Skip to main content

Did You Know? The Indian Boilers Act, 1923 Was Born Out of a Deadly Tragedy in West Bengal

Most of us know the Indian Boilers Act, 1923 as the backbone of boiler safety regulation in India – but very few know the tragic incident that forced its creation. In 1923, a catastrophic boiler explosion occurred at the Calcutta Electric Supply Corporation’s plant at Alipore, West Bengal (then Calcutta). The massive explosion completely destroyed the boiler and caused multiple fatalities and serious injuries. The scale of the disaster shocked the British Indian government and exposed the complete absence of uniform safety standards for steam boilers across the country.Prior to this, boiler regulation was scattered and inconsistent across provinces. The Alipore disaster became the tipping point. Within months, the Indian Boilers Act, 1923 was enacted and came into force on 1st January 1924.

Key highlights of the Act that emerged from this tragedy:
  • Mandatory registration and periodic inspection of all boilers
  • Certification of boiler operators and competent persons
  • Uniform safety standards across British India
  • Appointment of Chief Inspectors and Inspectors of Boiers
That single explosion in a Calcutta power plant literally saved thousands of lives in the decades that followed by making boiler safety non-negotiable.Next time you see a Form-IV certification or a boiler inspection stamp, remember – it all traces back to a dark day in West Bengal that changed industrial safety forever.

Subscribe to our newsletter for updates on DGMS mining engineering exam 

Study material for DGMS mining exams 

Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Geotechnical & Foundation Engineering (Solved Numerical Problems)

Numerical A 1000 cc core cutter weighing 946.80 g was used to find out the in-situ unit weight of an embankment. The weight of the core cutter filled with soil was noted to be 2770.60 g. Laboratory tests on the sample indicated a water content of 10.45 % and specific gravity of solids of 2.65. Determine the bulk unit weight, dry unit weight, void ratio, and degree of saturation of the sample. (AMIE Summer 2023, 8 marks) Solution Weight of soil in core cutter = 2770.60 - 946.80 = 1823.8 g Bulk unit weight, γ t = W/V = 1823.8/1000 = 1.82 g/ccDry unit weight, γ d = γ t /(1 + w) = 1.82/(1 + 0.1045) = 1.65 g/cc Void ratio, e = (G s γ w /γ d ) - 1 = (2.65 x 1.0/1.65) - 1 = 0.61 Degree of saturation, S = wG s /e = (0.1045 x 2.65)/0.61 = 0.4540 = 45.4% Numerical What is the theoretical height of the capillary rise and the capillary pressure in fine-grained soil with an effective size (D 10 ) of 0.002 mm? (AMIE Summer 2023, 4 marks) Solution D 10 = 0.002 mm;  Using the assumption that th...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...