Fluid Mechanics

Fluid Mechanics

Similarity - Velocity

Water at 20^oC (viscosity = 1 cp) flows through a smooth straight pipe A of inside diameter 4 cm at an average velocity of 50 cm/sec. Oil flows through another pipe B of inside diameter 10 cm. Assuming similarities, calculate the velocity of oil through pipe B. Specific gravity of oil is 0.8 and its viscosity is 2 cp.

Data:

A:

Dia of A = 4 cm

Velocity of fluid in A = 50 cm/sec

Density of fluid in A = 1 g/cc

Viscosity of fluid in A = 1 cp

Dia of B = 10 cm

Density of fluid in B = 0.8 g/cc

Viscosity of fluid in A = 2 cp

Formulae:

For dynamic similarity, ratio of forces to be equal

Inertial force_A / Inertial force_B = Viscous force_A / Viscous force_B

i.e. NRe_A = NRe_B

Calculations:

NRe_A = NRe_B

Therefore,

4 x 50 x 1 / 1 = 10 x V_B x 0.8 / 2

200 = 4 V_B

V_B = 200/4 = 50 cm/sec

Velocity of oil through B = 50 cm/sec

Similarity - Pressure Drop

A geometrically similar model of an air duct is built 1:30 scale and tested with water which is 50 times more viscous and 800 times more dense than air. When tested under dynamically similar conditions, the pressure drop is 2.25 atm in the model. Find the corresponding pressure drop in the full-scale prototype.

Data:

Dimensions of the model = 1/30 times the dimensions of prototype

Viscosity of fluid in the model = 50 times the viscosity of fluid in prototype

Density of fluid in the model = 800 times the density of the fluid in prototype

Formulae:

For dynamic similarity, ratio of forces to be equal

Inertial force_m / Inertial force_p = Pressure force_m / Pressure foce_p

= Viscous force_m / Viscous force_p � 1

Pressure force = (2frLv²/ D)D² (i.e. = frictional pressure drop x area) � 2

From 1,

NRe_m = NRe_p

Calculations:

NRe_m = NRe_p

1 x v_m x 800 / 50 = 30 x v_p x 1 / 1

v_m = 1.875 v_p

Pressure force ratio = (2f_m x 800 x 1 x v_m² / 1) x 1²/ ((2f_p x 1 x 30 x (v_m / 1.875)² / 30) x 30²)

= 1600f_mv_m² / 512f_pv_m²

=3.125f_m/f_p

f = 0.079 (Dvr/m)^-0.25

f_m/f_p = (1 x v_m x 800 / 50)^-0.25 / (30 x (v_m/1.875) x 1 / 1)^-0.25 = 16^-0.25/16^-0.25 = 1

pressure ratio = 3.125 x 30² / 1² = 2812.5 ( i.e. pressure = pressure force/area)

Pressure drop of the prototype = 2.25/2812.5 atm = 0.0008 atm = 81.06 N/m²

U - tube Manometer

A U-tube manometer filled with mercury is connected between two points in a pipeline. If the manometer reading is 26 mm of Hg, calculate the pressure difference between the points when (a) water is flowing through the pipe (b) air at atmospheric pressure and 20^oC is flowing in the pipe.
Density of mercury = 13.6 gm/cc Density of water = 1 gm/cc Molecular weight of air = 28.8

Data:

Manometer reading(h) = 26 mm Hg = 0.026 m Hg

Density of mercury(r_m) = 13.6 gm/cc

Density of water = 1 gm/cc

Molecular weight of air(MW) = 28.8

Temperature of air = 20^oC = 293 K

R = 8314 J/(kmol.K)

Formulae:

For simple U - tube manometer,

P₁ - P₂ = Dp = (r_m- r)gh.

