- #1
MonkSpeed
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Homework Statement
Hi there,
I am trying to complete a thermodynamics assignment. I am doing fine until I get to the following question.
"Hx = Specific enthalpy at dryness fraction = hf + x . hfg" - I am not quite sure where Hf came from, I guess they want a value for Saturated liquid from the Enthalpy tables but which one?
Homework Equations
I have two Hf's I have already calculated.
Hf @ Ti 63 kJ/kg
Hf @ Tb 419.2 kJ/kg
These are just the values for saturated water at 15°C and 100°C.
Other relevant equations:
X = Dryness fraction = Mev / Mw = 0.013kg / 1.805kg = 0.007
Hfg at boiling temperature = 2256.4
Hev = Hx - Hf @ Ti
Qev = Mw * Hx
The Attempt at a Solution
I would just like to add that I have calculated all of the above, they were not given to me.
When I put the numbers I assume is supposed to be there I get.
Hx = Hf + x * Hfg = 419.2 - 0.007 * 2256.4 = 435
Hev = Hx - Hf @ Ti = 435 - 63 = 372
Qev = Mw * Hx = 1805grams * 435 = 785175J
Energy delivered by kettle
Qk = Pk * t = 2100watt * 310.4 = 651840J
Efficiency of my kettle
Nev = Qev / Qk = 785175 / 651840 = 1.20 or 120% lol
That can't be 120% so I have gone wrong somewhere and I think it is to do with "Hx = Specific enthalpy at dryness fraction = hf + x . hfg"
If anyone could shed some light that would be great.
Many Thanks.
Sorry for the long post.