LA Steam Engineer| 91 QUESTIONS| WITH COMPLETE SOLUTIONS

LA Steam Engineer| 91 QUESTIONS| WITH COMPLETE
SOLUTIONS
 Course
 Steam Engineer

1. Boiler Efficiency Calculation

Question:
A boiler produces 10,000 lb/hr of steam at 150 psig with a feedwater temperature of 200°F. The
enthalpy of steam at 150 psig is 1191.6 Btu/lb, and the enthalpy of feedwater at 200°F is 168
Btu/lb. If the fuel input to the boiler is 12,500,000 Btu/hr, calculate the boiler efficiency.

Solution:
Boiler efficiency (%) = useful energy outputtotal energy input×100\frac{\text{useful energy
output}}{\text{total energy input}} \times 100total energy inputuseful energy output×100

Useful energy output = 10,000×(1191.6−168)10,000 \times (1191.6 - 168)10,000×(1191.6−168)
= 10,000×1023.610,000 \times 1023.610,000×1023.6
= 10,236,000 Btu/hr

Boiler efficiency = 10,236,00012,500,000×100\frac{10,236,000}{12,500,000} \times
10012,500,00010,236,000×100
= 81.9%



2. Steam Pipe Sizing

Question:
A plant requires 8,000 lb/hr of saturated steam at 100 psig. If the steam velocity should not
exceed 6,000 ft/min, what should be the minimum internal diameter of the pipe? (Assume steam
density at 100 psig is 0.26 lb/ft³).

Solution:
Volumetric flow rate = mass flow ratesteam density\frac{\text{mass flow rate}}{\text{steam
density}}steam densitymass flow rate
= 8,0000.26\frac{8,000}{0.26}0.268,000
= 30,769.2 ft³/hr

Convert to ft³/min:
= 30,769.260\frac{30,769.2}{60}6030,769.2
= 512.8 ft³/min

,Using the velocity equation:
A=QV=512.86,000A = \frac{Q}{V} = \frac{512.8}{6,000}A=VQ=6,000512.8
= 0.0855 ft²

Pipe diameter DDD = 2×Aπ2 \times \sqrt{\frac{A}{\pi}}2×πA
= 2×0.08553.14162 \times \sqrt{\frac{0.0855}{3.1416}}2×3.14160.0855
= 4 inches (minimum)



3. Safety Valve Sizing

Question:
A boiler generates 20,000 lb/hr of steam at 200 psig. The required safety valve capacity is 1.5
times the normal steam flow. What is the minimum safety valve capacity required?

Solution:
Required safety valve capacity = 1.5×1.5 \times1.5× steam flow
= 1.5×20,0001.5 \times 20,0001.5×20,000
= 30,000 lb/hr



4. Equivalent Evaporation

Question:
A boiler produces 8,000 lb/hr of steam from feedwater at 212°F and at a pressure of 100 psig.
The enthalpy of steam at 100 psig is 1,187 Btu/lb, and the enthalpy of water at 212°F is 180
Btu/lb. Determine the equivalent evaporation.

Solution:
Heat added per lb of steam = 1,187 - 180
= 1,007 Btu/lb

Equivalent evaporation (lb/hr) = total heat added970.3\frac{\text{total heat added}}
{970.3}970.3total heat added
= 8,000×1,007970.3\frac{8,000 \times 1,007}{970.3}970.38,000×1,007
= 8,306.3 lb/hr



5. Steam Turbine Power Output

Question:
A steam turbine receives steam at 400 psia and expands it to 50 psia. If the turbine efficiency is
85% and the steam enthalpies are:

,  At 400 psia: 1200 Btu/lb
 At 50 psia (ideal): 950 Btu/lb

Determine the actual power output if the steam flow is 5,000 lb/hr.

Solution:
Ideal work done per lb = 1200 - 950 = 250 Btu/lb

Actual work done per lb = 0.85 × 250 = 212.5 Btu/lb

Total power output = 5000 × 212. (to convert Btu/hr to HP)
= 417.3 HP



6. Blowdown Rate Calculation

Question:
A boiler operates at 100 psig with a steam flow of 15,000 lb/hr. The feedwater TDS is 300 ppm,
and the allowed boiler water TDS is 3,000 ppm. Determine the required blowdown rate.

Solution:
Blowdown rate (%) = Feedwater TDSBoiler water TDS×Steam flow\frac{\text{Feedwater
TDS}}{\text{Boiler water TDS}} \times \text{Steam flow}Boiler water TDSFeedwater TDS
×Steam flow
= 3003000×15,000\frac{300}{3000} \times 15,0003000300×15,000
= 1,500 lb/hr



7. Fuel Consumption of a Boiler

Question:
A boiler has an efficiency of 80% and needs to generate 10,000 lb/hr of steam with an enthalpy
rise of 1,000 Btu/lb. If the fuel has a heating value of 18,000 Btu/lb, determine the required fuel
consumption.

Solution:
Total energy required = 10,000×1,00010,000 \times 1,00010,000×1,000
= 10,000,000 Btu/hr

Fuel input required = 10,000,0000.80\frac{10,000,000}{0.80}0.8010,000,000
= 12,500,000 Btu/hr

Fuel consumption = 12,500,00018,000\frac{12,500,000}{18,000}18,00012,500,000
= 694.4 lb/hr

, 8. Condensate Recovery Savings

Question:
A plant recovers 5,000 lb/hr of condensate at 200°F instead of discharging it. The makeup water
temperature is 60°F. If the fuel cost is $10 per million Btu and the boiler efficiency is 80%,
determine annual savings assuming 8,000 operating hours per year.

Solution:
Energy saved per hour = 5,000×(200−60)5,000 \times (200 - 60)5,000×(200−60)
= 700,000 Btu/hr

Fuel savings per hour = 700,0000.80×1,000,000×10\frac{700,000}{0.80 \times 1,000,000} \
times 100.80×1,000,000700,000×10
= $8.75/hr

Annual savings = 8.75×8,0008.75 \times 8,0008.75×8,000
= $70,000



9. Pump Horsepower for Boiler Feedwater

Question:
A boiler feedwater pump handles 100 gpm at a total dynamic head of 150 ft. The pump
efficiency is 75%. Determine the required pump horsepower.

Solution:
Water horsepower = 100×150×8.343960\frac{100 \times 150 \times 8.34}
{3960}3960100×150×8.34
= 31.6 HP

Brake horsepower = 31.60.75\frac{31.6}{0.75}0.7531.6
= 42.1 HP



10. Steam Flow through an Orifice

Question:
A steam system has an orifice with a discharge coefficient of 0.97. If the upstream pressure is
150 psig, downstream pressure is atmospheric, and orifice diameter is 1 inch, determine the mass
flow rate of steam.