Solucionario Cap4

PROBLEM 4.1 4.1 An 8-ft3 tank contains air at an initial temperature of 80°F and initial pressure of 100 lb/in.2 The tank develops a small hole, and air leaks from the tank at a constant rate of 0.03 lb/s for 90 s until the pressure of the air remaining in the tank is 30 lb/in.2 Employing the ideal gas model, determine the final temperature, in °F, of the air remaining in the tank. KNOWN: Air at specified initial temperature and pressure leaks from rigid tank until a final specified pressure is attained by the air remaining in the tank. FIND: Final temperature of air remaining in tank. SCHEMATIC AND GIVEN DATA: Initial State - State 1 Process 1 → 2 Final State - State 2 ∆t = 90 s ṁout = 0.03 lb/s Air Air V1 = 8 ft3 V2 = V1 = 8 ft3 T1 = 80°F p2 = 30 lb/in.2 p1 = 100 lb/in.2 ENGINEERING MODEL: 1. The control volume is defined by the dashed line on the accompanying diagram. 2. Air can be modeled as an ideal gas. ANALYSIS: The ideal gas model can be applied to the final state, state 2, to determine the temperature of the air remaining in the tank. p2V2 = m2RT2 Solving for temperature yields T2 = p2V2/m2R Pressure and volume are known at state 2. The mass in the tank at state 2, m2, equals the initial mass in the tank, mi, less the mass that leaks from the tank. Since the mass flow rate, ṁout, is constant, the amount of mass that leaks from the tank is: mout ∆t = (0.03 lb/s)(90 s) = 2.7 lb The initial mass, mi, is obtained using the ideal gas equation of state PROBLEM 4.1 (Continued) mi = p1V1 / RT1 The gas constant, R, is the universal gas constant divided by the molecular weight of air. Temperature must be expressed on an absolute scale, T1 = 80°F = 540°R. Substituting values and applying the appropriate conversion factor yield mi = (100 lb/in.2) (8 ft3) 1545 ft·lbf (144 in.2) lbmol · °R (540°R) mi = 4.0 lb Collecting results: m2 = 4.0 lb - 2.7 lb = 1.3 lb Substituting m2 to solve for T2 yields T2 = (30 lb/in.2) (8 ft3) 1545 ft·lbf (144 in.2) lbmol·°R (1 ft2) T2 = 498.5°R = 38.5°F Note the need to convert the final temperature from °R to °F to provide the answer in the requested units. PROBLEM 4.2 KNOWN: An initially-empty storage tank is filled with liquid propane at a constant mass flow rate. State data and tank dimensions are provided. FIND: Determine the time, in minutes, to fill the tank. SCHEMATIC & GIVEN DATA: Move(t0) = 0 ṁin = 10 kg/s D = 4 m Initial condition in tank T = 20°C, p = 8 bar Engineering Model: 1. A control volume encircles the tank. There is a single inlet. 2. The mass flow rate is constant. ANALYSIS: For the single-inlet control volume, the mass rate balance reads dmcv/dt = ṁin dmc(t0) = ∫ ṁin dt = ṁin tf Mcv(tf) = ṁin tf At tf, the tank is filled with liquid propane at sat. T=20°C. mf = V/vf(T1) =((4/9)((4^2)(π)(9m))) = 159,160 kg Equation (1) gives (tf) = 159,160 kg/(10 kg/s) (1 min/60 s) = 265.2 min Analysis complete. PROBLEM 4.3 4.3 A 380 L tank contains steam, initially at 400°C, 3 bar. A valve is opened and steam flows out of the tank at a constant mass flow rate of 0.005 kg/s. During steam removal, a heater maintains the temperature within the tank constant. Determine the time, in s, at which 75% of the initial mass remains in the tank; and also determine the specific volume, in m³/kg, and pressure, in bar, in the tank at that time. KNOWN: Steam exits at a constant mass flow rate from a tank filled with steam at a constant temperature. FIND: Determine the time, in s, when 75% of the initial mass remains in the tank, and the specific volume, in m³/kg, and pressure, in bar, in the tank at that time. SCHEMATIC AND GIVEN DATA: V = 380 L pi = 3 bar Ti = 400°C mi = 0.75mi ENGINEERING MODEL: (1) The control volume is shown on the accompanying diagram. (2) The temperature in the tank remains constant. (3) The tank volume is constant. ANALYSIS: Applying the mass rate balance: dmi/dt = Σmi - Σme = -me = -0.005 kg/s Integrating over time and rearranging for t: t = (mi(t) - mi(0)) / -0.005 kg/s (1) From Table A-4 at Ti and pi; vi = 1.032 m³/kg. Mass values are as follows: PROBLEM 4.3 (Continued) mi(0) = V/vi = 380 [10⁻³ m³] / 1.032 m³/kg = 0.368 kg mi(t) = 0.75 × mi(0) = 0.75 × 0.368 kg = 0.276 kg Determine time, in s, using Eq. (1). t = (0.276 - 0.368) kg / -0.005 kg/s = 18.4 s To find p2, first calculate v2. v2 = V / mi(t) = 380 L [10⁻³ m³/L] / 0.276 kg = 1.377 m³/kg Interpolating in Table A-4 at Ti = 400°C, v2 = 0.1.377 m³/kg: p2 ≈ 2.5 bar PROBLEM 4.4 KNOWN: Data are provided for a crude oil storage tank. FIND: After 24 h, determine the mass and volume of oil in the tank. SCHEMATIC GIVEN DATA: Total volume = 2500 m³ (V) = 2 m³/min Initial volume of crude oil = 1.5(200π)³ Engr. Model: 1. As shown by the sketch, a control volume encloses the storage tank. 2. The specific volume of the oil is constant: v = 0.001 m³/kg (a) Mass rate balance: dmc/dt = mj - me, where mj = (A√y) / v * (2√m³/min * 60 min / 1 h) = 8 × 10⁴ kg me = π * Di² / 4 * Vi / v = π (0.1 - (Vt - 5)/V) = (0.001 m³*10³/1 h) = 8.36 × 10⁴ kg Integrating ⇒ mcv - mc(0) = (1.64 × 10⁴ kg) (2*24 h) = 39.36 × 10⁴ kg Lc: Vc = 4/3(8/5) & Vt = = 66.67 × 10⁴ kg So, mcv(24 h) = (C1.6+39.36) × 10⁴ kg = 1.06 × 10⁶ kg mcv (b) (v(24 h)) * v mcv(24 h) = (0.001 m³/kg) × (199 × 10⁴ kg) = 1590 m³ v PROBLEM 4.5 KNOWN: A kitchen-sink water tap leaks one drop per second. FIND: Determine on an annual basis the gallons of water wasted and the mass of the water, in lb. ANALYSIS: [Annual Volumetric Flow rate] = (1 drop/s) (1 gal/36,000 drop) (3,600 s/h) (1 year) / (8,760 h) = 6.336 gal/year [Annual Mass Flow Rate] = (6.336 gal/year)(62.3 lb/ft^3) (1.3346 ft^3/gal) = 52.740 lb/year Rgiven PROBLEM 4.6 * Figure P4.6 shows a mixing tank initially containing 3000 lb of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of 0.8 lb/s and the other delivering cold water at a mass flow rate of 1.6 lb/s. Water exits through a single exit pipe at a mass flow rate of 2.6 lb/s. Determine the amount of water, in lb, in the tank at time to+one hour. SCHEMATIC & GIVEN DATA: [m_dot = 2.6 lb/s] ENGR. MODEL: 1. The control volume hosts the inlets and one exit, as shown in the sketch. 2. The mass flow rates are constant. ANALYSIS: The mass rate balance, Eq. [2.7], reads [dm_cv/dt = m_dot_i + m_dot_j - m_dot_k] = (0.8 + 1.6 - 2.6) lb/s = 0.8 - 5 lb/s Integrating from t = 0 to t = 1h (3600s), [m_cv(1h) - m_cv(0) = (-0.5 lb/s)(3600s)] m_cv(1h) = m_cv(0) - 1,800 lb = 3,000 lb - 1,800 lb = 1,200 lb PROBLEM 4.7 KNOWN: Water enters and exits a tank. State data are provided. FIND: Determine the mass flow rate at the inlet and exit, each in kg/s. Also, determine the rate of change of mass contained within the tank, in kg/s. SCHEMATIC & GIVEN DATA: T(liquid) | | --- 1 bar 200°C --- 100 bar --- 199.9°C (Table A-3) V (tank) (t_tank = ambient) Liquid-Water in V1 = 200 m^3 A1 = 6.00 m^2 V1 = 60 m/s Steam-Water out Vg = 100 m^3 A2 = 10 m^2 V2 = 110 m/s Engineering Model: 1. The control volume is denoted by the dashed line on the figure. Analysis: With data from Table A-4, m_dot_i = A1*V1/V_s = (6.0 m^2 * 60 m/s)*1 kg/(0.1994 m^3/kg) With V_ffg(*T_) and data from Table A-2, m_dot_2 = A_2*V_2/V_fg = (10 m^2 * 110 m/s)*1 kg/(0.904 m^3/kg) A mass rate balance reads, dm_cv/dt = m_dot_i - m_dot_2 = 1 kg/s - 5.5 kg/s = -4.5 kg/s PROBLEM 4.8 KNOWN: Liquid water flows at 20°C through a duct at steady state. State data and duct dimensions are provided. FIND: At a duct exit, determine the mass flow rate, in kg/s, and velocity, in m/s. SCHEMATIC & GIVEN DATA: T = 20°C Liquid water @ 20°C d = 0.02 m d_1 = 0.04 m p_in = 50 bar V_2 = ? Analysis: At steady-state the mass rate balance reduces to m_dot_2 = m_dot_1 Mass inflow Engineering Model: 1. A control volume encloses the duct between the figure. 2. The control volume is at steady rate. 3. U(T_2*) = U(T_1) (Table A-4) With m_dot_2 = 0.8 m^3/s, [Analysis] Continuity Eq: A2*V2/A1*V1 = mass assumption Thus, m_dot_1 = A_1*V_1 (ρ*V_1^2/(P/T)*400=1/s) 12.54 kg/s Continuity Eq: A2*V2 = A1*V1 Thus, V2 = (A_1/A_2)*V1 = (π * 0.02^2)/gπ(0.04)^2)40 m^3)) = 10 m/s A simple duct such as the one under consideration, we do not expect pressure P2 to depart significantly from pressure P1.