# A boiler takes fuel oil of specific gravity 0.81 and calorif

A boiler takes fuel oil of specific gravity 0.81 and calorific value of 27912KJ/KG form a tank 6.0M in diameter. At the time of the maximum firing, the level of fuel oil in the tank is lowered 51mm in one hour, and the absolute boiler pressure is 1620Kpa.

What is the mass flow rate of steam from the boiler in KG/HR

Note: I can get to the point of finding the calories released per hour of the fuel, but I'm having some trouble translating that into an evaporation rate for the boiler. Any thoughts?

I need this to find the CSA of a safety valve.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Let say on the basis of one hour:

The amount (kg) of fuel oil consumed is: 0.051*pi*3^2*810 = 1167 (kg/hr)

The amount of energy by this oil:27912KJ/KG*1167 kg/hr = 32,585,044 (kJ/hr)

Here you should use the latent heat of vaporization of water, which is 2257 kJ/kg at 100C and 101kPa (see reference)

Therefore the mass flowrate of steam from the boiler will be:

32,585,044/2,257 = 14,437 (kg/hr)

Reference:

http://energy.sdsu.edu/testcenter/testhome/Test/solve/basics/tables/tablesPC/TSatH2O.html

https://brainmass.com/physics/heat-thermodynamics/thermodynamics-mass-flow-rate-of-steam-118297