[latex] \\Delta \\rm{energy = + in - out}[\/latex]<\/span><\/p>\r\n \r\n\r\nThe change in the total energy of a system during a process from states 1 to 2 can be expressed as\r\n [latex] \\Delta E = E_2 - E_1 ={}_{1}Q_{2} - {}_{1}W_{2}[\/latex]<\/p>\r\n \r\n\r\nIf the changes in the kinetic and potential energies of the system are negligible, i.e., [latex] \\Delta KE = \\Delta PE = 0[\/latex], then the first law of thermodynamics for a closed system can be simplified as\r\n [latex]\\Delta U = U_2-U_1 = {}_{1}Q_{2} - {}_{1}W_{2}[\/latex]<\/p>\r\n where<\/p>\r\n [latex] E[\/latex]: total energy stored in a system<\/p>\r\n [latex] U[\/latex]: internal energy of a system<\/p>\r\n [latex] Q[\/latex]: heat transfer in a process. A common sign convention: positive sign (+) for the heat transfer into a system, and negative sign (-) for the heat transfer out of a system. In short, the sign for heat transfer: in (+), out (-). See Figure 4.4.1<\/a>.<\/p>\r\n [latex] W[\/latex]: work done by or to a system. A common sign convention: positive sign (+) for the work output (work done by a system to its surroundings), and negative sign (-) for the work input (work done by the surroundings to the system). In short, the sign for work: in (-), out (+). See Figure 4.4.1<\/a>.<\/p>\r\n Subscripts 1 and 2 refer to the initial and final states of a <\/a>process.<\/p>\r\n \r\n\r\n[caption id=\"attachment_2614\" align=\"aligncenter\" width=\"400\"] Example 1<\/p>\r\n\r\n<\/header>\r\n [latex]\\because \\Delta U = 0=Q-W[\/latex]<\/p>\r\n [latex]\\therefore Q_L-Q_H-(-W_{in})=0[\/latex]<\/p>\r\n [latex]\\therefore Q_H=Q_L+W_{in}[\/latex]<\/p>\r\nNote the sign convention for heat: in (+), out (-) and for work: in (-), out (+). This relation can be interpreted as the total energy transferred out of the cycle remains the same as the total energy transferred into the cycle.\r\n\r\n<\/div>\r\n<\/div>\r\n Example 2<\/p>\r\n\r\n<\/header>\r\n A rigid tank has two rooms, both containing R134a at the following initial states.<\/p>\r\n Room A: m<\/em>= 2 kg, P<\/em>=200 kPa, v<\/em>=<\/span>0.132 m3<\/sup>\/kg<\/p>\r\n Room B: m<\/em>= 3 kg, P<\/em>=500 kPa, T<\/em>=100\u00b0C<\/p>\r\nA crack is developed in the partition between the two rooms, which allows R134a in the tank to mix. Assume the mixing takes place slowly until R134a in the whole tank reaches a uniform state at 50\u00b0C. Find the heat transfer during this process. The process can be treated as a quasi-equilibrium process.\r\n\r\n[caption id=\"attachment_2115\" align=\"aligncenter\" width=\"500\"] Because the volume of the tank remains constant, the boundary work during the mixing process is zero; therefore, from the first law of thermodynamics,<\/p>\r\n [latex]\\because \\Delta U = Q - W[\/latex]\u00a0\u00a0 and\u00a0\u00a0 [latex]W = 0[\/latex]<\/p>\r\n [latex]\\therefore \\Delta U = Q[\/latex]<\/p>\r\n The heat transfer during the process depends on the internal energies of the initial and final states.