Examples – Internal Flows

 
These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

Oil with a density \rho_{\rm oil} = 900 kg/m^3 and kinematic viscosity \nu_{\rm oil}=10^{-5} m^2/s, flows at 0.2 m^3/s through a 500 m length of 300 mm diameter cast iron pipe. The average roughness of the pipe’s surface is 0.26 mm. Calculate: (a) The average flow velocity in the pipe; (b) The Reynolds number of the flow and explain if the flow is laminar or turbulent; (c) The pressure drop and head loss along the pipe; (d) The minimum power of the pump needed to move the oil.

(a) The average flow velocity is calculated from the volume flow rate, i.e.,

    \[ V_{\rm av} = \frac{\dot{Q}}{A }=  \frac{4\dot{Q}}{\pi d^2} = \frac{4 \times 0.2}{\pi \times 0.3^2} = 2.83~\mbox{m/s} \]

(b) The Reynolds number of the flow in the pipe is

    \[ Re_d = \frac{\rho_{\rm oil} V_{\rm av} d}{\mu_{\rm oil} } =\frac{V_{\rm av} d}{\nu_{\rm oil}} = \frac{2.83\times 0.3 }{10^{-5}} = 84,900 \]

so in this case the pipe flow will be turbulent because the Reynolds number is greater than 2,000 and we will need to use the Moody chart to find the friction factor f.

(c) We now need the pressure drop and head loss along the pipe. To use the Moody chart we also need the relative surface roughness, which is

    \[ \frac{e}{d} = \frac{0.26}{300} = 0.00087 \]

From the Moody chart for a Reynolds number of 84,900 and a relative roughness of 0.00087 (using interpolation) we have f \approx 0.022.  Therefore the pressure loss over the length of pipe is

    \[ \Delta p  = \frac{1}{2} \rho_{\rm oil} V_{\rm av}^2 f \left( \frac{L}{d}\right) = 0.5\times 900\times 2.83^2 \times 0.022\times \frac{500}{0.3} = 132.15~\mbox{kPa} \]

The corresponding head loss over this pipe is

    \[ h  = \frac{\Delta p}{\rho_{\rm oil} g} = \frac{132.15\times 10^3}{900\times 9.81} = 14.97~\mbox{m} \]

(d) The pumping power required will be

    \[ P_{\rm req} = \dot{Q} \, \Delta p_{AB} = 0.2 \times 132.15 \times 10^3 = 26.34\mbox{kW} \]

Worked Example #2

You are tasked with the design of the fuel delivery system. The system requires a flow through a 200 m length of smooth pipe of 15 mm diameter. The required fuel flow rate is 125 kg hr^{-1}. The fluid properties of the fuel are given as: \rho = 800 kg m^{-3} and \mu = 0.00164 kg m^{-1} s^{-1}. All entrance effects can be disregarded. (a) What is the pressure drop along the length of the pipe? (b) What pressure capability (in terms of head) is required of the pump? (c) What are the pumping power requirements?

(a) The cross-sectional area of the pipe is

    \[ A_c = \frac{\pi D^2}{4} = \frac{\pi \times 0.015^2}{4} = 0.0001767~\mbox{m$^2$} \]

The mass flow rate \dot{m} is given as 125~kg hr^{-1} = 0.0347~kg s^{-1}, i.e.,

    \[ \dot{m} = \rho \, A_c V_{\rm av} = 0.0347~\mbox{kg s$^{-1}$} \]

so the average flow velocity in the pipe is

    \[ V_{\rm av} = \frac{\dot{m}}{\rho \, A_c} = \frac{0.0347}{800.0 \times 0.0001767} = 0.246~\mbox{m s$^{-1}$} \]

The Reynolds number based on pipe diameter is

    \[ R\!e = \frac{\rho V_d \, D}{\mu} = \frac{ 800.0 \times 0.246 \times 0.015}{0.00164} = 1,797 \]

Notice that this Reynolds number is the laminar regime (so-called Couette flow) so the friction factor f is given by

    \[ f = \frac{64}{R\!e} = \frac{64}{1,797} = 0.0356 \]

The pressure drop \Delta p is given by

    \[ \Delta p = \frac{1}{2} \rho V_{\rm av}^2 f \frac{L}{D} = \frac{1}{2} \times 800.0 \times 0.246^2 \times 0.0356 \times \frac{20}{0.015} = 11,457.4~\mbox{Pa} = 11.46~\mbox{k Pa} \]

