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SHELL AND TUBE HEAT EXCHANGERS

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tomato5

Mechanical
Joined
Jan 18, 2012
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US
COULD YOU HELP WITH A CHEMICAL ENGINEERING PROBLEM

1. GASEOUS FLOW
2. AS OPERATING PRESSURE RISES, GAS DENSITY INCREASES CORRECT?
3. PRESSURE DROP IS DIRECTLY PROPORTIONAL TO SQUARE OF MASS VELOCITY ?
4. INVERSELY PROPORTIONAL TO DENSITY ?

PLEASE BRIEFLY OUTLINE YOUR REASONING, ESPECIALLY FOR 4

THANK YOU
 
2. Yes. PV=nRT [ρ]=n/V [ρ]=P/RT
3. Yes. [Δ]P=f(L/D)([ρ]v2/2)
4. No. [Δ]P=f(L/D)([ρ]v2/2) DIRECTLY PROPORTIONAL TO DENSITY, but if the equation is written in terms of mass velocity ([ρ]v), it may appear to be so. [Δ]P=f(L/D)(([ρ]v)2/(2[ρ])


Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.
 
For various (not piping) geometries, for example a bank of tubes or a packed bed, the friction factor f is generally an inverse function of the Reynolds number [Re] to a "n" power (generally "n" being smaller than 1). By being the Re itself proportional to density [ρ], it may be that the proportionality of [Δ]P to the density is, in fact, to [ρ](1-n).
 
For a given mass flow, higher the fluid density (say for a liquid) lower the pressure drop. On the other hand, lower the density of the fluid (a gas/vapor) higher the pressure drop. So pressure drop is inversely proportional to density.
 
A1nee,

For gas flow in ducts with usual Reynolds number in the range 100,000 to 500,000 and friction factor f=1/200, the fricion drop is estimated from:

[Δ]Pf = (L[÷]100Dh)[ρ]v2

This empirical equation shows what happens at constant mass flow rate from the effects of a drop in density [ρ] and the resulting increased velocity v.
The pressure drop would increase, but the density is still proportional to [Δ]Pf.
 
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