A pump is to deliver 10, 000 kg/h of toluene at 1140C and 1.1 atm absolute pressure from the Reboiler of a distillation tower to the second distillation unit without cooling the toluene before it enters the pump. If the friction loss in the line between the Reboiler and the pump is 7 kN/m2. The density of toluene is 886 kg/m3. How far above the pump must the liquid be maintained to avoid cavitation

Respuesta :

Answer:

3.4093

Explanation:

NPSHa = hatm + hel + hf +hva

the elevation head is the hel

friction loss head is hf

NPSHa is the head of vapour pressure of fluid

atmospheric pressure head is hatm

log₁₀P* = [tex]A -\frac{B}{C+T}[/tex]

[tex]A, B, C are fixed[/tex]

log₁₀Pv = [tex]4.07827-\frac{1343.943}{387.15-53.773}[/tex]

= 4.07827 - 1343.943/333.377

=4.07827 - 4.0313009

= 0.0469691

we take the log

p* = 1.114218

we convert this value to get 111421.8

hvap = 111421.8 * 1/776.14 * 1/9.81

= 14.63

hatm = 1.1 *101325/1 * 1/9.81 *1/776.14

=14.64

hf = 7000/1 * 1/776.14 * 1/9.81

= 0.9193

NPSHa = 2.5

hel = 0.9193 + 2.5 + 14.63 - 14.64

hel = 3.4093

The NSPH values are used to calculate cavitation. The vapor pressure of the liquid is 1.114 atm.

The vapor pressure can be calculated by,

[tex]\mathrm {NPSH_A}= ( \frac {p_i}{\rho g} + \frac {V_i^2}{2g})- \frac {p_v}{\rho g}[/tex]

Where,

[tex]\mathrm {NPSH_A}[/tex] = available NPSH

[tex]p_i[/tex]     = absolute pressure at the inlet = 1.1 atm

[tex]V_i[/tex]     = average velocity at the inlet = 10, 000 kg/h

[tex]\rho[/tex] = fluid density = 886 kg/m3.  

g = acceleration of gravity = 9.8 m/s²

[tex]p_v[/tex] = vapor pressure of the fluid = ?

Put the values in the equation, we get

[tex]p_v = 1.114\ atm[/tex]

Therefore, the vapor pressure of the liquid is 1.114 atm.

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