I originally had 1.13m2 but that was due to a units error, the 724.5 x 105 is in KJ/hr which means when it comes to calculating the Area it is 724.5 x 108 as the coefficient is in W m-2 K-1. This balances the units. I think that's correct anyway.
My materials state this as the route to determine number of tubes.
qmh=ρAfu
where
qmh - mass flowrate of the liquid/gas = 50000/3600 kJ/s
ρ is the density of the liquid/gas = 1.108 kg m-3
Af is the flow area - (unknown calculated as 0.569 m2 using this equation)
u is the velocity of the...
As far as I am aware the A in the Φ=UAΔT equation gives you the surface area of the heat transfer. Which is what I'm after no?
That's what Steamking basically confirmed was ok to Mitch earlier? i.e Mitch said 'So, (1600-200)*1.15*.9*50,000=72450000kj/hr'
Steamking said, 'that looks much...
Φ =qmhCph (TH1-TH2)
= 50000 x 1.15(1600-200)
= 80.5x106 J/hr
with 10% heat loss = 80.5x106 x 0.9
=724.5 x 105 J/hr
Calculated ΔT as per MCtachyons post above (slightly different answer as i didn't round the 4.26 number)
Φ=UAΔT
724.5x105 = 54 x A(328.4)
A= 724.5x105/3600 / 54 x 328.4
= 1.13m2...
I am confused by this question as well.
When i calculate the area as per your last post MCtachyon, i end up with 1.13m2 ( i had a slightly different temp result than you - and 54x328.64 doesn't equal what you calculated its 17746.56)
But when i use this to calculate the number of tubes i end...