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Answer:
final volume V2 = 0.71136 m³
work done in process W = -291.24 kJ
heat transfer Q = 164 kJ
Explanation:
given data
mass = 1.5 kg
pressure p1 = 200 kPa
temperature t1 = 150°C
final pressure p2 = 600 kPa
final temperature t2 = 350°C
solution
we will use here superheated water table that is
for pressure 200 kPa and 150°C temperature
v1 = 0.95964 m³/kg
u1 = 2576.87 kJ/kg
and
for pressure 600 kPa and 350°C temperature
v2 = 0.47424 m³/kg
u2 = 2881.12 kJ/kg
so v1 is express as
V1 = v1 × m ............................1
V1 = 0.95964 × 1.5
V1 = 1.43946 m³
and
V2 = v2 × m ............................2
V2 = 0.47424 × 1.5
final volume V2 = 0.71136 m³
and
W = P(avg) × dV .............................3
P(avg) = =
= 400 × 10³
put here value
W = 400 × 10³ × (0.71136 - 1.43946 )
work done in process W = -291.24 kJ
and
heat transfer is
Q = m × (u2 - u1) + W .............................4
Q = 1.5 × (2881.12 - 2576.87) + 292.24
heat transfer Q = 164 kJ
To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.
The gas ideal law is given as,
Where,
P = Pressure
V = Volume
m = mass
R = Gas Constant
T = Temperature
Our data are given by
Note that the pressure to 38°C is 0.06626 bar
PART A) Using the ideal gas equation to calculate the mass flow,
Therfore the mass flow rate at which water condenses, then
Re-arrange to find
PART B) Enthalpy is given by definition as,
Where,
= Enthalpy of dry air
= Enthalpy of water vapor
Replacing with our values we have that
In the conversion system 1 ton is equal to 210kJ / min
The cooling requeriment in tons of cooling is 437.2.