Factor of evaporation in boiler definition
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This will be 30 pounds multiplied by the factor of evaporation for 70 pounds gauge pressure and 100 degrees feed temperature, or 1.1494. 30 × 1.1494 = 34.482, or approximately 34.5 pounds. Hence, oneboiler horse power is equal to an evaporation of 34.5 pounds of water per hour from and at 212 degrees Fahrenheit.
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