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igor_vitrenko [27]
3 years ago
6

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.

Engineering
1 answer:
masha68 [24]3 years ago
6 0

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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Explanation:

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Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin
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Multiple Choice
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Answer:

A geological engineer

Explanation:

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3 years ago
A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di
Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 \frac{m^{3} }{s}</h3>

Explanation:

Given :

Length of pipe A L_{A}  = 1500 m

Length of pipe B L_{B} = 2500 m

Flow rate through pipe A Q_{A}  = 0.4 \frac{m^{3} }{s}

Diameter of pipe D = 30 \times 10^{-2} m

Velocity from pipe A,

  V _{A} = \frac{Q_{A} }{A}

  V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2}  }

  V_{A}  = 5.66 \frac{m}{s}

Here, head loss is same because height is same.

    h_{a} = h_{b}

L_{A} V_{A} ^{2} = L_{B}  V_{B} ^{2}

V_{B} = \sqrt{\frac{1500}{2500}}    (5.66)

V_{B} = 4.38 \frac{m}{s}

Now rate of flow from pipe B is,

Q_{B}  = V_{B} A

Q_{B}  = \frac{\pi }{4}  (0.3)^{2} \times 4.38

Q_{B} = 0.3094 \frac{m^{3} }{s}

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