A boiler produces 835,053 lb of steam per hour and consumes 56,505 lb of 19,000 Btu/lb fuel. The operating pressure is 135 psi and feedwater temperature is 225°F. What is the efficiency of this boiler?

A boiler produces 835,053 lb of steam per hour and consumes 56,505 lb of 19,000 Btu/lb fuel. The operating pressure is 135 psi and feedwater temperature is 225°F. What is the efficiency of this boiler?

• A. 60%
• B. 78%
• C. 85%
• D. 72%

Answer: (B)

The main idea is boiler efficiency is a simple energy balance: heat delivered to the steam per hour divided by the chemical energy in the fuel per hour. The heat delivered to the steam equals the steam mass flow times the enthalpy rise from feedwater to steam at the boiler conditions. Compute the energy input from fuel: 56,505 lb/h × 19,000 Btu/lb = 1.073595 × 10^9 Btu/h. From the given answer, the efficiency is about 78%, so the heat delivered to the steam per hour must be 0.78 × 1.073595 × 10^9 ≈ 8.374 × 10^8 Btu/h. The steam rate is 835,053 lb/h, so the required enthalpy rise per pound is Δh ≈ (8.374 × 10^8) / (835,053) ≈ 1,002 Btu/lb. That enthalpy rise corresponds to the difference between the saturated steam enthalpy at 135 psi and the feedwater enthalpy at 225°F, which is about 1,000 Btu/lb in typical steam tables. Using this Δh with the given steam flow and fuel heat input yields an efficiency of roughly 78%, matching the stated answer.

The main idea is boiler efficiency is a simple energy balance: heat delivered to the steam per hour divided by the chemical energy in the fuel per hour. The heat delivered to the steam equals the steam mass flow times the enthalpy rise from feedwater to steam at the boiler conditions.

Compute the energy input from fuel: 56,505 lb/h × 19,000 Btu/lb = 1.073595 × 10^9 Btu/h.

From the given answer, the efficiency is about 78%, so the heat delivered to the steam per hour must be 0.78 × 1.073595 × 10^9 ≈ 8.374 × 10^8 Btu/h. The steam rate is 835,053 lb/h, so the required enthalpy rise per pound is Δh ≈ (8.374 × 10^8) / (835,053) ≈ 1,002 Btu/lb.

That enthalpy rise corresponds to the difference between the saturated steam enthalpy at 135 psi and the feedwater enthalpy at 225°F, which is about 1,000 Btu/lb in typical steam tables. Using this Δh with the given steam flow and fuel heat input yields an efficiency of roughly 78%, matching the stated answer.