Question

Olive oil with a density of 830 kg/m^3 ,in a food factory needs to be pumped in a 30mm diameter pipe to a tank 10m above the pump with a rate of 18m^3/hr. The depth of fluid in the tank is 2.75m, what pressure must be added to lift the oil to the upper tank. Friction losses as 5kpa using this equation
P1+p v1^2 + h1 pg + trianglePe = P2+p v2^2 + h2 pg + trianglePf
2 2
trianglePe= pumping energy trianglePf=frictional losses
given that Power= pressure=volume flow

Answer 1

Answer on Question #39754, Engineering, Other

Question

Olive oil with a density of $830\ \mathrm{kg/m^3}$, in a food factory needs to be pumped in a 30 mm diameter pipe to a tank 10 m above the pump with a rate of $18\ \mathrm{m^3/hr}$. The depth of fluid in the tank is 2.75 m, what pressure must be added to lift the oil to the upper tank.

Friction losses as 5kpa using this equation

$$P_1 + \frac{\rho v_1^2}{2} + \rho g h_1 + \Delta P_e = P_2 + \frac{\rho v_2^2}{2} + \rho g h_2 + \Delta P_f$$

Solution

The Bernoulli equation states that,

$$P + \frac{\rho v^2}{2} + \rho g h = \text{constant}$$

In the above equation, $P$ is pressure, which can be either absolute or gage, but should be in the same basis on both sides, $\rho$ represents the density of the fluid, assumed constant, $v$ is the velocity of the fluid at the inlet/outlet, and $h$ is the elevation about a datum that is specified.

We use location 1 for "in" and location 2 for "out."

We have Bernoulli equation

$$P_1 + \frac{\rho v_1^2}{2} + \rho g h_1 + \Delta P_e = P_2 + \frac{\rho v_2^2}{2} + \rho g h_2 + \Delta P_f$$

Now list all the known information at the two locations.

$P_1 = 0$ gage (Open to atmosphere)

$v_1 = 0$ (Large cross-sectional area)

$h_1 = 0$ (By choice of datum)

From the given discharge rate $Q = 18\ \mathrm{m^3/hr} = 18/3600\ \mathrm{m^3/s}$ and the diameter of the pipe $d = 30\ \mathrm{mm}$ at the upper tank inlet, we can calculate $v_2$.

$$Q = v _ {2} A = v _ {2} \frac {\pi d ^ {2}}{4}$$ $$v _ {2} = \frac {4 Q}{\pi d ^ {2}}$$

where A is cross sectional area of a pipe

$$v _ {2} = \frac {4 Q}{\pi d ^ {2}} = \frac {4 \cdot 1 8}{3 . 1 4 \cdot (3 0 \cdot 1 0 ^ {- 3}) ^ {2} \cdot 3 6 0 0} = 7. 0 8 \mathrm {m / s}$$

p = 0 gage (Absorber is at atmospheric pressure)

$v_{2} = 7.08 \, \text{m/s}$ (from specified data)

$h_2 = 10 - 2.75 = 7.25 \, \text{m}$ (specified)

Substituting some of the known information into the above equation, we obtain

$$0 + \frac {\rho \cdot 0}{2} + \rho g \cdot 0 + \Delta P _ {e} = 0 + \frac {\rho v _ {2} ^ {2}}{2} + \rho g h _ {2} + \Delta P _ {f}$$ $$\Delta P _ {e} = \frac {\rho v _ {2} ^ {2}}{2} + \rho g h _ {2} + \Delta P _ {f}$$ $$\Delta P _ {e} = \frac {8 3 0 \cdot 7 . 0 8 ^ {2}}{2} + 8 3 0 \cdot 9. 8 1 \cdot 7. 2 5 + 5 \cdot 1 0 ^ {3} = 8 4 8 3 4. 1 \mathrm {P a} = 8 4. 8 \mathrm {k P a}$$

Answer. $\Delta P_{e} = 84.8\mathrm{kPa}$