Download Solved Assignment 2 - Applied Thermodynamics | M E 320 and more Assignments Thermodynamics in PDF only on Docsity!
2.32) Air contained within a piston-cylinder assembly is slowly heated. As shown in Fig. P2.32, during this process the pressure first varies linearly with volume and th4en remains constant. Determine the total work, in kJ.
๐ 2
๐ 1 Since we have a plot on a p-V diagram, this integral is most easily evaluated by finding the area under the curve.
Braking up the diagram into 3 simple figures as follows:
S
S
S
๐ = 0.015๐^3 100 ๐๐๐ + 0.5 0.015๐^3 50 ๐๐๐ + (0.025๐^3 )(150๐๐๐)
๐ = 1.5๐^3 ๐๐๐ + 0.375๐^3 ๐๐๐ + (3.75๐^3 ๐๐๐)
๐ = 5.625 ๐^3 ๐๐๐ ๐ฅ
1 ๐/๐^2
๐ = 5.625 ๐๐ฝ ANS
๐๐ก =^ ๐ โ ๐
Since we are at steady-state there is no change in the energy with respect to time, therefore:
๐ = ๐
Since there are no work terms in this system, we are only dealing with heat transfers, thus:
๐ = 0
๐๐๐ โ ๐๐๐ข๐ก = 0
๐๐ฅ โ โ๐๐ = 0
โ๐๐ = ๐๐ฅ
The negative sign on the Qout term comes from our convention that heat out is a negative quantity. Substituting in Newtonโs law of cooling for Qc and Fourierโs law for Qx we have:
Note that when the book introduces Newtonโs law of cooling it assumes the heat going out is positive. Since the normal convention is to have heat out as negative we introduce a negative sign into Newtonโs law of cooling to conform to convention.
Solving this equation yields:
(b) For Temperature in Kelvin, with To = 200 K and Ts = 250 K, a sketch of Ti vs. B looks like:
(c) The minimum value of B to satisfy these parameters will occur when Ti exactly equals 60 oC. Plugging this into our equation from (a) yields:
50 100 150 200 250 300
B
Ti
2.56) As shown in Fig. P2.56, 5 kg of steam contained within a piston-cylinder assembly undergoes an expansion from state 1, where the specific internal energy is u 1 = 2709.9 kJ/kg, to state 2, where u 2 = 2659.6 kJ/kg. During the process, there is heat transfer to the steam with a magnitude of 80 kJ. Also, a paddle wheel transfers energy to the steam by work in the amount of 18.5 kJ. There is no significant change in the kinetic or potential energy of the steam. Determine the energy transfer by work from the steam to the piston during the process, in kJ.
Using the First Law of Thermodynamics we have:
โ๐พ๐ธ + โ๐๐ธ + โ๐ = ๐ โ ๐
0 + 0 + ๐ข 2 ๐ โ ๐ข 1 ๐ = ๐ โ (๐๐๐๐๐๐๐ + ๐๐๐๐ ๐ก๐๐ )
๐๐ 5 ๐๐^ โ^ 2709.^
๐๐ 5 ๐๐^ =^80 ๐๐ฝ^ โ^ (โ18.5^ ๐๐ฝ^ +^ ๐๐๐๐ ๐ก๐๐^ )
๐๐๐๐ ๐ก๐๐ = 350 ๐๐ฝ ANS
2.60) As shown in Fig. P2.60, the outer surface of a transistor is cooled convectively by a fan-induced flow of air at a temperature of 25 oC and a pressure of 1 atm. The transistorโs outer surface area is 5 x 10-4^ m^2. At steady state, the electrical power to the transistor is 3 W. Negligible heat transfer occurs through the base of the transistor. The convective heat transfer c coefficient is 100 W/m^2 โข K. Determine (a) the rate of heat transfer between the transistor and the air, in W, and (b) the temperature at the transistorโs outer surface, in oC.
(a) Since we are at steady state there is no change in the system with respect to time. Using this we may use the energy balance equation to determine the rate of heat transfer between the transistor and the air. ๐๐ธ ๐๐ก
Since there is no heat transfer at the base of the transistor, the only heat change is the heat transferred from the transistor to the air. The only work in the system is the electrical work supplied to the transistor. Therefore we may write:
๐๐๐๐ = ๐๐๐๐๐
๐๐๐๐ = โ 3 ๐ ANS
Note that since Work into a system is negative and heat out of a system is negative, this statement does indeed conform to our sign conventions.
Supplemental Question 1) How would the Ideal Gas Law change if the linear kinetic energy of the gas molecules was equal to 5/2 k T?
From the lecture notes where we derived the ideal gas law, we have:
Pressure = (N/3) (m s^2 / V)
At this point we made use of the relation that the linear kinetic energy of the gas molecules was equal to 5/2 k T. If this was changed to 5/2 k T, the Ideal Gas Law would change as follows:
1/2 m s^2 = 5/2 k T
๐ ^2 =
๐ 5 ๐๐^ ๐
๐ ๐ ๐๐^ ๐
๐๐ = 53 ๐๐
๐ ANS
Supplemental Question 2) What is the difference between the equation for energy conservation in a system and the First Law of thermodynamics?
The first law of thermodynamics is an experimentally derived principle that posits that energy in conserved in all thermodynamic processes. The equation for energy conservation in a system is derived from the first law and accounts for all of the energy transfers and processes that occur in a system. The balance relation is set up as such to be in accordance with the first law of thermodynamics.