(b) The equivalent head loss h will be

    \[ h = \frac{\Delta p}{\rho \, g} = \frac{11,457.4}{800.0 \times 9.81} = 1.46~\mbox{m} \]

(c) The pumping power P can be determined from

    \[ P = \frac{\dot{m} \Delta p}{\rho} = \frac{0.0347 \times 11357.4}{800.0} = 0.497~\mbox{W} \]

Worked Example #3

Oil at 20 ^{\circ}C (where the density and viscosity are \rho = 888 kg/m^3 and \mu = 0.800 kg/m/s, respectively) is flowing steadily through a 6 cm diameter 40 m long pipe. The pressure at the pipe inlet and outlet are measured to be 745 kPa and 97 kPa, respectively. Determine the flow rate of oil through the pipe.

This pressure drop is a direct consequence of the action of viscous effects. The fluid is oil (dense and thick) so we expect the flow to be at very low Reynolds number (i.e., laminar), so from laminar pipe flow theory we have that

    \[ \Delta p = p_1 - p_2 = \frac{8 \mu \, L \, V_{\rm avg}}{R^2} = \frac{32 \, \mu \, L \, V_{\rm avg}}{D^2} \]

Notice that for a given length of pipe the pressure drop is proportional to the viscosity of the fluid and to the flow speed (so the faster we try to move a given fluid the larger the pressure drop will be) but inversely proportional to the square of the pipe diameter. The average flow velocity is

    \[ V_{\rm avg} = \frac{\left( p_1 - p_2 \right) D^2}{32 \, \mu \, L} = \frac{\left(745 - 97 \right) \times10^3 (0.06)^2}{(32) \, (0.8) \, (40)} = 2.28~\mbox{m/s} \]

We can now check to see that the flow is laminar by calculating the pipe Reynolds number, i.e.,

    \[ Re = \frac{\rho V_{\rm avg} D}{\mu} = \frac{(888.0) (2.28) (0.06)}{0.80} = 151.8 \]

so it is very low and the expectation of laminar flow is verified confirming the use of the correct equation for the pressure drop. The volume flow rate through the pipe will be

    \[ \dot{Q} = A \, V_{\rm avg} = \frac{\pi D^2}{4} V_{\rm avg} = \frac{\pi \left( p_1 - p_2 \right) D^4}{128 \, \mu \, L} \]

Substituting in the known values gives

    \[ \dot{Q} = \frac{\pi \left( p_1 - p_2 \right) D^4}{128 \, \mu \, L} = \frac{\pi \left(745 - 97 \right) \times10^3 (0.06)^4}{(128) \, (0.8) \, (40)} = 0.00644~\mbox{m$^3$/s} \]

The corresponding mass flow rate of oil will be

    \[ \dot{m} = \rho \dot{Q} = (888.0)(0.00644)= 5.72~\mbox{kg/s} \]

Worked Example #4

A coolant type of fluid used in the air-conditioning system of an aircraft is flowing through a smooth 0.12 inch diameter 30 ft long horizontal pipe steadily at an average flow speed of 3~ft/s. The fluid has a temperature of 40^{\circ}F, \rho = 1.93~slug ft^{-3}, and \mu = 3.326\times10^{-5}~slug ft^{-1}s^{-1}. Determine for this pipe flow: (a) The Reynolds number of the flow based on pipe diameter and whether the flow is laminar or turbulent; (b) The pressure drop along the length of the pipe; (c) The pumping power requirement to overcome this pressure drop; (d) Because the pump manufacturer provides pumps measured in units of “head of inches of water,” what is the minimum pump head would you need in this case?

The pressure losses associated with the pipe flow are determined using the equation

    \[ \Delta p_L = \frac{1}{2}\rho V_{\rm avg}^2 f \left( \frac{L}{D} \right) \]

where f is the Darcy-Welsbach friction factor. The friction factor f depends on the flow Reynolds number and the roughness of the pipe material. Hence, the pressure loss in a pipe is also a function of the Reynolds number and the roughness of the pipe. Remember that this pressure loss is caused by the frictional effects of viscosity, and loss is irrecoverable and appears as heat.

