kpa

h_f = 31.21 KJ/Kg

h_fg = 209.90 KJ/Kg

so h=h_f + x.h_fg

h= 31.21+0.157*209.90

h=64.2 KJ/Kg

d. Volume occupied by the vapour phase

we know

Dryness fraction x= m_g / mt

hence m_g = x*m_t = 0.157*8

m_g=1.256 Kg

volume occupied by vapour phase = m_g * v_g

=1.256*0.12348

V =0.15509 m³ = 155.09 L

Ans-b

First Part

Given,

mass m =100Kg

Saturated Temperature T =90⁰ C

From Pressure tables,

at 90⁰ C

P_sat = 70.183 KPa

v_sat = v_f = 0.001036 as it is saturated liquid

so specific volume v = Volume/mass

Volume V = v * m

=0.001036*100

So Volume V=0.1036 m³

Second Part

Given,

Pressure = 0.5 MPa = 5 bar

Enthalapy h = 2890KJ/Kg

From Pressure tables at 5 bar

h_f = 640.23

h_g = 2748.7

Since h>h_g , it is super heated vapour

From Superheated tables at P=5 bar,

2939.9

By Interpolation

(2855.4-2890)/(2939.9-2855.4) = (200-T)/(240-200)

-16.37 = (200 - T)

therefore Temperature of Water, T = 216.37 ⁰C

T (0 C)	h (KJ/Kg)
200	2855.4
240	2939.9