Ideal Gas law

PV = n RT

Molal density = n/V = P/(RT)

Mass density(r) = Molecular weight x molal density

Calculations:

(a) Water is flowing through the pipe:

Dp = (r_m- r)gh = (13600 - 1000) x 9.812 x 0.026 = 3214.4 N/m²

(b) Air at atmospheric pressure and 20^oC is flowing in the pipe:

r = 28.8 x 101325/(8314 x 293) = 1.2 kg/m³

Dp = (r_m- r)gh = (13600 - 1.2) x 9.812 x 0.026 = 3469.2 N/m²

Manometer Connected between Two pipes

A U - tube differential mercury manometer is connected between two pipes X and Y. Pipe X contains carbon tetra chloride (Sp.gr. 1.59) under a pressure of 103 kN/m² and pipe Y contains oil (Sp.gr. 0.8) under a pressure of 172 kN/m². Pipe X is 2.5 m above pipe Y. Mercury level in the limb connected to pipe X is 1.5 m below the centerline of pipe Y. Find the manometer reading as shown by a centimeter scale attached to it.

Data:

Fluid in pipe X = carbon tetra chloride

Density of Carbon tetra chloride = 1.59 x 1000 kg/m³

P_x = 103 kN/m²

Fluid in pipe Y= oil

Density of Carbon tetra chloride = 0.8 x 1000 kg/m³

P_y = 172 kN/m²

Formula:

P = rgh

Principle: Pressure at the same level in a continuous body of static fluid is equal.

Calculations:

Equating the pressure at the two legs of the manometer at OO':

For the left hand leg pressure at O = P_o = P_x + (2.5 + 1.5) x 1590 x 9.812 + a x 13600 x 9.812 N/m²

For the right hand leg pressure at O' = P_o' = P_y + (1.5 + a) x 800 x 9.812 N/m²

P_x + (2.5 + 1.5) x 1590 x 9.812 + a x 13600 x 9.812 = P_y + (1.5 + a) x 800 x 9.812

Substituting for P_x and P_y,

103000 + 62404.32 + 133443.2 x a = 172000 + 7849.6 x a + 11774.4

-18370.08 = - 125593.6 x a

a = 0.146 m = 14.6 cm

Manometer Reading = 14.6 cm of Hg

Unit II

Flow Direction

The pressures at two sections of a horizontal pipe are 0.3 kgf/cm² and 0.6 kgf/cm² and the diameters are 7.5 cm, and 15 cm respectively. Determine the direction of flow if water flows at a rate of 8.5 kg/sec. State your assumptions.

Data:

P₁= 0.3 kgf/cm²

P₂ = 0.6 kgf/cm²

D₁= 7.5 cm

D₂ = 15 cm

Mass flow rate = 8.5 kg/sec

Formulae:

Equation of continuity
r₁A₁v₁ = r₂A₂v₂
Bernoulli's equation
For the flow direction from 1 to 2,
Mass flow rate = volumetric flow rate x density

Calculations:

Volumetric flow rate = 8.5/1000 = 8.5 x 10^-3 m³/sec

V₁ = 8.5 x 10^-3/(pD₁²/4) = 8.5 x 10^-3/(p x 0.075²/4) = 8.5 x 10^-3/0.00441= 1.924 m/sec

V₂ = 8.5 x 10^-3/(pD₂²/4) = 8.5 x 10^-3/(p x 0.15²/4) = 8.5 x 10^-3/0.01767 = 0.481 m/sec

P₁ = 0.3 kgf/cm² = 0.3 x 9.812 N/cm² = 2.9436 x 10⁴ N/m²

P₂ = 0.6 kgf/cm² = 0.6 x 9.812 N/cm² = 5.8872 x 10⁴ N/m²

Assuming the flow direction is from 1 to 2:

2.9436 x 10⁴/1000 + 1.924²/2 = 5.8872 x 10⁴/1000 + 0.481²/2 + h + w -q

29.436 + 1.851 = 58.872 + 0.116 + h + w -q

In the given problem work done by fluid (w) and pump work on fluid (q) are zero.

So to balance the above equation, the quantity h has to have negative values. This is not possible.

The above equation will be a correct one, if the flow is from 2 to 1.

i.e. 58.872 + 0.116 = 29.436 + 1.851 + h

Therefore the flow direction is from the end at which pressure is 0.6 kgf/cm² and diameter is 15 mm to the end at which pressure is 0.3 kgf/cm² and diameter is 7.5 mm.

