Posted by Janus on Tuesday, February 16, 2010 at 10:30am.
A centrifugal pump compresses 3000 liters/min of water from 98 KPa to 300KPa. the inlet and outlet temp are 25°C. The inlet and discharge pipe are on the same level but the diameter of the inlet piping is 15cm whereas that of the discharge piping is 10cm. Determine the pump work in kW
- Thermodynamics - drwls, Tuesday, February 16, 2010 at 1:34pm
Calculate 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.
= 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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