Question

An oil- fired boiler takes in feed water at 75°C and produces wet steam at a pressure of 5 bars. The steam flow rate 1.50 tons/hr with a dryness fraction of 0.89. The fuel consumption rate is 6.10kg/min and its net calorific value is 41MJ/Kg. Determine the thermal efficiency of the boiler.

Answer 1

Answer on Question #79807, Physics / Molecular Physics | Thermodynamics

An oil- fired boiler takes in feed water at $75{}^{\circ}\mathrm{C}$ and produces wet steam at a pressure of 5 bars. The steam flow rate 1.50 tons/hr with a dryness fraction of 0.89. The fuel consumption rate is $6.10\mathrm{kg/min}$ and its net calorific value is $41\mathrm{MJ/Kg}$. Determine the thermal efficiency of the boiler.

Solution:

Finding enthalpy of feed water, $h_1$ and output steam $h_2$

Use the table http://materias.df.uba.ar/f4aa2015c1/files/2015/03/Tableswater.pdf

$$h_1 = h_f \text{ at } 75{}^{\circ}\mathrm{C}$$ $$h_1 = 313.93 \, \mathrm{kJ/Kg}$$ $$h_2 = h_f + \frac{\kappa h_{fg}}{k}$$ $$h_2 = 640 \, \mathrm{kJ/Kg} + 0.89 \times 2108 \, \mathrm{kJ/Kg} = 2516.12 \, \mathrm{kJ/Kg}$$

Finding steam flow rate in kg/s

$$m_s = (1.5 \times 10^3) / 60^2 = 0.42 \, \mathrm{kg/s}$$

Finding power rating of boiler.

$$BPR = (h_2 - h_1) \times m_s$$ $$BPR = (2516.12 \, \mathrm{kJ/Kg} \times 2 - 313.93 \, \mathrm{kJ/Kg}) \times 0.42 \, \mathrm{kg/s} = 924.9 \, \mathrm{kJ} = 0.925 \, \mathrm{MJ}$$

Finding thermal efficiency of boiler

$$\eta = \frac{(h_2 - h_1) \times m_s}{CV m_f}$$

Where $m_f = 0.102 \, \mathrm{kg/s}$, $CV = 41 \, \mathrm{MJ/Kg}$

$$\eta = \frac{0.925 \, \mathrm{MJ}}{41 \, \mathrm{MJ/Kg} \times 0.102 \, \mathrm{kg/s}} = 0.22 = 22\%$$

Answer: 22 %

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