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

[Bradracer18]

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

 

[Bradracer18]

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.

 

[Blueshift5]

it is important to approach problems like this systematically and use the appropriate equations and principles. In this case, the problem involves a pump, which is a device that transfers energy to a fluid (in this case, water) in order to increase its pressure. The key equation to use here is the Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid at different points in a system.

To start, we can use the given information to determine the pressures at points 1 and 2. Point 1 is where the water enters the pump and point 2 is where it exits. We know that the pressure at point 1 is 100 kPa and the pressure at point 2 is 800 kPa. Since the flow velocities at these points are the same, we can assume that the kinetic energy effects can be neglected. This simplifies the Bernoulli's equation to:

P1 + (rho*g*z1) = P2 + (rho*g*z2)

Where P is pressure, rho is the density of water, g is gravity, and z is elevation. We can rearrange this equation to solve for the elevation difference between points 1 and 2:

z2 - z1 = (P1 - P2)/(rho*g)

Plugging in the given values, we get:

z2 - z1 = (100 kPa - 800 kPa)/(1000 kg/m^3 * 9.81 m/s^2) = -0.071 m

This tells us that point 2 is 0.071 m lower in elevation than point 1. Next, we can use the fact that the flow rate is steady to determine the work done by the pump, which is equal to the power input. The equation for work is:

W = m*delta(h)

Where m is the mass flow rate and delta(h) is the change in enthalpy between points 1 and 2. We can determine the enthalpy at each point using the specific enthalpy of water at the given temperature of 20°C:

h1 = 83.93 kJ/kg
h2 = 84.03 kJ/kg

Plugging these values into the work equation, we get:

W = (0.035 m^3/s)(1000 kg/m^3)(84.03 kJ/kg - 83.93 kJ/kg) =