Siphon

A siphon consisting of a 3 cm diameter tube is used to drain water from a tank. The outlet end of the tube is 2 m below the water surface in the tank. Neglecting friction, calculate the discharge. If the peak point of the siphon is 1.4 m above the water surface in the tank, estimate the pressure of fluid at the point of siphon.

Data:

Dia of tube = 3 cm = 0.03 m

Formula:

Bernoulli's equation for frictionless flow is

Discharge(Q) = cross sectional area x velocity

Calculations:

Applying Bernoulli's equation for the points 1 and 3, ( i.e. comparing the energy levels for the fluid in the tank surface and at the discharge point of tube)

p₁= 0 N/m²(g)

p₃= 0 N/m²(g)

z₁ = 0 m

z₃ = -2 m

Since the rate of fall of liquid level in the tank is almost negligible,

v₁ = 0 m/sec.

Therefore,

0 + 0 + 0 = 0 + (v₃² / 2g) - 2

v₃ = (2 x 2g)^0.5 = 6.265 m/sec

Q = (p/4)D² v = (p/4) x 0.03² x 6.265 = 0.00443 m³/sec = 15.94 m³/hr

Applying Bernoulli's equation for the points 1 and 2, ( i.e. comparing the energy levels for the fluid at the tank surface to the peak point of siphon)

p₁= 0 N/m²(g)

z₂ = 1.4 m

v_{2 =}v₃= 6.265 m/sec (since the cross sectional area of sections 2 and 3 are the same)

0 + 0 + 0 = p₂ / (rg) + 6.265² / (2g) + 1.4

p₂ / (rg) = -3.4 m

p₂= -3.4 x 1000 x 9.812 N/m²(g) = -33360.8 N/m²(g)

Absolute pressure at point 2 = 101325 - 33360.8 = 67964.2 N/m²(a)

Capillary Tube Viscosity Measurements

A capillary tube 0.2 cm in diameter and 10 cm long discharge one liter of a liquid in ten minutes under a pressure difference of 5 cm mercury. Find the viscosity of the liquid using the following data:
Density of oil = 850 kg/m³, Density of mercury = 13600 kg/m³
Data:

Dia of tube(D) = 0.2 cm = 0.002 m

Length of pipe(L) = 10 cm = 0.1 m

Density of mercury = 13600 kg/m³

Density of oil(r) = 850 kg/m³

Pressure drop(Dp) = 5 cm Hg = 0.05 x 13600 x 9.812 N/m² = 6672.16 N/m²

Flow rate(Q) = 1 litre/10 min = 0.001/(10 x 60) = 1.667 x 10^-6 m³/sec

Formula:

Hagen-Poiseuille law (pressure drop for laminar flow is related to the flow parameters)

Calculations:

In capillary tube viscosity measurements the flow is maintained in laminar conditions.

m = (p x 6672.16 x 0.002⁴)/(1.667 x 10^-6 x 128 x 0.1) = = 0.01572 kg/(m.sec) = 15.72 cp

Frictional Losses

2.16 m³/h water at 320 K is pumped through a 40 mm I.D. pipe through a length of 150 m in a horizontal direction and up through a vertical height of 12 m. In the pipe there are fittings equivalent to 260 pipe diameters. What power must be supplied to the pump if it is 60% efficient? Take the value of fanning friction factor as 0.008. Water viscosity is 0.65 cp, and density = 1 gm/cc.

Data:

Flow rate(Q) = 2.16 m³/h = 2.16/3600 m³/sec = 0.0006 m³/sec

Dia of pipe(D) = 40 mm = 0.04 m

Length of pipe in horizontal direction = 150 m

Length of pipe in vertical direction(Dz) = 12 m

Equivalent length of fittings = 260 pipe diameters

Friction factor(f) = 0.008

Efficiency of pump(h ) = 0.6

Viscosity of fluid(m) = 0.65 cp = 0.00065 kg/(m.sec)

Density of fluid(r) = 1 gm/cc = 1000 kg/m³

Formulae:

Bernoulli's equation

Frictional losses per unit mass of flowing fluid

Power required for pumping = (mass flow rate x head developed by pump)/ h
= (volumetric flow rate x pressure developed by pump)/ h

Calculations:

