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September 1, 2014

September 1, 2014

Posted by **Janus** on Tuesday, February 16, 2010 at 10:30am.

- Thermodynamics -
**drwls**, Tuesday, February 16, 2010 at 1:34pmCalculate the inlet and outlet velociies V1 and V2 using the contiuity equation

V * A = 3000 liters/min = 3*10^6 cm^3/min

A1 = pi*(15)^2 = 706.9 cm^2

A2 = pi*(10)^2 = 314.2 cm^2

V1 = 3*10^6/706.9

= 4244 cm/s = 42.44 m/s

V2 = 95.48 m/s

The steady state energy equation says that

Mdot*(U + P*v + KE + PE)out = Mdot*(U + P*v+ KE+ PE)in + (Power)in

Mdot is the mass flow rate

U is the internal energy, which does not change because inlet and outlet temperatures are equal. KE(per mass) = V^2/2. PE is potential energy which does not change since the inlet and outlet are at the same elevation. P v

is pressure times specific volume, which is the same as pressure divided by density.

The energy equation simplifies to:

Power in = Mdot*[P/density + V^2/2]out -Mdot*[P/density + V^2/2]in

For water, density = constant = 1000 kg/m^3

Mdot = 3 m^3/min * 1000 kg/m^3 = 3000 kg/min = 50 kg/s

First let us calculate the power needed to increase the pressure.

Mdot*(Pout-Pin)/density

= 50 kg/s*(202*10^5 N/m^2)/1000 kg/m^3

= 60.6*10^6 J/s = 1.01 Megawatts

You need to add to that the additional power needed to increase the kinetic energy

Mdot*[V2^2/2 - V1^2/2]

Take it from there and check my numbers.

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