<\/p>\r\n [latex]\\begin{align*} \\Delta U &= U_3 - (U_1 + U_2) \\\\&= (m_1 + m_2)u_{3} - (m_{1}u_{1}\u00a0 + m_{2}u_{2}) \\end{align*}[\/latex]<\/p>\r\n where the subscripts 1, 2, and 3 represent the initial states of R134a in rooms A and B, and the final state of R134a in the whole tank, respectively.<\/p>\r\nSecond, find the specific internal energies, u1<\/sub>, u2 <\/sub>, <\/em>and u3.<\/sub><\/em>\r\n Room A at the initial state: P<\/em> =200 kPa, v<\/em> = 0.132 m3<\/sup>\/kg<\/p>\r\n From Table C1<\/a>, at T<\/em> = -10\u00b0C, Psat<\/sub><\/em> = 200.6 kPa, vg<\/sub><\/em> = 0.09959 m3<\/sup>\/kg. Since v > vg<\/sub><\/em>, R134a at this state is a superheated vapour.<\/p>\r\n From Table C2<\/a>, at P<\/em> = 200 kPa, T<\/em> = 60\u00b0C, v<\/em> = 0.132057 m3<\/sup>\/kg \u2248 0.132 m3<\/sup>\/kg<\/p>\r\n [latex] u_1 = 427.51 \\ \\rm{kJ\/kg} [\/latex]<\/p>\r\n [latex] \\mathbb{V}_A = m_{1}v_{1} = 2 \\times 0.132 = 0.264 \\ \\rm{m^3} [\/latex]<\/p>\r\n Room B at the initial state: P<\/em> =500 kPa, T<\/em> = 100\u00b0C<\/p>\r\n From Table C1<\/a>, at T<\/em> = 100\u00b0C, P<\/em>sa<\/em>t<\/sub> = 3972.38 kPa.\u00a0 Since P < Psat<\/sub><\/em>, R134a at this state is a superheated vapour.<\/p>\r\n From Table C2<\/a>, at P<\/em> =500 kPa, T<\/em> = 100 \u00b0C<\/p>\r\n [latex] u_2 = 459.65 \\ \\rm{kJ\/kg} [\/latex]<\/p>\r\n [latex] v_2 = 0.058054 \\ \\rm{m^3\/kg} [\/latex]<\/p>\r\n [latex] \\mathbb{V}_B = m_{2}v_{2} = 3 \\times 0.058054 = 0.1742 \\ \\rm{m^3} [\/latex]<\/p>\r\n The final state of R134a in the whole tank: T<\/em> = 50\u00b0C<\/p>\r\n [latex] v_3 = \\dfrac{\\mathbb{V}_{tot}}{m_{tot}} = \\dfrac{\\mathbb{V}_A + \\mathbb{V}_B}{m_1 + m_2} =\\dfrac{0.264 + 0.1742}{2+3} = 0.0876 \\ \\rm{m^3\/kg}[\/latex]<\/p>\r\n From Table C1<\/a>, at T<\/em> = 50\u00b0C, v<\/em>g<\/em>\u00a0<\/sub>= 0.015089 m3<\/sup>\/kg. Since v3<\/sub> > v<\/em>g<\/sub>, R134a at the final state is a superheated vapour.<\/p>\r\n From Table C2<\/a>,<\/p>\r\n At P<\/em> = 200 kPa, T<\/em> = 50\u00b0C, v<\/em> = 0.127663 m3<\/sup>\/kg, u<\/em> = 419.29 kJ\/kg<\/p>\r\n At P<\/em> = 300 kPa, T<\/em> = 50\u00b0C, v<\/em> = 0.083723 m3<\/sup>\/kg, u<\/em> = 418.19 kJ\/kg<\/p>\r\n Use linear interpolation,<\/p>\r\n [latex]\\because\\dfrac{P_{3}-200}{300-200} =\\dfrac{0.0876-0.127663}{0.083723-0.127663} =\\dfrac{u_{3} - 419.29 }{418.19-419.29}[\/latex]<\/p>\r\n [latex] \\therefore P_3 = 291.2 \\ \\rm{kPa}[\/latex]\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0 [latex]u_3 = 418.287 \\ \\rm{kJ\/kg}[\/latex]<\/p>\r\nLast, substitute u1<\/sub>, u2<\/sub><\/em> and u3<\/sub><\/em> into the simplified first law,\r\n [latex] \\begin{align*} Q &= \\Delta U = (m_1 + m_2)u_{3} - (m_{1}u_{1}\u00a0 + m_{2}u_{2}) \\\\& = 5 \\times 418.287 - (2 \\times 427.51 + 3 \\times 459.65) \\\\&= -142.535 \\ \\rm{kJ}\\end{align*}[\/latex]<\/p>\r\nDuring the mixing process, the heat is transferred from the tank to the surroundings; therefore, the sign for the heat transfer is negative.