A feed water pump operates with a flow rate of 350 kg/s. The water enters at 420

Question

A feed water pump operates with a flow rate of 350 kg/s. The water enters at 420 K. The inlet pressure is 0.7 MPa and the outlet pressure 20 MPa. Determine the power needed to drive the pump if the pump isentropic efficiency is 87%. Also determine the outlet temperature of the water. Plot the process on T-s coordinates. Show the saturated liquid line. Compressed liquid properties for water are given below. Note: Isentropic efficiency is defined differently for pumps and for turbines. Refer equations 7.27 and 7.28. (20)

Pressure

(MPa)

Temperature

(K)

Specific volume

(m3/kg)

Enthalpy

(kJ/kg)

Entropy

(kJ/kg

Pressure

(MPa)

Temperature

(K)

Specific volume

(m3/kg)

Enthalpy

(kJ/kg)

Entropy

(kJ/kg

Explanation / Answer

My values are slightly different from the ones given, possibly because of a difference in the constant used for density of water.

At 420 K, (147 C) density of water is about 920 kg/m^3 (http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html)

We can apply Bernoulli's equation to the system:

p1/? + V12/2g +z1+hp= p2/? + V22/2g + z2+hL

P is pressure,? is specific weight, V is velocity, g is gravity, z is elevation, hpis pump head and hLis head loss. V1and V2are equivalent assuming the area of the pipe doesn't change. z1and z2can also assumed to be constant, and head loss can be neglected.

P1/? + hp = P2/? hp= (p2/? - p1/?)

Work done by a pump = Q?hp

Where Q is the flow of the water and equals density times the mass flow rate:

Q= (350 kg/s)/920 kg/m3= .3804 m3/s

Work = (.3804 m3/s)(?)(p2/? - p1/?) =(.3804 m3/s)(p2- p1) =(.3804 m3/s)(20-.7 MPa) = 7.34 MW

7.24 MW is the theoretical work done by the pump, but more energy will always be required in a real life situation, so efficiency must be taken into account.

Actual work = theoretical work/efficiency = 7.34 MW/ (0.87) = 8.44 MW

Using 8.368 MW for temperature change:

specific heat of water is 4.31 kJ/(kg*k), and 8.368 MJ/s are applied to the water at a rate of 350 kg/s

?T= (8368 kJ/s)/(350 kg/s * 4.31 kJ/kg*K * 2) = 2.77 K

Add 2.77 K to original 420 K = 422.7 K

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