Respuesta :
Answer:
T2=26C
Φ2=74%
V2=17.4 m/s
Explanation:
Determine first mass flow rate
mass flow rate=V1*r^2*pi/v1
V1 is the entrance velocity and v1 is the specific volume
mass flow rate=1.41 kg/s
using mass flow rate calculate h2
h2=h1-q/mass flow rate
h2=67.23 kJ/kg
Now using psychrometric chart calculate exist temperature
T2=26 C
V2=v2*V1/v1
V2=0.87*0.18/0.9
V2=17.4 m/s
The exit temperature, the exit relative humidity of the air, and the exit velocity are;
A) T2 = 26°
B) Φ2 = 74%
C) v2 = 17.4 m/s
Thermodynamics
We are given;
- Diameter of pipe; d = 30 cm = 0.3 m
- Radius; r = d/2 = 0.3/2 = 0.15 m
- Pressure at inlet; P1 = 1 atm
- Temperature at inlet; T1 = 35°C = 308 K
- Relative humidity at inlet; Φ1 = 45%
- Velocity at inlet; v1 = 18 m/s
- Rate of heat removal; Q_out = 750 kJ/min = 12.5 kJ/s
Now, psychrometric chart is a chart that shows the graphical representation of the physical and thermal properties of atmospheric air at particular pressure values.
Now, from online psychrometric chart,
At Initial condition of P1 = 1 atm and Φ1 = 45%, we have;
- Enthalpy at inlet; h1 = 76.1 kJ/kg
- Absolute humidity of dry air; ω = 0.016 /kg of dry air
- Specific volume at inlet; υ1 = 0.9 m³/kg
- Formula for mass flow rate is;
m' = (v1 × A1)/υ1
m' = (18 × π × 0.15²)/0.9
m' = 1.41 kg/s
- Enthalpy at the outlet would be gotten from the formula; h2 = h1 - Q_out/m'
h2 = 76.1 - (12.5/1.41)
h2 = 67.43 kJ/kg
A) From psychrometric chart online,
At h2 = 67.43 kJ/kg, ω = 0.016 /kg, the temperature at the outlet which is the exit temperature is; T2 = 26°
B) From psychrometric chart online,
At h2 = 67.43 kJ/kg, ω = 0.016 /kg, the relative humidity at outlet which is the exit relative humidity is; Φ2 = 74%
C) From psychrometric chart online,
At h2 = 67.43 kJ/kg, ω = 0.016 /kg, the specific volume at outlet is; υ2 = 0.87 m³/kg
Formula for the exit velocity is;
v2 = (υ2 × v1)/υ1
v2 = (0.87 × 18)/0.9
v2 = 17.4 m/s
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