“A classroom has a design sensible cooling load of 36,000 Btu/hr and a latent load of 12,000 Btu/hr at peak conditions. The room setpoint is 75 °F and the supply air temperature is 55 °F. Using standard air properties at sea level and the relations $Q_s=1.08\times \text{CFM}\times \Delta T$ and $1\text{ ton}=\mathrm{12,000}\text{ Btu/hr}$, determine: (1) the required supply airflow (CFM) to handle the sensible load, (2) the total cooling capacity (tons), and (3) the moisture removal rate (lb/hr) implied by the latent load (use $h_{\mathrm{fg}}\approx 1061\text{ Btu/lb}$).”

(1) Supply airflow (CFM) from sensible load

$\Delta T=75-55=20\text{ °F}$

$Q_s=1.08\times \text{CFM}\times \Delta T\to \text{CFM}=\frac{\mathrm{36,000}}{1.08\times 20}=\frac{\mathrm{36,000}}{21.6}\approx \mathrm{1,667}\text{ CFM}$

(2) Total cooling capacity (tons)

$Q_{\text{total}}=Q_s+Q_{\ell }=\mathrm{36,000}+\mathrm{12,000}=\mathrm{48,000}\text{ Btu/hr}$

$\text{Tonnage}=\frac{\mathrm{48,000}}{\mathrm{12,000}}=4\text{ tons}$

(3) Moisture removal rate from latent load

$\dot{m}=\frac{Q_{\ell }}{h_{\mathrm{fg}}}=\frac{\mathrm{12,000}}{1061}\approx 11.3\text{ lb/hr}$

Answer: Required supply airflow ≈ 1,667 CFM; total capacity = 4 tons; moisture removal ≈ 11.3 lb/hr.