(a) The Reynolds number of the flow can be determined using the equation

    \[ Re = \frac{\rho_{} V_{\rm avg} D}{\mu_{}} = \frac{(1.93) (3) (0.12/12)}{3.326 \times 10^{-5}} = 1,740 < 2,300 \mbox{~~~~i.e., laminar Flow} \]

(b) The pressure loss can be determined using the equation

    \[ \Delta p_L = \frac{1}{2}\rho V_{\rm avg}^2 f \left( \frac{L}{D} \right) \]

where in this case the friction factor for the laminar flow in a smooth pipe is

    \[ f = \frac{64}{Re} \]

Therefore

    \[ \Delta p_L = \frac{1}{2}\rho V_{\rm avg}^2 \frac{64}{Re} \left( \frac{L}{D} \right) = 32\rho V_{\rm avg}^2 \left( \frac{L}{D} \right) \frac{\mu}{\rho V_{\rm avg}D} = \frac{32\mu V_{\rm avg}L}{D^2} \]

and so

    \[ \Delta p_L = \frac{32 (3.326 \times 10^{-5})(3)(30)}{(0.01)^2} = 957.89 \mbox{~lb/ft$^2$} \]

(c) The pumping power required to overcome this pressure drop is

    \[ P = \dot{Q} \, \Delta p_L \]

where the volume flow rate is

    \[ \dot{Q} = A_pV_{\rm avg} = \frac{\pi}{4}D^2V_{\rm avg} = \frac{\pi(0.01)^2(3)}{4} = 2.356 \times 10^{-4}~\mbox{ft$^3$/s} \]

Therefore

    \[ P = 2.536 \times 10^{-4} \times 957.89 = 0.257~\mbox{lb ft/s} \]

(d) The head loss can be determined using

    \[ h_L = \frac{\Delta p_L}{\rho_{\rm water}g} = \frac{957.89}{1.94 \times 32.17} = 15.35~\mbox{ft} = 184.15~\mbox{in~H}_2\text{O} \]

So, we would likely need to specify a pump that has a head capability of at least 185~\mbox{in~H}_2\text{O}.

Worked Example #5

In 1865, Van Syckle revolutionized petroleum oil delivery by using an oil pipeline. His oil company transported petroleum oil over a 5 statute mile long wrought iron pipe with a 2~inch diameter. If the company delivered 40 barrels per hour, what is the minimum power (in horsepower) required to pump the oil along the full length of the pipe? Assume \rho_{\rm oil} = 1.6046 slug ft^{-3} and \mu_{\rm oil} = 8.4627 \times 10^{-5} slug ft^{-1}s^{-1}. The roughness \epsilon = 0.045 mm for a wrought iron pipe. Assume 1 barrel = 5.615 ft^3.

To calculate the power we need to find the pressure loss \Delta p_L. To use the appropriate equation for the pressure loss, we need to know whether the flow is turbulent or laminar. For which, we need to know the Reynolds number

    \[ Re = \frac{\rho_{\rm oil} V_{\rm avg} D}{\mu_{\rm oil}} \]

For the above equation, the average flow velocity can be calculated based on the flow rate of the oil. Flow rate of the oil is

    \[ \dot{Q} = V_{\rm avg} \, \rm{Area} = V_{\rm avg} \frac{\pi}{4}D^2 = 40 \text{ barrels/hr} = \frac{40\times 5.615}{3600} = 0.062 \mbox{ ft}^3\mbox{/s} \rightarrow V_{\rm avg} = \frac{4\dot{Q}}{\pi D^2} = 2.86 \mbox{ ft/s} \]

Now the Reynolds number can be calculated, i.e.,

    \[ Re = \frac{\rho_{\rm oil} V_{\rm avg} D}{\mu_{\rm oil}} = \frac{(1.6046 \times 2.86)(2/12)}{8.4627 \times 10^{-5}} = 9,037 \]

This value of Reynolds number is larger than 4,000. Hence, the flow is turbulent. For a turbulent flow, the pressure loss can be determined using the equation

    \[ \Delta p_L = \frac{1}{2}\rho_{\rm oil}V_{\rm avg}^2f\frac{L}{D} \]

