A pump is driven by a 2-kW motor with water at 150 kPa, 10 Cesius entering. Find the maximum water flow rate you can get with an exit pressure of 1 MPa and negligible kinetic energies. The exit flow goes through a small hole in a spray nozzle out to the atmosphere at 100 kPa. Find the spray velocity.

Answer :

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Answer:

42.4m/s

Explanation:

To develop the problem it is necessary to apply the concepts related to thermodynamic work and Bernoulli's principle in which the behavior is described of a liquid moving along a stream.

The work of an incomprehensible liquid is given by the equation,

[tex]\dot{W} = \dot{m}w = -\dot{m}\upsilon (P_e-P_i)[/tex]

Where,

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

[tex]\upsilon[/tex] = Specific Volumen

P = Pressure

Our values are given by,

[tex]\dot{W}[/tex]= 2kW

[tex]P_i = 150kPa\\P_f = 1000kPa\\P_e = 100kPa[/tex]

[tex]\upsilon = -0.001000 m^3/kg \rightarrow[/tex] Table 1 for saturated water in 10°C

We need to find the mass flow, then re-arrange for [tex]\dot{m}[/tex]

[tex]\dot{m} = \frac{\dot{W}}{-\upsilon(P_f-P_i)}[/tex]

[tex]\dot{m} = \frac{-2}{-0.001000(1000-150)}[/tex]

[tex]\dot{m} = 2.356kg/s[/tex]

To find the Spray velocity, we apply Bernoulli equation, because at the Nozzle there is not Work or Heat transfer related. Then,

[tex]\frac{1}{2}V^2 = \upsilon \Delta P[/tex]

[tex]V = \sqrt{2*\upsilon(P_e-P_i)}[/tex]

[tex]V = \sqrt{2*0.001003(1000-100)}[/tex]

[tex]V = 42.4m/s[/tex]

Therefore the spray velocity is 42.4m/s

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