Thermodynamics- determining mechanical power

In summary, the conversation discusses a problem where water is pumped from a lower reservoir to a higher reservoir, and the goal is to determine the mechanical power that is converted to thermal energy due to frictional effects. After working through the problem, it is discovered that the incorrect density of water was used, leading to an incorrect answer. After correcting the mistake, the correct answer is found to be 6.76 kW lost to heat.
  • #1

MacLaddy

Gold Member
291
11

Homework Statement



Hello all,

I've worked through the problem below, but I feel my answer is incorrect. I would appreciate it if someone could check my work.

Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20kW of shaft power. The free surface of the upper reservoir is 45m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m^3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.


Homework Equations



[itex]e_m = ke =(\frac{P}{\rho})(\frac{1}{2}\vec{V}^2)(gz)[/itex]

[itex]\dot{E}=\dot{m}e_m[/itex]

[itex]\dot{m}=\rho\dot{V}[/itex]

[itex]\eta=\frac{energy-out}{energy-in}[/itex]


The Attempt at a Solution



[itex]e_m = ke =(\frac{P}{\rho})(\frac{1}{2}\vec{V}^2)(gz)=gz=(9.81\frac{m}{s^2})(45m)=441.45\frac{m^2}{s^2}=441.45\frac{J}{kg}[/itex]

[itex]\dot{m}=\rho\dot{V}=(1 \frac{kg}{m^3})(0.03\frac{m^3}{s})=0.03\frac{kg}{s}[/itex]

[itex]\dot{E}=\dot{m}e_m=(0.03\frac{kg}{s})(441.45\frac{J}{kg})=13.24\frac{J}{s}=13.24W = 13.24*10^{-3}kW[/itex]

[itex]\eta=\frac{13.24*10^{-3}kW}{20kW}=.66*10^{-3}=0.066[/itex]%

[itex]20kW-20\eta= 19.99 kW[/itex]


So 19.99 kW out of 20 are wasted to heat in this problem? doesn't seem correct.
 
Physics news on Phys.org
  • #2
You have the density of water as 1 kg / m^3. This is some really, really light water.
 
  • Like
Likes baldbrain
  • #3
Ha, how did I miss that? Thanks.

How's the rest of my method look?
 
  • #4
You have the wrong value for the density of water. [edit: nevermind, Steam King beat me to it]
 
  • #5
Going through it with the proper density, I come up with 6.76 kW lost to heat. Sound about right?
 
Last edited:
  • #6
Hello! How did you get 6.76 kW? I am with you until you get the .066%. Even with the correct density of water that should just eliminate your 10^ -3.

upload_2018-1-17_14-43-52.png
= .662

20 kW- .662

This is where I am stuck! Please help. Thank you so much.
 

Attachments

  • upload_2018-1-17_14-43-52.png
    upload_2018-1-17_14-43-52.png
    912 bytes · Views: 2,619
  • #7
AbbeyC172 said:
Hello! How did you get 6.76 kW? I am with you until you get the .066%. Even with the correct density of water that should just eliminate your 10^ -3.

View attachment 218603= .662

20 kW- .662

This is where I am stuck! Please help. Thank you so much.
20 kW- 13.24 kW = 6.76 kW
 
  • #8
Chestermiller said:
20 kW- 13.24 kW = 6.76 kW
Thank you! I feel stupid for missing that.
 

1. What is thermodynamics and why is it important?

Thermodynamics is the branch of physics that deals with the relationship between heat, energy, and work. It is important because it helps us understand how energy is transferred and transformed in different systems, and allows us to predict and control the behavior of these systems.

2. What is mechanical power and how is it measured?

Mechanical power is the rate at which work is done or energy is transferred. It is typically measured in watts (W) or horsepower (hp). One watt is equal to one joule per second, and one horsepower is equal to 550 foot-pounds per second.

3. How is mechanical power calculated using thermodynamics?

The mechanical power of a system can be calculated by multiplying the force applied to an object by its velocity or by the amount of work done over a certain period of time. In thermodynamics, mechanical power is often calculated using the first law of thermodynamics, which states that energy cannot be created or destroyed but can be converted from one form to another.

4. What are some real-world applications of thermodynamics in determining mechanical power?

Thermodynamics is used in a variety of real-world applications, such as calculating the power output of engines and turbines, designing refrigeration and air conditioning systems, and determining the efficiency of power plants. It is also used in designing renewable energy systems such as solar panels and wind turbines.

5. How does thermodynamics play a role in understanding the efficiency of mechanical systems?

Thermodynamics is essential in understanding the efficiency of mechanical systems because it helps us analyze the energy input and output of these systems. By applying the laws of thermodynamics, we can determine the maximum theoretical efficiency of a system and compare it to the actual efficiency, allowing us to identify areas for improvement and optimization.

Suggested for: Thermodynamics- determining mechanical power

Replies
2
Views
1K
Replies
2
Views
2K
Replies
3
Views
629
Replies
3
Views
2K
Replies
3
Views
1K
Replies
9
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
Back
Top