Length of pipe with fittings = 150 + 12 + 260 x 0.04 m = 172.4 m

Velocity = volumetric flow rate/flow cross sectional area

= 0.0006/((p/4) x 0.04²) = 0.477 m/sec

h_f = 2 x 0.008 x 172.4 x 0.477² / 0.04 = 15.69 m²/sec²

frictional losses per unit weight of fluid(h) = h_f / g = 15.69/9.812 = 1.6 m

pump head(q) = Dz + h = 12 + 1.6 = 13.6 m

pressure developed by pump = 13.6 x 1000 x 9.812 = 133443.2 N/m²

power required for pumping = 0.0006 x 133443.2 / 0.6 = 133.4 watt = 133.4/736 HP = 0.181 HP

Frictional Losses - with fittings

Water is pumped from a reservoir to a height of 1000 m from the reservoir level, through a pipe of 15 cm I.D. at an average velocity of 4 m/s. If the pipeline along with the fittings is equivalent to 2000 m long and the overall efficiency is 70%, what is the energy required for pumping? Friction factor f = 0.046 Re^-0.2.

Data:

Diameter of pipe(D) = 15 cm

Average Velocity(n) = 4 m/s

Equivalent Length of pipe with fittings = 2000 m

efficiency of pump(h) = 0.7

Formula:

Bernoulli equation:
Frictional losses:
Friction factor Relation:
f = 0.046 Re^-0.2

Calculations:

Re = Dnr/m

= 0.15 x 4 x 1000/0.001 = 600000

f = 0.046 x (600000)^-0.2 = 0.003215

h_f = 2 x 0.003215 x 2000 x 4²/0.15 = 1371.73 m²/sec²

Frictional losses per unit weight (h)= h_f/g = 1371.73/9.812 = 139.8 m

Power required for pumping = mass flow rate x g x h / efficiency of pump

Mass flow rate = volumetric flow rate x density

= velocity x cross sectional area x density

= 4 x (pD²/4) x r

= p x 0.15² x 1000 = 70.686 kg/sec

power required = 70.686 x 9.812 x 139.8/0.7 = 96961 watt = 138515.7 watt = 138.5 KW

Unit III

Venturi - Power required

A horizontal venturi meter having a throat diameter of 4 cm is set in a 10 cm I.D. pipeline. Water flows through the system and the pressure differential across the venturi meter is measured by means of a simple U-tube manometer filled with mercury. Estimate the flow rate when the manometer reading is 30 cm. Assume C_v= 0.98. If 10% of the pressure differential is permanently lost, calculate the power consumption of the meter.

Data:

Dia of pipe (D_a) = 10 cm = 0.1 m

Dia of throat (D_b) = 4 cm = 0.04 m

Manometer reading (h_m) = 30 cm of Hg

Venturi coefficient (C_v) = 0.98

Permanent pressure loss = 10 % of the pressure differential measured by the manometer.

r_m = 13.6 g/cc = 13.6 x 10^-3 kg/m³

r = 1 g/cc = 1000 kg/m³

Formulae:

Flow rate = Velocity at the throat x cross sectional area of throat

Velocity at the throat

Where b = D_b / D_a

The pressure difference measured by the manometer

P_a - P_b = (r_m - r<)gh_m

Power consumption of the meter = volumetric flow rate x permanent pressure loss

Calculations:

P_a - P_b = (13600 - 1000) x 9.812 x 0.3 = 37089.4 N/m²

b = 0.04/0.1 = 0.4

V_b = (0.98/(1-0.4⁴)^0.5) x (2 x 37089.4 / 1000)^0.5 = 0.993 x 8.613 = 8.553 m/sec

Cross sectional area of throat = pD_b²/4 = p x 0.04² / 4 = 0.00126 m²

Volumetric flow rate = 8.553 x 0.00126 = 0.01078 m³/sec = 10.78 lit/sec

Permanent pressure loss = 0.1 x 37089.4 = 3708.94 N/m²

Power consumption of the meter = 0.01078 x 3708.94 = 40 watt

Venturi - Coefficient:

The rate of flow of water in a 150 mm diameter pipe is measured with a venturi meter of 50 mm diameter throat. When the pressure drop over the converging section is 100 mm of water, the flow rate is 2.7 kg/sec. What is the coefficient of the meter?

