Mechanical Thermodynamics Help for a Mechanical Freshie

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SUMMARY

This discussion revolves around solving mechanical thermodynamics problems related to heat transfer and entropy changes. The first question involves a vessel containing saturated liquid water and vapor at 0.1 MPa, requiring the calculation of heat transfer to fill the vessel with saturated vapor. The second question focuses on a piston-cylinder system with nitrogen, where heat transfer during compression at varying pressures and temperatures must be determined. The third question addresses the change of entropy when mixing equal masses of liquid at different temperatures, emphasizing the necessity of a positive entropy change.

PREREQUISITES
  • Understanding of the First Law of Thermodynamics
  • Familiarity with steam tables for saturated water properties
  • Knowledge of specific heat capacity equations, particularly for nitrogen
  • Ability to interpret P-v and T-v diagrams in thermodynamic processes
NEXT STEPS
  • Study the use of steam tables for calculating properties of saturated water and vapor
  • Learn how to apply the First Law of Thermodynamics in closed systems
  • Explore the derivation and application of specific heat capacity equations
  • Investigate the concepts of entropy and its calculations in thermodynamic processes
USEFUL FOR

Mechanical engineering students, thermodynamics learners, and anyone seeking to understand heat transfer and entropy in thermodynamic systems.

elly92
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Homework Statement


Hi, I have this coursework due in 2 days and I have no clue how to start it! Please Help any hints will be very nice. I tried attempting it several times but I can't understand it the question properly! My Thermo lecturer isn't making the subject clear and I have no clue how to approach the question.

Q1. A vessel with volume 5m3 contains 0.05m3 of saturated liquid water and
4.95m3 of saturated water vapour at 0.1MPa. Heat is transferred until the
vessel is filled with saturated vapour. Determine the heat transfer for this
process

Q2. A cylinder fitted with a piston has an initial volume of 0.1m3 and contains
nitrogen at 150kPa, 25oC . The piston is moved, compressing the nitrogen
until the pressure is 1MPa and the temperature is 150oC . During this
compression process, heat is transferred from the nitrogen, and the work done
on the nitrogen is 20kJ. Determine the amount of this heat transfer. For the
nitrogen, you may assume that
c(p) = 1110 = 480(T/1000) + 960(T/1000)^2 -420(T/1000)^3
where T is measured in K, and C(p) is measured in J/KgK

Q3. A mass of a liquid, m, at temperature, T1, is mixed with an equal mass of the
same liquid at temperature, T2. The system is thermally insulated. Show that
the change of entropy of the Universe is:

2mc(p)ln[ ({T(1) + T(2)}/2)/[(Square root)T1*T2]]
and prove that this is necessarily positive.
[Do not specifically consider the mixing of the fluids.].

Thanks in advance!
 
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elly92 said:

Homework Statement


Hi, I have this coursework due in 2 days and I have no clue how to start it! Please Help any hints will be very nice. I tried attempting it several times but I can't understand it the question properly! My Thermo lecturer isn't making the subject clear and I have no clue how to approach the question.

Q1. A vessel with volume 5m3 contains 0.05m3 of saturated liquid water and
4.95m3 of saturated water vapour at 0.1MPa. Heat is transferred until the
vessel is filled with saturated vapour. Determine the heat transfer for this
process

Q2. A cylinder fitted with a piston has an initial volume of 0.1m3 and contains
nitrogen at 150kPa, 25oC . The piston is moved, compressing the nitrogen
until the pressure is 1MPa and the temperature is 150oC . During this
compression process, heat is transferred from the nitrogen, and the work done
on the nitrogen is 20kJ. Determine the amount of this heat transfer. For the
nitrogen, you may assume that
c(p) = 1110 = 480(T/1000) + 960(T/1000)^2 -420(T/1000)^3
where T is measured in K, and C(p) is measured in J/KgK

Q3. A mass of a liquid, m, at temperature, T1, is mixed with an equal mass of the
same liquid at temperature, T2. The system is thermally insulated. Show that
the change of entropy of the Universe is:

2mc(p)ln[ ({T(1) + T(2)}/2)/[(Square root)T1*T2]]
and prove that this is necessarily positive.
[Do not specifically consider the mixing of the fluids.].

Thanks in advance!
Welcome to Physics Forums. (And hopefully you have read the Private Message I sent you.)

If you're honestly totally lost, here is what I can tell you to get started. Energy can be added to (or taken from) a system in either of two ways, possibly a combination of both ways: (1) doing work on the system (or having the system do work), or (2) heating or cooling of the system. So there is a relation between work done W, heat exchanged Q, and the change in the energy of the system -- this should be in your class notes or textbook.

