Question

A house has an electric heating system that consists of a 150-W fan and an electric resistance heating element placed in a duct. Air flows steadily through the duct at a rate of 0.5 kg/s and experiences a temperature rise of 5°C. The rate of heat loss from the air in the duct is estimated to be 250 W. Determine the power rating of the electric resistance heating element in kW?

Answer 1

The work input to the fan is given as:

$W_{fan} = -150 \;W$

Negative sign indicates that work is done on the system.

The mass flow rate of air in the duct is:

$m_{air} = 0.5 \;kg/s$

The rise in temperature of the air in the duct is:

∆T = 5

The amount of heat lost to the surrounding is:

Q = -250 W

Apply energy balance equation for the conservation of energy.

$E_{in} = E_{out}$

$Q + m_{in}h_{in} = W + m_{out}h_{out}$

Apply mass balance:

$m_{in} = m_{out} = m_{air}$

Therefore, the energy equation becomes:

$Q = W_{fan} + W_{elec} + m_{air}(h_{out} - h_{in})$

Substituting the values in above equation

Substitute $c_pT$ for $h$ and take specific heat of air $c_p = 1.005 \;kJ/kgC$

$-0.25 = -0.15 + W_{elec} + 0.5 \times 1.005 \times 5$

$W_{elec} = -2.6125\; kW$

Negative sign indicates that the work is done on the system.

Therefore, the power rating of the electric resistance is 2.61 kW