91Ó°ÊÓ

We know that heat transfer through convection is given as \(Q = h \cdot A \cdot \Delta T\). In this case, the convection on the motor housing is given by \(Q_{h} = h_{h} \cdot A_{h} \cdot (T_{h} - T_{\infty})\).

Heat transfer from the rotating shaft to the ambient air by convection is given by: \(Q_{s} = h_{s} \cdot \pi \cdot D \cdot L \cdot (-\Delta T_{sh})\) where \(\Delta T_{sh} = T_{h} - T_{\infty}\).

The heat transfer through the pad is given by conduction, and can be calculated as \(Q_{p} = -k_p \cdot \frac{W^2}{t} \cdot \Delta T_{ph}\), where \(\Delta T_{ph} = T_{h} - T_{\infty}\).

The energy balance equation can be written as: \(P_{\text{elec}} = Q_{h} + Q_{s} + Q_{p} + P_{\text{mech}}\), because the electric power is producing heat as well as delivering mechanical power to the pump.

Plugging the values of \(Q_h\), \(Q_s\) and \(Q_p\) into the energy balance equation from Step 4, we can solve for \(P_{\text{elec}}\) and \(P_{\text{mech}}\): \(P_{\text{elec}} = h_{h} \cdot A_{h} \cdot (T_{h} - T_{\infty}) + h_{s} \cdot \pi \cdot D \cdot L \cdot (-\Delta T_{sh}) - k_p \cdot \frac{W^2}{t} \cdot \Delta T_{ph} + P_{\text{mech}}\) Now, we have the expression for the electric and mechanical power: \(P_{\text{elec}} = h_{h} \cdot A_{h} \cdot (T_{h} - T_{\infty}) + h_{s} \cdot \pi \cdot D \cdot L \cdot (-\Delta T_{sh}) - k_p \cdot \frac{W^2}{t} \cdot \Delta T_{ph}\) \(P_{\text{mech}} = P_{\text{elec}} - h_{h} \cdot A_{h} \cdot (T_{h} - T_{\infty}) - h_{s} \cdot \pi \cdot D \cdot L \cdot (-\Delta T_{sh}) + k_p \cdot \frac{W^2}{t} \cdot \Delta T_{ph}\) These are the expressions for electric power \(P_{\text{elec}}\) and mechanical power \(P_{\text{mech}}\) in terms of the given parameters.