Fluid Flow Questions involving Bernoulli Equation?

Question

Water at 15°C is pumped from a large reservoir to an elevated water storage tank through a 8 cm diameter pipe.

The water surface in the tank is located at an elevation of 26 m above the water surface in the reservoir.

Energy losses due to friction and pipe fittings are equivalent to a total of 18 m of water.

Determine how much energy must be added to the system (in J/kg) so that a constant pumping rate of 0.040 m3/s is maintained.

Answer:

a pipe with a diameter of 8cm and has a flow rate of 0.040 m^3/s has a velocity of 7.9577 m/s.

and you know the energy conservation equation is (v^2)/2 + g*h + P/rho

(7.9577^2)/2 + (18+26)*9.81 = 463.302 J/kg .... Theres no pressure difference

=========BUT, that's not the right answer...===============

So, what am I doing wrong?

Bomba · Accepted Answer

That is a huge  head loss for such a short run which I suppose is about 26-30 meters. Also the velocity is likewise huge. Perhaps that is why the head loss is so high.
Those (Euler) energy terms J/kg are : elevation head+ velocity head   + head loss + pressure head
 26  +  7.96^2 / 2*9.9   +   18  +    0  =  47 J/kg hydraulic energy

If you also assume 60% pump efficiency the energy applied to it is 47/0.6 = 78 J/kg
You can also add about 5% for motor effieiency loss.

Anonymous · Answer

I dont' have an entire answer yet perchance this can get you entering into the right route. the quantity of work being performed on each and each and every kg of water will be equivalent to lifting that bite of water 40 4 meters top plus friction loss. The rigidity utilized by making use of a million kg in earths gravitational container will be 9.8 m/s^2 * a million kg = 9.8 N the top is 40 4 m . 9.8 N * 40 4 m = N*m = joules. for the reason that foundation is a million kg that would nicely be joules consistent with kg of water.

alpha b · Answer

Just a hint.

If you have a answer just verify.

As per data given your actual head loss is ,

head loss (h) = -29.227573358308 meter


Bernoulli's equation is 

P_1 + 0.5*p*(v_1)^2 + p*g*(y_1) = P_2 + 0.5*p*(v_2)^2 + p*g*y_2 ,

P_1/p + 0.5*(v_1)^2 + g*(y_1) = P_2/p + 0.5*(v_2)^2 + g*y_2

0.5*(v_1)^2 + g*(y_1) = 0.5*(v_2)^2 + g*y_2

g*(y_1) = 0.5*(v_2)^2 + g*y_2 ...................v_1 is zero, Pressure difference is zero.

=>0.5*(v_2)^2  + g*(y_2 - y_1) + g*( head loss)

=>0.5*(7.9577)^2  + 9.81*(26) + 9.81*( 18 )

=>31.66 + 255.06 + 176.58

=>463.3 .......m^2/sec^2 ......... ( kg m^2/sec^2 /kg =J /kg)

Density = Mass / Volume 

=>Density*Volume =mass

=>mass=1000*0.040= 40 kg

where P_1 and P_2 are initial and final pressures, respectively, 

p is the density of the water, 

v_1 and v_2 are initial and final velocities, respectively, 

and y_1 and y_2 are initial and final heights, respectively. Each height is measured from the center of the pipe.

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Fluid Flow Questions involving Bernoulli Equation?

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