Here, the friction factor f is a function of relative roughness \epsilon/D and Reynolds number. For the wrought iron pipe, the relative roughness

    \[ \frac{\epsilon}{D} = \frac{0.045}{25.4 \times 2} = 0.00088 \]

Consulting the Moody chart for \epsilon/D = 0.00088 and Re = 9,037, we can find the friction factor, which is about f = 0.033. The pressure drop in the pipe can now be determined calculated, i.e.,

    \[ \Delta p_L = \frac{1}{2}\rho_{\rm oil}V_{\rm avg}^2f\frac{L}{D} = \frac{1}{2}\times 1.6046 \times (2.86)^2 \times 0.033 \left(\frac{5 \times 5280}{(2/12)}\right) = 3.43 \times 10^4 \mbox{lb/ft$^2$} \]

Required power to overcome this pressure loss is

    \[ \mbox{Power} = \Delta p_L \, \dot{Q} = \left(3.43 \times 10^4\right) (0.062) = 2140 \mbox{ lb ft/s} = 2140/550 = 3.89 \mbox{ hp} \]

Worked Example #6

Air enters a 50 ft long part of an air-conditioning duct made of smooth galvanized steel with a square cross-section of side 2 ft. The equivalent roughness of the duct is 0.00015 ft. The air is pushed through the duct with a fan at a volume flow rate of 1,800 ft^3/minute. (a) Determine the pressure drop and head loss along this part of the duct. (b) Determine the power needed to overcome the pressure losses over this part of the duct. Hints: 1. Disregard all entrance effects. 2. Consult the Moody chart.

(a) In this case we are dealing with a duct that has a rectangular cross-section, so we first need to find the hydraulic diameter D_h, which is

    \[ D_h = \frac{4 A_c}{p} \]

where A_c is the cross-sectional area and p is the perimeter of the section. We have a square cross-section of side a so the equivalent hydraulic diameter is

    \[ D_h = \frac{4 A_c}{p} = \frac{4 a^2}{4a} = a = 2\mbox{~ft} \]

We are given a hint to consult the Moody chart so the flow through the pipe is likely going to be turbulent. The equation for the pressure loss through a pipe is

    \[ \Delta p_L = \frac{1}{2} \rho V_{\rm avg}^2 f \left( \frac{L}{D} \right) \]

where f is the friction factor, still to be determined, which depends on the Reynolds number and pipe roughness and its value will come from the Moody chart.

The average flow velocity V_{\rm avg} can be found from the flow rate and cross-sectional area. The pipe area A_c is

    \[ A_c = a^2 = 2^2 = 4.0~\mbox{ft$^2$} \]

so the average flow velocity is

    \[ V_{\rm avg} = \frac{\dot{Q}}{A_c} = \frac{1,800}{60 \times 4.0} = 7.5~\mbox{ft/s} \]

To use the Moody chart we need the corresponding Reynolds number, which in this case is

    \[ Re = \frac{\rho V_{\rm avg} D_h}{\mu} = \frac{0.002378 \times 7.5 \times 2}{3.737 \times 10^{-7}} = 95,451 \]

where we find the density and viscosity of air are at ISA conditions.

Because in this case Re >10^3 then the flow through the duct is going to be in the fully turbulent regime. We also need the relative pipe roughness, which is

    \[ \frac{e}{D_h} = \frac{0.00015}{2} = 0.000075 = 7.5 \times 10^{-5} \]

Therefore, from the Moody chart the friction factor is

    \[ f \approx 0.018 \]

Returning to the calculation of the pressure drop then

    \begin{eqnarray*} \Delta p & = & \frac{1}{2} \rho V_{\rm avg}^2 \, f \left( \frac{L}{D_h} \right) \\ & = & \frac{1}{2} \times 0.002378 \times 7.5^2 \times 0.018 \times 50.0 / 2.0 \\ & = & 0.03~\mbox{lb ft$^{-2}$} \end{eqnarray*}

(b) The power required to move the air through this part of the duct and overcome this pressure drop is

    \[ P = \dot{Q} \, \Delta p_L \]

so

    \[ P = 1,800/60.0 \times 0.03 = 0.9~\mbox{ft lb / s} \]

Worked Example #7

You are asked to design an irrigation system in which water is pumped through a long horizontal pipe that is made of a smooth plastic tubing. The pipe is 150 m long and has a diameter of 10 cm. The pump output must provide a maximum volume flow rate of 0.09 m^3/s. For satisfactory operation, the sprinklers along the entire length of the pipe must operate at 205 kPa or higher pressure. (a) Find the pressure drop along the length of the pipe. (b) Calculate the minimum pressure p_1 that needs to be produced at the pump. (c) What pressure capability for the pump (in terms of head) would you specify? Notes: Assume turbulent flow. The density of water \rho_{\rm H_2O} = 1,000 kg/m^3 and the dynamic viscosity of water \mu_{\rm H_2O} = 1 \times 10^{-3} kg/m/s.

