(SP 1) DLLL. Given: In a closed rigid tank,
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1 (SP 1) Given: In a closed rigid tank, State 1: m 1,ice = 1, m 1,g = 0.05 P1= kpa, T1= -30 o C State 2: the liquid vapor equilibrium line, either saturated liquid or saturated vapor Find: (a) The volume of the tank (b) P1 in bar (c) a saturated liquid or a saturated vapor (d) T2 (e) P-T and P-V diagrams System sketch: State 1 State water vapor P1= kpa T1= -30 o C 1 ice DLLL Q 1.05 of Saturated liquid or Saturated vapor Assumptions: - Water vapor and ice are in equilibrium at state 1 - Pure saturated liquid or pure saturated vapor at state 2 Solution: (a) Vtank = m 1 v 1 = m 1,ice v 1,ice + m 1,g v 1,g
2 From Table A-6, m3 3 v 1,ice = v 1,g = 2943 m3 Then, the equation we wrote can be, Vtank = (1 )( m3 The volume of the tank is m 3 m3 ) + (0.05 ) (2943 ) = m3 (b) P1 = ( kpa)( 1 bar 10 5 Pa Pa ) ( kpa ) = bar (c) v 2 = v 1 = V tank m 1 = V tank m 2 = m = m3 which is greater than v crit, m3. Then, the state is a saturated vapor. (d) From Table A-2, v g = m3 at 5 o C v g = m3 at 6 o C Interpolate to know temperature at v g is m3. This provides T2 = o C. (e)
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4 (SP 2) Given: Two-phase R-134a State 1: T1= 50 o F State 2: superheated vapor at P2= 4 psia Find: (a) min and max of T2 (b) x1,min (c) P-h diagram of x1=0.9 (d) plot x1 vs T2 System sketch: Assumptions: 1. Steady-state, Steady-flow 2. KE = PE = 0 3. Adiabatic 4. No work Basic equations:
5 Solution: Apply our assumptions into the energy balance equation, The equation above is now, m 1h 1 = m 2h 2 Because of m 1 = m 2, h 1 = h 2 (a) Using the properties we know, establish an equation with the above equation. x 1 and T 2 are independent value and dependent value. h 1 (T 1, x 1 ) = h 2 (T 2, P 2 ) Use EES to solve (EES code is attached). T2,max= 29.5 o F for x1=1 and T2,min= -60 o F for x2=1 Maximum T2 when x1 occurs at 1.0 (sat. vapor at the throttle inlet). Minimum T2 when x2 occurs at 1.0 (sat. vapor at the throttle outlet).
6 (b) Minimum x1 occurs for T2= T2,min. See the attached EES code. x1,min= (c) and (d) EES code is attached.
7 "Specified inlet temperature and outlet pressure" T[1] = 50 [F] P[2] = 4 [psia] "Maximum outlet temperature where device works is when X[1] = 1 (saturated vapor)" P[1] = pressure(r134a,t=t[1],x=1.0) h_1_max = enthalpy(r134a,t=t[1],x=1.0) h_2_max = h_1_max T_2_max = temperature(r134a,p=p[2],h=h_2_max) "Minimum outlet temperature where device works is when X[2] = 1 (saturated vapor)" T_2_min = temperature(r134a,p=p[2],x=1.0) h_2_min = enthalpy(r134a,p=p[2],x=1.0) h_1_min = h_2_min X_1_min = quality(r134a,p=p[1],h=h_1_min) "Data for process line on P-h diagram with X[1] = 0.9" x[1] = 0.9 h[1] = enthalpy(r134a,t=t[1],x=x[1]) h[2] = h[1] T[2] = temperature(r134a,p=p[2],h=h[2]) "Data for plot of x_1 versus T_2" h_2 = enthalpy(r134a,t=t_2,p=p[2]) h_1 = h_2 x_1 = quality(r134a,t=t[1],h=h_1)
8
9 (SP 3) Given: 4L of water inside rigid body cooker from State 1 to State 2 State 1: 2L of saturated liquid P 1 = 200 kpa 2L of saturated water P 1 = 200 kpa State 2: 4L of saturated water P 2 = 200 kpa T stove = 1000, T 0 = 20, Q loss = 300 W, Q in = constant, t = 1 hour Find: (a) The total mass of the steam () entering the environment. (b) The rate of heat transfer (kw) from the source. (c) The total entropy production (kj/k) in one hour. (d) The total exergy destroyed (kj) in one hour. System: open system with saturated water inside 4L pressure cooker State1 2L saturated liquid water + 2L saturated water vapor at P= 200 kpa State 2 4L saturated water vapor at P= 200 kpa Assumptions: 1. Steady flow (No steady state!) 2. No change in PE and KE 3. No work 4. Temperature and pressure are uniform throughout the cooker
10 5. Constant properties 6. A half of the volume is saturated liquid water and a half of the volume is saturated water vapor inside the cooker at state 1 7. Water vapor inside the cooker at state 2 8. Water vapor leaves the cooker at state 2 Basic Equations: dm cv = m in m out de cv ds cv = Q cv = Q cv T b W cv + m in (h + PE + KE) in m out (h + PE + KE) out + m ins in m outs out + σ CV E d = T 0 σ cv Solution: (a) Because pressure 200 kpa is given for the saturated water, we can find properties of saturated water from table A-3. Tsat = vf = m 3 / uf = kj/ hf = kj/ sf = kj/k vg = m 3 / ug = kj/ hg = kj/ sg = kj/k First, mass balance equation for open system is dm cv = m in m out no inlet Integrate the equation w.r.t. time and then the equation can be written as m = m 2 m 1 = m out The mass at state 1 is m 1 = m f + m g = V f vf + V g vg = 2L m 3 / + 2L m 3 /
11 = 0.002m m3 = = m m3 The mass at state 2 is m 2 = V v g = 4L m 3 / = 0.004m m3 = Thus, m = m 2 m 1 = = m out = m out = Therefore, of steam enters into the environment (b) Now, the energy balance equation for open system is de cv = Q cv W cv + m in (h + PE + KE) in m out (h + PE + KE) out Apply no work, no KE, no PE, and one outflow, then, the equation above can be de cv = Q cv W cv + m in (h + PE + KE) in m out (h + PE + KE) out Rewrite the equation,
12 Integrate the equation w.r.t. time, de cv = Q cv m outh out E cv = Q cv m out h out Because no KE and no PE, the equation above can be written as m 2 u 2 m 1 u 1 = Q in Q loss m out h out kJ/ ( kJ/ kJ/) = Q in 0.3kJ/s 3600s kJ/ The rate of heat transfer is Q in = Q in t = kj 3600 s Q in = kj = kw Therefore, kw of the rate of heat transfer is entering into the cooker from the stove. (c) The entropy balance equation for open system is ds cv = Q cv T b + m ins in m outs out + σ CV Apply assumptions and integrate the equation w.r.t. time, S cv = Q cv T b m out s out + σ cv Rewrite, m 2 s 2 m 1 s 1 = Q in Q loss m T stove T out s out + σ cv 0 where m 1 s 1 = m f s f + m g s g Once you substitute all known values, you will get one unknown,
13 kj kj ( K K kj 0.3 kw 3600 s = 1273 K 293 K σ cv = = kj/k kj K ) kj K + σ cv Therefore, kj/k of the total entropy production is generated in one hour. (d) The total exergy destruction can be calculated by E d = T 0 σ cv = 293 K kj = kj K Therefore, kj of the total exergy is destroyed during one hour.
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