What Is the Minimum Power Input to the Pump in This Thermodynamics Problem?

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SUMMARY

The minimum power input to the pump in the thermodynamics problem is determined using the energy balance equation: 0 = -W + ṁ((h1-h2) + g(z1-z2)). Given the parameters, including inlet conditions of T1=20°C, p1=100 kPa, and a mass flow rate of 0.035 m³/s, the correct answer is 34 kW. The calculations involve converting pressure units and applying gravitational potential energy changes, with the exit height at 6.1 m above the inlet.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically energy balance equations.
  • Familiarity with fluid mechanics concepts, including mass flow rate and pressure conversions.
  • Knowledge of specific enthalpy values for water at given temperatures and pressures.
  • Ability to perform unit conversions, particularly between kPa and MPa.
NEXT STEPS
  • Study the application of the energy balance equation in pump systems.
  • Learn how to use steam tables and thermodynamic property tables for water.
  • Explore the calculation of specific enthalpy for water at various states.
  • Research the impact of gravitational potential energy in fluid flow problems.
USEFUL FOR

Students preparing for engineering exams, particularly those focused on thermodynamics and fluid mechanics, as well as professionals involved in pump system design and analysis.

Bradracer18
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Homework Statement


HI guys, this is an exam review question from the Fundamentals Engr exam. Not sure how to do it, please help me along so I can learn it too!

Water enters a pump steadily at T1=20°C, p1=100 kPa at a rate of 35 L/s and leaves at p2=800 kPa. The flow velocities at the inlet and the exit are the same, but the pump exit where the discharge pressure is measured at 6.1 m above the inlet section. Neglect Kinetic energy effect. The minimum power input to the pump is:

(a) 34 kW (b) 22 kW (c) 27 kW (d) 52 kW (e) 44 kW

Homework Equations



0 = -W + m\dot{}((h1-h2) + g(z1-z2))

The Attempt at a Solution



Well, I simplified the equation by eliminating Q(heat transfer) and KE. I converted kPA to Mpa by multiplying by .100.

So, p1= .1 MPa, p2= .8MPa. The mass flow rate(m)= converted 35L/s to 0.035m^3/s by multiplying by .001. g is gravity which is 9.81m/s^2. z1= 0m and z2= 6.1m

Basically, not sure how to start this problem out. Not sure how I find my h1 and h2. Maybe I'm not even using the correct equation, but I think I am.

Thank you,
Brad
 
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Okay, I found something else out...but have yet to figure out how to use the tables correctly. I think the water is classified as a compressed liquid, but that's as far as I've gotten.
 

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