Data:

Dia of pipe (D_a) = 150 mm = 0.15 m

Dia of throat (D_b) = 50 mm = 0.05 m

Pressure drop (P_a - P_b) = 100 mm of water = (0.1/10.33) x 101325 N/m² = 980.9 N/m²

Mass flow rate = 2.7 kg/sec

Formulae:

Velocity at the throat

Where b = D_b/D_a

Calculations:

Volumetric flow rate = mass flow rate / density = 2.7 / 1000 = 0.0027 m³/sec

Velocity at the throat = Volumetric flow rate / cross sectional area of throat

= 0.0027 / (pD_b²/4) = 0.0027 / (p x 0.05²/4) = 1.375 m/sec

Therefore,

C_v = 1.375 x ( 1 - (0.05/0.15)⁴)^0.5 / (2 x 980.9/1000)^0.5 = 0.976

Orifice Size

Brine of specific gravity 1.2 is flowing through a 10 cm I.D. pipeline at a maximum flow rate of 1200 liters/min. A sharp edged orifice connected to a simple U-tube mercury manometer is to be installed for the purpose of measurements. The maximum reading of the manometer is limited to 40 cm. Assuming the orifice coefficient to be 0.62, calculate the size of the orifice required.

Data:

Density of brine (r) = 1.2 x 1000 kg/m³ = 1200 kg/m³

Dia of pipe (D_a) = 10 cm = 0.1 m

Maximum flow rate (Q) = 1200 liters/min = 1.2 m³/min = 0.02 m³/sec

Maximum manometer reading (h_m) = 40 cm = 0.4 m of Hg

Density of manometric fluid (r_m) = 13600 kg/m³

Orifice coefficient (C_o) = 0.62

Formulae:

Velocity at the orifice

Where b = D_b/D_a

D_b = dia of orifice

D_a = dia of pipe

For the U-tube manometer,

P_a - P_b = (r_m - r<)gh_m

Calculations:

P_a - P_b = (13600 - 1200) x 9.812 x 0.4 = 48667.5 N/m²

The quantity (1 - b⁴) will be approximately 1.

Therefore,

V_b = 0.62 x ( 2 x 48667.5 / 1200 )^0.5 = 5.584 m/sec

Cross sectional area of orifice = volumetric flow rate / velocity at the orifice = 0.02 / 5.584 = 0.00358 m²

Dia of orifice (D_b) = (0.00358 x 4/p)^0.5 = 0.0675 m = 6.75 cm

Pitot tube

A Newtonian fluid having a viscosity of 1.23 poise, and a density of 0.893 gm/cm³, is flowing through a straight, circular pipe having an inside diameter of 5 cm. A pitot tube is installed on the pipeline with its impact tube located at the center of the pipe cross section. At a certain flow rate, the pitot tube indicates a reading of 8 cm of mercury. Determine the volumetric flow rate of the fluid.

Data:

Viscosity of fluid (m) = 1.23 poise = 0.123 kg/(m.sec)

Density of fluid (r) = 0.893 gm/cm³= 893 kg/m³

Dia of pipe (D) = 5 cm = 0.05 m

Manometer reading (h_m) = 8 cm of Hg = 0.08 m of Hg

Formulae:

Velocity at the center of pipe

V_o = ( 2 x (P_s - P_o ) / r )^0.5 = ( 2 x h_m x (r_m - r<) x g / r )^0.5

Calculations:

V_o = ( 2 x 0.08 x (13.6 - 0.893) x 9.812 / 0.893 )^0.5 = 4.73 m/sec

NRe = D x V_o x r / m = 0.05 x 4.73 x 893 / 0.123 = 1717

Since the flow is laminar (NRe < 2100), the average velocity is given by

V_avg = V_o / 2 = 4.73 / 2 = 2.365 m/sec (velocity at the centerline is the maximum velocity)

Volumetric flow rate = V_avg x p x D² / 4 = 2.365 x p x 0.05² / 4 = 0.00464 m³/sec = 4.64 lit/sec

Rotameter

A rotameter calibrated for metering has a scale ranging from 0.014 m³/min to 0.14 m³/min. It is intended to use this meter for metering a gas of density 1.3 kg/m³ with in a flow range of 0.028 m³/min to 0.28 m³/min. What should be the density of the new float if the original one has a density of 1900 kg/m³? Both the floats can be assumed to have the same volume and shape.

