A centrifugal ar compressor has a rotor inrer diameter of \(D_{1}=2.0\) in.., a rotor oater diameter of \(D_{2}=6.5\) in, a rotor depth of 10 in. and a rotor rotational speed of 3600 rpm. The fluid relative velocities \((W)\) are purely radial. The compressor delivers an air mass flow rate of \(1.0 \mathrm{lom} / \mathrm{s}\) with \(T_{1}=70^{\circ} \mathrm{F}, p_{1}=4.7 \mathrm{psia}, T_{2}=\) \(240^{\circ} \mathrm{F},\) and \(p_{2}=33.1\) psia. Find the power transferied by the rotor to the air and the inlet relative velocity \(W\)

Given that the air mass flow rate is 1.0 lbf/s (where lbf represents pound-force), it must be converted to lbm/sec (where lbm represents pound-mass). Using the conversion factor, \(1 \, \text{lbf} = 32.174 \, \text{lbm} \, \text{ft/sec}^2\), the mass flow rate becomes \(1.0 \, \text{lbf/s} \times \frac{1}{32.174 \, \text{lbm} \, \text{ft/sec}^2} = 0.0311 \, \text{lbm/sec}\).

Knowing the mass flow rate, the energy equation can be applied next. The energy equation here is \( \text{Power} = \dot{m}(h_2 - h_1 + \frac{W_1^2}{2}-\frac{W_2^2}{2})\), where, \( \dot{m}\) is the mass flow rate. The enthalpy (h) at states 1 and 2 could be looked up in the property tables for air. To calculate W, we consider W to be purely radial and hence, \(W = u =R \cdot \omega\) (with ω the angular velocity and R the radius). The radius \(R= \frac{D}{2}\) can be considered as halfway between the inner and outer diameters of the rotor. Hence \( R = \frac{D_1 + D_2}{4} = \frac{2+6.5}{4} = 2.125\) in. Convert \( R\) to feet, this gives \( R = 2.125/12 = 0.177\) ft. The rotational speed \( \omega\) in rad/sec is then \( \omega = \frac{2\pi \times 3600}{60} = 377\) rad/sec. Hence, \(W = u = 377 \times 0.177 = 66.7\) ft/sec. With the \(W_1\) and \( W_2\) (considered to be equal) and \( h_1\) and \( h_2\) known, the Power can be calculated with the energy equation.

After substituting the values in the energy equation, calculate the power transferred. However, to calculate the power, the change in enthalpy is needed. In order to compute the enthalpy change, the air specifics should be considered (specific heat capacity and the temperature difference). Once enthalpy, \( \Delta h = c_p (T_2 - T_1)\), is calculated, substitute everything into the equation for Power.