Consider a pump operating adiabatically at steady state. Liquid water enters at 20◦C, 100 kPa with a mass flow rate of 53 kg/min
. The pressure at the pump exit is 5 MPa. The pump isentropic efficiency is 75%. Assumer negligible changes in kinetic and potential energy and the water behaves as an incompressible substance. Determine the power required by the pump, in kW.
1 answer:
Answer:
5778.86W
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2. find the specific weights of the water and its density using T = 20C, using thermodynamic tables
note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
3. use the Bernoulli equation to find the height of the pump's power, taking into account that the potential and kinetic energy changes are insignificant
4. find the ideal pump power
5. find the real power of the pump using the efficiency equation
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Answer:
hello your question is incomplete attached below is the complete question
answer :
Slopes : B = 180 mm , C = 373 mm
Deflection: B = 0.0514 rad , C = 0.077 rad
Explanation:
Given data :
I = 500(10^6) mm^4
E = 70 GPa
The M / EI diagram is attached below
<u><em>Deflection angle at B</em></u>
∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI
= 1800 / ( 500 * 70 ) = 0.0514 rad
<u><em>slope at B </em></u>
ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI
= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm
<u><em>Deflection angle at C </em></u>
∅C = ∅CA = [ 1800 + 300*3 ] / EI
= 2700 / ( 500 * 70 )
= 2700 / 35000 = 0.077 rad
<u><em>Slope at C </em></u>
ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]
= 13050 / 35000 = 373 mm
