Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
1 answer:
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Answer:
Part a: The volume of vessel is 4.7680 and total internal energy is 3680 kJ.
Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.
Explanation:
Part a:
As per given data
m=2 kg
T=80 °C =80+273=353 K
Dryness=70% vapour =0.7
<em>From the steam tables at 80 °C</em>
Specific volume of saturated vapours=v_g=3.40527
Specific volume of saturated liquid=v_f=0.00102
Now the relation of total specific volume for a specific dryness value is given as
Substituting the values give
Now the volume of vessel is given as
So the volume of vessel is 4.7680.
Similarly for T=80 and dryness ratio of 0.7 from the table of steam
Pressure=P=47.4 kPa
Specific internal energy is given as u=1840 kJ/kg
So the total internal energy is given as
The total internal energy is 3680 kJ.
So the volume of vessel is 4.7680 and total internal energy is 3680 kJ.
Part b
Volume of vessel is given as 1.6
mass is given as 2 kg
Pressure is given as 0.2 MPa or 200 kPa
Now the specific volume is given as
So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives
Temperature=T=120 °C
Quality=x=0.903 ≈ 90.3%
Specific internal energy =u=2330 kJ/kg
The total internal energy is given as
So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.
Answer:
The mass flow rate of bleed steam to be supplied is; m'_s = 1.85 Kg/s
Explanation:
From the first table i attached, at temperature of 70°C, the enthalpy of the cold water feed is;
h_c = 293 KJ/Kg
Also from the same table, at a temperature of 160°C and by interpolation, the enthalpy of the steam is;
h_s = 2790 KJ/Kg
Now, the exiting water is leaving the water heater at a combined rate of m'_c + m'_s and we know that it is a saturated liquid at Pressure of 200 KPa.
Where m'_c is mass flow rate of cold water feed while m'_s is mass flow rate of bleed steam.
Thus, from the second table attached and at 200KPa,the enthalpy of the saturated liquid is;
h_e = 504.71 KJ/kg
Now, from energy balance equation, we have;
E'_e = E'_c + E'_s
(m'_c + m'_s)h_e = (m'_c)h_c + (m'_s)h_s
Plugging in the relevant values, we have;
(20 + m'_s)504.71 = (20)293 + (m'_s)2790
10094.2 + 504.71m'_s = 5860 + 2790m'_s
2790m'_s - 504.71m'_s = 10094.2 - 5860
2285.29m'_s = 4234.2
m'_s = 4234.2/2285.29
m'_s = 1.85 Kg/s
