Answer:
a) volume flow rate of air at the inlet is 0.0471 m³/s
b) the velocity of the air at the exit is 8.517 m/s
Explanation:
Given that;
The electrical power Input W_elec = -1400 W = -1.4 kW
Inlet temperature of air T_in = 22°C
Inlet pressure of air p_in = 100 kPa
Exit temperature T_out = 47°C
Exit area of the dyer is A_out = 60 cm²= 0.006 m²
cp = 1.007 kJ/kg·K
R = 0.287 kPa·m3/kg·K
Using mass balance
m_in = m_out = m_air
W _elec = m_air ( h_in - h_out)
we know that h = CpT
so
W _elec = m_air.Cp ( T_in - T_out)
we substitute
-1.4 = m_air.1.007 ( 22 - 47 )
-1.4 = - m_air.25.175
m_air = -1.4 / - 25.175
m_ air = 0.0556 kg/s
a) volume flow rate of air at the inlet
we know that
m_air = P_in × V_in
now from the ideal gas equation
P_in = p_in / RT_in
we substitute our values
= (100×10³) / ((0.287×10³)(22+273))
= 100000 / 84665
P_in = 1.18 kg/m³
therefore inlet volume flowrate will be;
V_in = m_air / P_in
= 0.0556 / 1.18
= 0.0471 m³/s
the volume flow rate of air at the inlet is 0.0471 m³/s
b) velocity of the air at the exit
the mass flow rate remains unchanged across the duct
m_ air = P_in.A_in.V_in = P_out.A_out.V_out
still from the ideal gas equation
P_out = p_out/ RT_out ( assume p_in = p_out)
P_out = (100×10³) / ((0.287×10³)(47+273))
P_out = 1.088 kg/m³
so the exit velocity will be;
V_out = m_air / P_out.A_out
we substitute our values
V_out = 0.0556 / ( 1.088 × 0.006)
= 0.0556 / 0.006528
= 8.517 m/s
Therefore the velocity of the air at the exit is 8.517 m/s
