A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. The exit line enters a pipe that goes up to an elevation 20 m above the pump and river, where the water runs into an open channel. Assume the process is adiabatic and that the water stays at 10oC. Find the minimum pump power required.

Respuesta :

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

[tex]\Delta U = Q-W[/tex]

[tex]\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)[/tex]

Where

[tex]\dot{m} =[/tex] Mass flow

[tex]h_1 =[/tex]Specific Enthalpy (IN)

[tex]h_2 =[/tex] Specific Enthalpy (OUT)

[tex]g =[/tex] Gravity

[tex]z_{1,2} =[/tex] Heigth state (In, OUT)

[tex]V_{1,2} =[/tex]Velocity (In, Out)

Our values are given by,

[tex]T_i = 10\°C[/tex]

[tex]P_1 = 100kPa[/tex]

[tex]\dot{m} = 5kg/s[/tex]

[tex]z_2 = 20m[/tex]

For this problem we know that as pressure, temperature as velocity remains constant, then

[tex]h_1 = h_2[/tex]

[tex]V_1 = V_2[/tex]

Then we have that our equation now is,

[tex]\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}[/tex]

[tex]\dot{W} = \frac{(5)(9.81)(0-20)}{1000}[/tex]

[tex]\dot{W} = -0.98kW[/tex]