Data:

Old:

Density of float (r_f)= 1900 kg/m³

Density of gas (r) = 1.3 kg/m³

Q_min = 0.014 m³/min

New:

Density of gas (r) = 1.3 kg/m³

Q_min = 0.028 m³/min

Formula:

Where Q = volumetric flow rate

A₂ = area of annulus (area between the pipe and the float)

A₁ = area of pipe

A_f = area of float

V_f = volume of float

C_D = rotameter coefficient

Calculations:

From the equation, for the same float area and float volume and the pipe geometry,

Q = k (r_f - r<)^0.5

where k is a proportionality constant

Q_new / Q_old = ( (r_fnew - r<) / (r_fold - r<) )^0.5

i.e. 0.028 / 0 .014 = ( (r_fnew- 1.3) / (1900 - 1.3) )^0.5

2 = (r_fnew- 1.3)^0.5 / 43.574

(r_fnew- 1.3)^0.5 = 87.15

(r_fnew- 1.3) = 7595.1

r_fnew = 7596 kg/m³

Rectangular Weir

A rectangular weir 0.75 m high and 1.5 m long is to be used for discharging water from a tank under a head of 0.5 m. Estimate the discharge (i) when it is used as a suppressed weir (ii) when it is used as a contract weir. Use Rehbock equation for estimating C_d in both cases.

Data:

Weir height (P) = 0.75 m

Width of weir (B) = 1.5 m

Head (H) = 0.5 m

Formulae:

Rehbock equation

H and P in meter

Suppressed weir

Contracted weir

Where n = number of contractions

Q = flow rate

Calculations:

Suppressed weir:
C_d = 0.605 + 1 / (1000 x 0.5) + 0.08 x 0.5 / 0.75 = 0.66

Q = 0.66 x (2/3) x 1.5 x (2 x 9.812)^0.5 x 0.5^3/2 = 1.034 m³/sec
Contracted weir
Q = 0.66 x (2/3) x (1.5 - 0.1 x 2 x 0.5) x (2 x 9.812)^0.5 x 0.5^3/2 = 0.965 m³/sec

Unit IV

Frictional Losses through a Packed Column

Figure shows a water softener in which water trickles by gravity over a bed of spherical ion-exchange resin particles, each 0.05 inch in diameter. The bed has a porosity of 0.33. Calculate the volumetric flow rate of water.

Calculations:

Data:

m = 1 cp = 1 x 6.72x10^-4 lb/(ft.sec)

e = 0.33

r = 62.3 lb/ft³

Dx = 1 ft

g = 32.2 ft/sec²

Formula:

Applying Bernoulli's equation from the top surface of the fluid to the outlet of the packed bed and ignoring the kinetic-energy term and the pressure drop through the support screen, which are both small, we find

g(Dz) = h_f

h_f = Dp/r

For laminar flow, (Blake-Kozeny Equation)

Calculations:

Therefore, V_s = 32.2 x 1.25 x (0.05/12)² x 0.33³ x 62.3 / ( 150 x (1 x 6.72x10^-4) x (1 - 0.33)² x 1)

= 0.035 ft/sec = 0.011 m/sec.

Q = AV_s = (2/12)² x (p/4) x 0.035 = 0.00075 ft3/sec = 21 cm³/sec.

Before accepting this as the correct solution, we check the NRe_m.

NRe_m= (0.05/12) x 0.035 x 62.3 / (1 x 6.72x10^-4 x (1 - 0.33) ) = 20.2

This is slightly above the value of 10 (up to which the Blake-Kozeny Equation can be used), for which we can safely use without appreciable error.

Air flow through packed bed

Calculate the pressure drop of air flowing at 30^oC and 1 atm pressure through a bed of 1.25 cm diameter spheres, at a rate of 60 kg/min. The bed is 125 cm diameter and 250 cm height. The porosity of the bed is 0.38. The viscosity of air is 0.0182 cP and the density is 0.001156 gm/cc.