(a) The volume flow rate, \dot{Q} is given by

    \[ \dot{Q} = V_{\rm av} \, A \]

The average flow velocity can be calculated using

    \[ V_{\rm av} = {\dot{Q}}{A} = \frac{\dot{Q}}{\displaystyle{\frac{\pi D^2}{4}}} = \frac{4\dot{Q}}{\pi D^2} \]

    \[ V_{\rm av} = \frac{4 (0.09)}{\pi (0.1)^2} = 11.46~\mbox{m/s} \]

Calculating the Reynolds number gives

    \[ Re = \frac{\rho V_{\rm av} D}{\mu} = \frac{1000 \times 11.46 \times 0.1}{0.001} = 1.1 \times 10^6 \]

For a smooth pipe the roughness factor, \epsilon = 0.0015 mm and \epsilon/D = 1.5 \times 10^{-5}. Obtaining the friction factor from the Moody chart gives f = 0.01. So, pressure drop along the pipe can be calculated using

    \[ \Delta p = p_1 - p_2 = \frac{1}{2} \rho V_{\rm av}^2 f \frac{L}{D} \]

and substituting values gives

    \[ \Delta p = \bigg(\frac{1000 \times 11.46^2}{2}\bigg) \times 0.01 \bigg(\frac{150}{0.1}\bigg) = 985.0~\mbox{kPa} \]

(b) The pressure at the pump located at point 1 is then the required pressure at the last sprinkler p_2 plus the pressure drop

    \[ p_1 = p_2 + \Delta p \]

Substituting values in the above equation gives

    \[ p_1 = 205 + 985.0 = 1,190~\mbox{kPa} \]

(c) The corresponding head loss can be determined using

    \[ h_L = \frac{\Delta p}{\rho_{\rm H_2O } \, g} = \frac{1,190 \times 10^3}{1,000 \times 9.81} = 121.3~\mbox{m~H}_2\text{O} \]

So, we would likely need to specify a pump that has a head capability of at least 121.3~\mbox{m~H}_2\text{O}, which would be a fairly large pump.

Worked Example #8

Cooling water to a wind tunnel heat exchanger flows at a rate of 0.075 m^{3}/s through an asphalted cast-iron pipe that is 30 m long and 15 cm in diameter. Assume for water that its density \rho = 1,000 kg/m^3 and its kinematic viscosity \nu = 10^{-6}m^2/s, and the pipe material has an equivalent roughness = 0.12 mm. Hint: Consult the Moody Chart. (a) Using the basic principles of pipe flows, show how to calculate the pressure loss and head loss (i.e., frictional losses) along the length of the pipe. (b) Determine the pumping power required.

(a) The general equation for the pressure drop is

    \[ \Delta p = \frac{1}{2} \rho V_{\rm avg}^2 \, f \left( \frac{L}{D} \right) \]

where f is the friction factor, which depends on the Reynolds number and pipe roughness. The average flow velocity V_{\rm av} can be found from the flow rate and pipe area. The pipe area A_p is

    \[ A_p = \pi \left( \frac{D^2}{4} \right) = \pi \left( \frac{0.15^2}{4} \right) = 0.0177~\mbox{m$^2$} \]

so the average flow velocity is

    \[ V_{\rm av} = \frac{\dot{Q}}{A_p} = \frac{0.075}{0.0177} = 4.24~\mbox{m/s} \]

To use the Moody chart we need the Reynolds number, which in this case is

    \[ Re = \frac{V_{\rm av} D}{\nu} = \frac{4.24 \times 0.15}{10^{-6}} = 6.36\times 10^5 \]

We also need the relative roughness, which in this case is

    \[ \frac{e}{D} = \frac{0.00012}{0.15} = 0.0008. \]

