SHELL AND TUBE HEAT EXCHANGERS

[tomato5]

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

 

[Latexman]

2. Yes. PV=nRT [ρ]=n/V [ρ]=P/RT
3. Yes. [Δ]P=f(L/D)([ρ]v[sup]2[/sup]/2)
4. No. [Δ]P=f(L/D)([ρ]v[sup]2[/sup]/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)[sup]2[/sup]/(2[ρ])


Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[25362]

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][sup](1-n)[/sup].

 

[A1nee]

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.

 

[25362]

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]P[sub]f[/sub] = (L[÷]100D[sub]h[/sub])[ρ]v[sup]2[/sup]​

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]P[sub]f[/sub].