Data:

Mass flow rate of Air = 60 kg/min = 1 kg/sec

Density of Air (r) = 0.001156 gm/cc = 1.156 kg/m³

Viscosity of Air (m) = 0.0182 cP = 0.0182 x 10^-3 kg/(m.sec)

Bed porosity (e) = 0.38

Diameter of bed (D)= 125 cm = 1.25 m

Length of bed (L) = 250 cm = 2.5 m

Dia of particles (D_p)= 1.25 cm = 0.0125 m

Sphericity (F _s) = 1 (sphere)

Formulae:

NRe_PM= D_pV_or/(m (1 - e) )

For laminar flow (i.e. NRe_PM < 10) pressure drop is given by Blake-Kozeny equation.

For turbulent flow (i.e. NRe_PM > 1000) pressure drop is given by Burke-Plummer equation.

For intermediate flows pressure drop is given by Ergun equation

Superficial velocity V_o = Volumetric flow rate/ cross-sectional area of bed

Calculations:

Volumetric flow rate = mass flow rate / density = 1 / 1.156 = 0.865 m³/sec

Superficial velocity V_o = 0.865 / ( (p/4) D² ) = 0.865 / ( (p/4) 1.25² ) = 0.705 m/sec

NRe_PM= 0.0125 x 0.705 x 1.156 / (0.0182 x 10^-3 x ( 1- 0.38 ) ) = 903

We shall use Ergun equation to find the pressure drop.

i.e. Dp x 0.0125 x 0.38³ / ( 2.5 x 1.156 x 0.705² x ( 1 - 0.38 ) ) = 150 / 903 + 1.75

Dp x 7.702 x 10^-4 = 1.92

Dp = 1.92 / 7.702 x 10^-4 = 2492.92 N/m²

Regenerative Heater packed with cubes

A regenerative heater is packed with a bed of 6 mm cubes. The cubes are poured into the cylindrical shell of the regenerator to a depth of 3.5 m such that the bed porosity was 0.44. If air flows through this bed entering at 25^oC and 7 atm abs and leaving at 200^oC, calculate the pressure drop across the bed when the flow rate is 500 kg/hr per square meter of empty bed cross section. Assume average viscosity as 0.025 cP and density as 6.8 kg/m³.

Data:

Mass flow rate of Air / unit area = 500 kg/(hr.m²) = 0.139 kg/(sec.m²)

Density of Air (r) = 6.8 kg/m³

Viscosity of Air (m) = 0.025 cP = 0.025 x 10^-3 kg/(m.sec)

Bed porosity (e) = 0.44

Length of bed (L) = 3.5 m

Dia of particles (D_p)= 6 mm = 0.006 m

Formulae:

Sphericity (F _s) = 6 v_p / (D_pS_p)

v_p = volume of particle = D_p³ (for cube)

S_p = surface area of particle = 6 x D_p² (for cube)

NRe_PM= D_pV_or/(m (1 - e) )

For laminar flow (i.e. NRe_PM < 10) pressure drop is given by Blake-Kozeny equation.

For turbulent flow (i.e. NRe_PM > 1000) pressure drop is given by Burke-Plummer equation.

For intermediate flows pressure drop is given by Ergun equation

Superficial velocity V_o = Volumetric flow rate/ cross-sectional area of bed

Calculations:

Superficial velocity V_o = mass flow rate per unit area / density = 0.139 / 6.8 = 0.0204 m/sec

(F _s) = 6 v_p / (D_pS_p) = 6 x 0.006³ / (0.006 x 6 x 0.006²) = 1

NRe_PM= 0.006 x 0.0204 x 6.8 / (0.025 x 10^-3 x ( 1- 0.44 ) ) = 59.45

We shall use Ergun equation to find the pressure drop.

i.e. Dp x 0.006 x 0.44³ / ( 3.5 x 6.8 x 0.0204² x ( 1 - 0.44 ) ) = 150 / 59.45 + 1.75

Dp x 0.0712 = 4.273

Dp = 4.273 / 0.0712 = 60 N/m²

Berl Saddle packing

7000 kg/hr of air, at a pressure of 7 atm abs and a temperature of 127^oC is to be passed through a cylindrical tower packed with 2.5 cm Berl saddles. The height of the bed is 6 m. What minimum tower diameter is required, if the pressure drop through the bed is not to exceed 500 mm of mercury?
For Berl saddles, D p = (1.65 x 10⁵ Z V_s^1.82 r ^1.85 )/D_p^1.4
where D p is the pressure drop in kgf/cm², Z is the bed height in meter, r is the density in g/cc, D_pis nominal diameter of Berl saddles in cm, V_sis the superficial linear velocity in m/sec.