Therefore, from the Moody chart it can be found that

    \[ f \approx 0.019 \]

Returning to the calculation of the pressure drop then

    \begin{eqnarray*} \Delta p & = & \frac{1}{2} \rho V_{\rm av}^2 \, f \left( \frac{L}{D} \right) \\ & = & \frac{1}{2} \times 1,000 \times 4.24^2 \times 0.019 \times \frac{30}{0.15} \\ & = & 32.36~\mbox{kN/m$^2$} \end{eqnarray*}

The corresponding head loss h_L can be determined using

    \[ h_L = \frac{\Delta p}{\rho_{\rm H_2O} \, g} = \frac{39.55 \times 10^3}{1,000 \times 9.81} = 3.3~\mbox{m} \]

(b) The pumping power required to overcome this pressure drop is

    \[ P = \dot{Q} \, \Delta p \]

where the volume flow rate is

    \[ \dot{Q} = 0.075~\mbox{m$^{3}$/s} \]

Therefore

    \[ P = 0.075 \times 39.55 \times 10^3 = 2.427 \times 10^3~\mbox{W} = 2.43~\mbox{kW} \]

 

Worked Example #9

Consider a pump pushing water steadily through a section of cast iron pipe that is 200 ft long, as shown in the figure below. The rectangular cross-section of the pipe has a height h = 3 in and a width w = 6 in. The internal surface roughness is \epsilon = 0.006 inches. The volumetric flow rate through the pipe is 0.75 ft^3 s^{-1}. Assume that the density of the water \rho_{\rm w} is 1.940 slug ft^{-3} and its viscosity \mu is 2.1\times 10^{-5} slug ft^{-1} s^{-1}. Also assume that entrance effects can be ignored.

    1. Determine the hydraulic diameter D_h of the pipe.
    2. Determine the average flow velocity V_{\rm av} though the pipe.
    3. Determine the Reynolds number Re_{D_h} of the flow in the pipe.
    4. Is the flow in the pipe laminar or turbulent? Explain.
    5. Determine the friction factor f of the pipe.
    6. Determine the pressure drop \Delta p and head loss \Delta h through the pipe.

1. The cross-sectional area of the rectangular pipe is

    \[ A = w \times h = (3.0/12.0)(6.0/12.0) = 0.125~\mbox{m$^2$} \]

so the hydraulic diameter is

    \[ D_h = \frac{4 \times \mbox{Area}}{\mbox{Perimeter}} = \frac{4A}{2 w h } =  \frac{2A}{w h }  = \frac{2.0 \times 0.125}{(3.0/12.0 + 6.0/12.0)} = 0.33~\mbox{ft} \]

2. The average flow velocity V_{\rm av} in the pipe is determined from the volumetric flow rate and the cross-sectional area, i.e.,

    \[ V_{\rm av} = \frac{\dot{Q}}{A} = \frac{0.75}{0.125} = 6.0~\mbox{ft/s} \]

3. The Reynolds number based on the hydraulic diameter is

    \[ Re_{D_h}  = \frac{\rho_{\rm w} \, V_{\rm av} \, D_h}{\mu_{\rm w}} = \frac{1.940 \times 6.0 \times 0.33}{2.1\times 10^{-5}} = 184,762 \approx 185,000 \]

4. Because the Reynolds number is greater than 4,000, the flow in the pipe will be turbulent.

5. The relative surface roughness is

    \[ \frac{\epsilon}{D_h} = \frac{0.006/12.0}{0.33} = 0.0015 \]

Using the Moody chart, the friction factor for this Reynolds number and relative roughness is about 0.023. Some small error in reading the chart is acceptable.

6. The pressure drop over a pipe length of 200 ft is

    \[ \Delta p = \frac{1}{2} \rho_ w \, V_{\rm av}^2 \, f \left(\frac{L}{D_h}\right) \]

and inserting the known values gives

    \[ \Delta p = 0.5 \times 1.94 \times  6.0^2 \times 0.0023 \left( \frac{200.0}{0.33} \right) = 480.0~\mbox{lb ft$^{-2}$} \]

The corresponding head loss is

    \[ \Delta h = \frac{\Delta p}{\rho_w \, g} = \frac{480.0}{1.94 \times 32.17} = 7.69~\mbox{ft} \]

 

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