For Q1, usually a question like this is done using a table for saturated water, which should be in your textbook, probably in an Appendix in the back of the book. Can you find the table?
 
Additional tips:
Anytime you start a problem classify it as an open or closed system (if you're unclear what I mean by this I can elaborate).

Then try to classify the type of process if possible based on the information given. For example, it could be an isobaric, isometric, isothermal, polytropic, etc. process.

Finally, draw a graph of the process on a P-v and/or T-v diagram with the vapor dome, label the points. If you do these 3 things and know all of your formulas you can do any problem.

Always remember how to calculate work for for a closed system in general: ∫pdV
as well as an open system in general: -∫vdP
you also need to know how to derive for specific process, such as the ones I mentioned above, isothermal, isobaric, etc.
recall the first law, at least one equation that relates quality.
 
I've attached a picture that is essentially the setup for the problem. There's really only one other major obstacle to overcome and that is that the states are not fixed. I.e. state 1 all you know that is obvious is that P1=0.1Mpa=100kpa but you need to know something else to be able to start doing some calculations. My advice is to open your book or notes and look for the basic definition of quality. Once you get the quality, with knowing the pressure the state is fixed and you can easily calculate anything you could want (including internal energy, u1). State 2 also needs to be fixed, they give you x2=1, you have to read the problem carefully to gather this information. You need to know more information about state 2 in order to fix it, so that you can calculate other information about state 2, such as u2. You can actually calculate v1 and you know that v1=v2 because it's a closed system, and I assume the vessel to be rigid, therefore V1=V2 and v1=v2. Now you know x2=1 and v2=v1 which you can calculate v1 if you know x1. Now you've got state 2 fixed and you can calculate the quality at state 2, then you can calculate the internal energy at state 2. Then by utilizing the first law, you just need to determine what the work is, look at my post above for the general formula for a closed system for work. Notice that this is an isometric system (constant volume) so what does that mean about dV?
 

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Moderator's note: we should let elly92 respond with an attempt at solving the problem before offering further help.
 
Hey RedBelly and Reverks,

Thanks this is really helpful! I spent days revising thermo from scratch but i was not able to understand the question properly. Its really nice of you all to help me in this. Thank you. I will post my final answer here soon.

Thank you guys!
 
what would happen if I integrate mc(delta)t?
 
elly92 said:
what would happen if I integrate mc(delta)t?

Integrate it over what, since your variables are m, c and T, is it dm, dc or dt? I know that formula but I don't actually have any experience using it other than I recall for some stuff in a physics course I took two semesters ago...and I'm 99% sure it's not super accurate because it uses specific heat as a constant, which is only really accurate for very small temperature ranges.
 
Last edited by a moderator:
  • #10
i have the exact same problem as elly92 and have so far got to this, please tell me if I am right and if so where to go or where i may have made a mistake

using values from a steam table i have determined that 1.693 * Dryness Fraction = Specific Volume at (x) - 1/1000. In the knowledge that heat is transferred until the vessel is filled with vapour then the Dryness Fraction 2 becomes 1 and the specific volume at x for the change becomes equal to the specific volume of the vapour.

But where do i go from here?

I know the first law of thermodynamics but don't know how to use it?
 
  • #11
chrisgauntlet said:
i have the exact same problem as elly92 and have so far got to this, please tell me if I am right and if so where to go or where i may have made a mistake

using values from a steam table i have determined that 1.693 * Dryness Fraction = Specific Volume at (x) - 1/1000. In the knowledge that heat is transferred until the vessel is filled with vapour then the Dryness Fraction 2 becomes 1 and the specific volume at x for the change becomes equal to the specific volume of the vapour.

But where do i go from here?

I know the first law of thermodynamics but don't know how to use it?

What did you get for the value of quality, x?
Once you have the value of x you can easily calculate almost anything you could want, let's say you want entropy, s1=sf + x1sfg
let's say you want internal energy, u1=uf + x1sfg
and so on, enthalpy, specific volume, etc. etc.

In terms of using the first law, some people like to write it different ways, some like to automatically knock terms off that are 0 for that specific problem, but a good starting point for the first law for a closed system (which this is in part a) is this:
1Q2 - 1W2 = m(u2 - u1)

And then some sign conventions come into play, if there is work done BY the system, then you plug a positive value into W, don't change the negative sign in front of it. If there is work done ON the system, then work is negative, so plug a negative in for work, and now you've got two negatives so it becomes positive. I hope this has clarified things for you, I'm starting up a forum that is focused on thermodynamics, if you have any questions feel free to check it out, it's in my profile, thermoforums.com
 

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