\r\n\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n Example 3<\/p>\r\n\r\n<\/header>\r\n [latex]\\because \\Delta U = Q - W [\/latex]<\/p>\r\n [latex]\\therefore Q = W + \\Delta U =\u00a0 W + m(u_2 - u_1) [\/latex]<\/p>\r\nSecond, consider the boundary work in a polytropic process. The specific volumes are unknowns\r\n [latex] W = \\dfrac{P_{2}\\mathbb{V}_{2} - P_{1}\\mathbb{V}_{1}}{1-n} = \\dfrac{m\\left({P_{2}{v}_{2} - P_{1}{v}_{1}}\\right)}{1-n} [\/latex]<\/p>\r\nThird, find the specific volumes and specific internal energies at both initial and final states.\r\n At the initial state: P1<\/sub><\/em> = 100 kPa, T1<\/sub><\/em> = 0\u00b0C. From Table B1<\/a>, Psat<\/sub><\/em> = 429.39 kPa at 0\u00b0C. Since P1\u00a0<\/sub>< P<\/em>sa<\/em>t<\/sub>, ammonia is a superheated vapour.<\/p>\r\n From Table B2<\/a>,<\/p>\r\n v<\/em>1<\/em>\u00a0<\/sub>= 1.31365 m3<\/sup>\/kg,\u00a0\u00a0 u1<\/sub> <\/em>= 1504.29 kJ\/kg.<\/p>\r\n At the final state P2<\/sub> = 150 kPa. The process is polytropic with n<\/em> =1.25.<\/p>\r\n [latex] \\because P_1v_{1}^n = P_2v_{2}^n [\/latex]<\/p>\r\n [latex]\\therefore v_2=v_1\\left(\\dfrac{P_1}{P_2}\\right)^{1\/n}=1.31365\\times\\left(\\dfrac{100}{150}\\right)^{1\/1.25} = 0.94974 \\ \\rm{m^3\/kg} [\/latex]<\/p>\r\n From Table B1<\/a>: at T<\/em> = -25 \u00b0C and P<\/em> = 151.47 kPa \u2248 150 kPa, vg<\/sub><\/em> = 0.771672 m3<\/sup>\/kg. Since v2<\/sub> > vg<\/sub><\/em>, ammonia at the final state is a superheated vapour.<\/p>\r\n From Table B2<\/a>,<\/p>\r\n At P<\/em> = 150 kPa, T<\/em> = 20\u00b0C, v<\/em> = 0.938100 m3<\/sup>\/kg, u<\/em> = 1535.05 kJ\/kg<\/p>\r\n At P<\/em> =150 kPa, T<\/em> = 30\u00b0C, v<\/em> = 0.972207 m3<\/sup>\/kg, u<\/em> = 1551.95 kJ\/kg<\/p>\r\n Use linear interpolation,<\/p>\r\n [latex]\\because\\dfrac{T_{2}-20}{30-20} =\\dfrac{0.94974-0.938100}{0.972207-0.938100} =\\dfrac{u_{2} - 1535.05 }{1551.95-1535.05}[\/latex]<\/p>\r\n [latex] \\therefore T_2 = 23.4 \\ ^ \\rm{{\\circ}} \\rm{C}[\/latex]\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0 [latex]u_2 = 1540.82 \\ \\rm{kJ\/kg}[\/latex]<\/p>\r\nLast, the heat transfer in this process can now be found from\r\n [latex]\\begin{align*} Q &= W + \\Delta U = m\\dfrac{P_{2}{v}_{2} - P_{1}{v}_{1}}{1-n} + m(u_2 - u_1) \\\\&= 0.5 \\left[ \\dfrac{150 \\times 0.94974 - 100 \\times 1.31365}{1-1.25} + (1540.82 - 1504.29)\\right] \\\\&= - 3.928 \\ \\rm{kJ}\\end{align*}[\/latex]<\/p>\r\nDuring this process heat is rejected to the surroundings; therefore, the sign for heat transfer is negative.\r\n\r\n<\/div>\r\n<\/div>\r\n Example 4<\/p>\r\n\r\n<\/header>\r\n A piston-cylinder device contains steam initially at 200o<\/sup>C and 200 kPa. The steam is first cooled isobarically to saturated liquid, then isochorically until its pressure reaches 25 kPa.