1. The problem is to compute the efficiency of a centrifugal blower. 2. Efficiency ($\eta$) of a centrifugal blower is defined as the ratio of the useful power output to the power input supplied to the blower. 3. The formula for efficiency is: $$\eta = \frac{\text{Output Power}}{\text{Input Power}} \times 100\%$$ 4. For a centrifugal blower, the output power is the power imparted to the air, which can be calculated as: $$\text{Output Power} = \rho \times Q \times g \times H$$ where: - $\rho$ is the density of air (kg/m$^3$), - $Q$ is the volumetric flow rate (m$^3$/s), - $g$ is acceleration due to gravity (9.81 m/s$^2$), - $H$ is the total pressure head developed by the blower (m). 5. The input power is the mechanical power supplied to the blower shaft, usually measured by a power meter or calculated from motor specifications. 6. Therefore, efficiency calculation steps: - Measure or obtain $\rho$, $Q$, $H$, and input power. - Calculate output power using the formula above. - Calculate efficiency using: $$\eta = \frac{\rho Q g H}{\text{Input Power}} \times 100\%$$ 7. Important notes: - Ensure units are consistent. - Efficiency is always less than or equal to 100%. - This formula assumes no losses other than mechanical and aerodynamic losses. Final answer: $$\boxed{\eta = \frac{\rho Q g H}{\text{Input Power}} \times 100\%}$$