Data:

Mass flow rate = 7000 kg/hr = 1.944 kg/sec

Height of bed (Z) = 6 m

D_p = 2.5 cm

760 mm Hg = 1 kgf/cm² = 1 atm

D p = 500 mm Hg = (500/760) x 1 kgf/cm² = 0.65 kgf/cm²

Formula:

Ideal gas law:

PV = nRT

Formula given,

D p = (1.65 x 10⁵ Z V_s^1.82 r ^1.85 )/D_p^1.4

Calculations:

r = Mn/V = MP/(RT) = 29 x 7 x 1.01325 x 10⁵ / (8314 x (273 + 127) ) = 6.185 kg/m³ = 6.185 x 10^-3 g/cc

D p = (1.65 x 10⁵ Z V_s^1.82 r ^1.85 )/D_p^1.4

0.65 = (1.65 x 10⁵ x 6 x V_s^1.82 x (6.185 x 10^-3 )^1.85 ) / 2.5^1.4

V_s^1.82 = 4.603 x 10^-5

V_s = 0.029 m/sec.

Volumetric flow rate = mass flow rate/density = 1.944/6.185 = 0.3144 m³/sec

Cross sectional area = Volumetric flow rate / V_s = 0.3144 / 0.029 = 10.84 m²

Required Minimum Diameter (D) = (A x (4/p) )^0.5 = (10.84 x 4/p)^0.5 = 3.715 m.

Pressure drop through catalyst tower

A catalyst tower 50 ft high and 20 ft in diameter is packed with 1-in. diameter spheres. Gas enters the top of the bed at a temperature of 500^oF and leaves at the same temperature. The pressure at the bottom of the bed is 30 lb_f/in.² abs. The bed porosity is 0.40. If the gas has average properties similar to propane and the time of contact (based on superficial velocity of gas) between the gas and the catalyst is 10 s, what is the inlet pressure?

Data:

Dia of tower (D) = 20 ft = 20 x 0.3038 m = 6.096 m

Height of tower (L) = 50 ft = 15.24 m

Dia of packing (D_p) = 1 inch = 2.54 cm = 0.0254 m

Temperature of gas (T) = 500^oF = (500 - 32) x 5/9 ^oC = 260^oC = (273 + 260) K = 533 K

Pressure at the bottom = 30 lb_f/in.² abs = 30/14.7 atm(a) = 2.04 atm(a) = 206785.7 N/m²(a)

Bed porosity (e) = 0.4

Time of contact = 10 sec

Molecular weight of gas = 44 (molecular weight of propane (C₃H₈) )

Formulae:

Density of gas (r) = PM/(RT)

Ergun equation:

NRe_PM= D_pV_or/( ( 1 - e )m )

Calculations:

Density of the leaving gas,

r = 206785.7 x 44 / (8314 x 533) = 2.053 kg/m³

V_o = Height / time = 15.24 / 10 = 1.524 m/sec

Taking the viscosity of gas as that of air (m = 0.025 x 10^-3 kg/(m.sec) )

NRe_PM= 0.0254 x 1.524 x 2.053 / ( (1 - 0.4) x 0.025 x 10^-3 ) = 5298

For these NRe_PM Burke-Plummer equation can be used.(i.e. Turbulent part of the Ergun's equation)

Dp x 0.0254 x 0.4³ / ( 15.24 x 2.053 x 1.524² x (1-0.4) ) = 1.75

Dp = 46937.5 N/m²

Pressure at the inlet of the column = 206785.7 + 46937.5 = 253723.2 N/m²(a) = 36.81 lb_f/in²(a)

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