<\/p>\r\n\r\n [latex] \\because\\Delta u = q - w[\/latex]<\/p>\r\n [latex] \\therefore q= \\Delta u + w = (u_3 - u_1) + w [\/latex]<\/p>\r\nSecond, analyze the processes.\r\n The process is isobaric from state 1 to state 2, then isochoric from state 2 to state 3. The specific work is the shaded area of the rectangle shown in the [latex]P - v [\/latex] diagram; therefore,<\/p>\r\n [latex] w = P_1(v_3 - v_1) [\/latex]\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0 [latex] v_2 = v_3 [\/latex]<\/p>\r\nThird, determine the specific volumes and specific internal energies at states 1 and 3.\r\n At state 1, P1<\/sub><\/em> = 200 kPa and T1<\/sub><\/em> = 200o<\/sup>C.<\/p>\r\n![\"A](\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/07\/Heat-and-work-to-closed-systems.png\")
<\/a> Figure 4.4.1<\/strong><\/em>\u00a0A common sign convention for heat and work transfer to a closed system<\/em>[\/caption]\r\n\r\nThe following procedure may be followed when solving problems with the first law of thermodynamics.\r\n\r\n \t
![\"Vapor](\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2022\/01\/Fig.-6-6_revised-300x246.png\")
<\/a> Figure 4.4.e1<\/strong><\/em> Vapor compression refrigeration cycle consisting of a compressor, condenser, expansion device, and evaporator<\/em>[\/caption]\r\n\r\nSolution<\/span><\/em>:\r\n\r\nThe vapour compression refrigeration cycle can be regarded as a closed system with the initial and final states being identical; therefore, [latex]\\Delta U = 0[\/latex].\r\n![\"A](\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/08\/4.5.1-e1629312480473-1024x371.png\")
<\/a> Figure 4.4.e2 <\/em><\/strong>A rigid tank with two rooms<\/em>[\/caption]\r\n\r\nSolution:<\/em><\/span>\r\n\r\nFirst, set the whole rigid tank as the closed system.\r\n![\"Ammonia](\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/08\/NH3-in-a-Piston-Cylinder-300x249.png\")
<\/a> Figure 4.4.e3<\/strong> Ammonia in a piston-cylinder device<\/em>[\/caption]\r\n\r\nSolution:<\/em><\/span>\r\n\r\nFirst, set ammonia in the piston-cylinder as the closed system. From the first law of thermodynamics,\r\n\r\n \t
![\"Steam](\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/07\/Steam-in-a-Piston-Cylinder.png\")
<\/a> Figure 4.4.e4<\/strong> Steam in a piston-cylinder device<\/em>[\/caption]\r\n\r\nSolution:<\/em><\/span>\r\n\r\n1. [latex]P - v [\/latex] and [latex]T - v [\/latex] diagrams\r\n\r\n[caption id=\"attachment_2145\" align=\"aligncenter\" width=\"588\"]![\"P-v](\"https:\/\/pressbooks.bccampus.ca\/thermo1\/wp-content\/uploads\/sites\/499\/2021\/08\/4.5.3-300x106.png\")
<\/a> Figure 4.4.e5<\/strong> P-v and T-v diagrams of the whole process<\/em>[\/caption]\r\n\r\n2. Calculate the specific heat transfer\r\n\r\nFirst, set the steam in the piston-cylinder device as a closed system. From the first law of thermodynamics,\r\n