Recent content by Lewishio

Thanks for your help on this, really appreciated.

No there isn't another part to this question.

Thanks for your help, am I right in saying the power rating is 12MW?

1,500kg/hour x 40,000,000Jkg / 3600 = 16,666,666.67 J/sec 12,002,886,3J/sec / 16,666,666.67J/sec x 100 = 72.01%

Thanks Chester, I can't seem to get the right answer for this: Efficiency = Q out / Q in x 100 (4.17 x 3,176,280) ÷ (4.17 x 297,890) x 100 = 1066.26%

4.17kg/sec x (3,176,280J/kg - 297,890J/kg) = 12,002,886.3J/sec

Would this be: 4.16kg/sec x 297,890J/kg = 1,239,222.4J/sec

Thanks for your help, how does this look: In this problem what is the rate at which liquid water enters the boiler ˙minm˙in? 4.16kg/sec In this problem, what is the rate at which water (steam) exits the boiler ˙moutm˙out? 4.16kg/sec From your steam tables, what is the enthalpy per unit mass of...

I’m not sure if this is correct but thought I’d have a go: Q - W + Σmi x Hi - Σme x He = 0 Q (Heat transfer): McΔT 15,000kg x 4.19KJ/kg°c x (400°c - 70°c) = 20,740,500 KJ°c p/h W (Work done): Consumption rate x calorific value 1,500kg x 40,000KJkg = 60,000,000 KJ p/h Heat of vaporization: Q...

Equation assuming negligible change in kinetic and potential energy: Q - W + Σmi x Hi - Σme x He = 0

Is this the correct formula: Q - W + Σm in (Hi + V1^2/2 + gzi) - Σm exit (He + V1^2/2 + gze)

Hi Chester, Do I need to find the volume of the mass entering the system? If so, there are 15 tonnes entering, so this would be 15,000L. Would I then multiply this by the Water Enthalpy at 70°c at 60 bar which is 297.89KJ/kg? This would be 4,468,350 KJ. Would I then multiply the mass by the...

I will do some research on open systems for the first law of thermodynamics and I can now see I haven't included the heat of vaporization which I believe is 2,260KJ/kg. I am not too sure which equations are to be used to figure this out.

I'm not sure

Steam Enthalpy at 400°c at 60 bar = 3176.28KJ/kg Water Enthalpy at 70°c at 60 bar = 297.89KJ/kg