KNOWN: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe.

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Oscar Tyler
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1 4.3 Steam enters a.6-cm-diameter pipe at 80 bar and 600 o C with a velocity of 50 m/s. Determine the mass flow rate, in kg/s. KNOWN: Pressure, temperature, and velocity of steam entering a.6-cm-diameter pipe. FIND: Mass flow rate, in kg/s. SCHEMATIC AND GIVEN DATA: Steam p = 80 bar T = 600 o C V = 50 m/s D =.6 cm ENGINEERING MODEL:. Flow is one-dimensional. ANALYSIS: The governing equation for one-dimensional flow in terms of specific volume is m AV v At p = 80 bar and T = 600 o C, the steam is superheated vapor. Obtaining specific volume of the steam from Table A-4: v = m 3 /kg. Cross-sectional area of the flow based on pipe diameter is A D 4 m (.6 cm) cm m Solving for the mass flow rate yields m ( m )(50 m/s) = 0.6 kg/s m /kg

2 4.6 Air enters a one-inlet, one-exit control volume at 6 bar, 500 K, and 30 m/s through a flow area of 8 cm. At the exit, the pressure is 3 bar, the temperature is K, and the velocity is 300 m/s. The air behaves as an ideal gas. For steady-state operation, determine (a) the mass flow rate, in kg/s. (b) the exit flow area, in cm. KNOWN: Air flows through a one-inlet, one-exit control volume with known pressure, temperature, and velocity at the inlet and exit. FIND: Determine the mass flow rate and exit flow area. SCHEMATIC AND GIVEN DATA: p = 6 bar T = 500 K V = 30 m/s Air p = 3 bar T = K V = 300 m/s ENGINEERING MODEL:. The control volume shown on the accompanying figure is at steady state.. The ideal gas model applies for the air. ANALYSIS: (a) The mass rate balance for one-inlet, one-exit, steady flow is m = m m For the inlet, state, the mass flow rate can be determined from given data and the ideal gas equation of state. A V A V p m = v R / M T Substituting values yields m = m 8 cm 30 6 bar s N m 834 kmol K 500 K kg 8.97 kmol 5 N 0 m bar m 4 0 cm = 0.35 kg/s

3 (b) The exit flow area can be determined from given data and the ideal gas equation of state. m = AV v AV p R / M T Solving for area A = m R / M T V p = N m 834 kg 0.35 kmol K K s kg 8.97 kmol m bar s bar N 0 m 0 4 cm 5 m = 5. cm

4 4.36 Nitrogen, modeled as an ideal gas, flows at a rate of 3 kg/s through a well-insulated horizontal nozzle operating at steady state. The nitrogen enters the nozzle with a velocity of 0 m/s at 340 K, 400 kpa and exits the nozzle at 00 kpa. To achieve an exit velocity of m/s, determine (a) the exit temperature, in K. (b) the exit area, in m. KNOWN: Nitrogen flows through a nozzle. FIND: (a) the exit temperature, in K, and (b) the exit area, in m. SCHEMATIC AND GIVEN DATA: Nitrogen T = 340 K p = 400 kpa V = 0 m/s m 3 kg/s Nozzle p = 00 kpa V = m/s ENGINEERING MODEL:. The control volume shown with the schematic is at steady state.. For the control volume, W cv 0, Q cv 0, and pe = Model nitrogen as an ideal gas. ANALYSIS: (a) The energy rate balance 0 = Q W cv cv m [(h h ) + ½ (V V ) + g(z z )] simplifies to 0 = [(h h ) + ½ (V V )] Since enthalpy values for nitrogen are provided on a molar basis in Table A-3, the energy rate balance is expressed in terms of molar enthalpy (h h ) [ + ½ (V V )] M 0 Solving for exit molar enthalpy, which is a function of only temperature, gives M h h (V V ) From Table A-3, h 9888 kj/kmol From Table A- (nitrogen): M = 8.0 kj/kmol

5 Substituting values and solving for exit molar enthalpy give h 9888 kj kmol 8.0 kg kmol m s h = 6683 kj/kmol m 0 s kj 000 N N m kg m s From Table A-3, the temperature that corresponds to the exit molar enthalpy is T = 30 K. (b) From the mass rate balance Substituting v RT p A AV m m v from the ideal gas equation of state and solving for the area give m RT A V p kj 8.34 (3 kg/s) kmol K (30 K) kg 8.0 kmol m (478.8 )(00 kpa) s A = m 000 N m kpa kj N 000 m

8 4.00 Separate streams of air and water flow through the compressor and heat exchanger arrangement shown in Fig. P4.00. Steady-state operating data are provided on the figure. Heat transfer with the surroundings can be neglected, as can all kinetic and potential energy effects. The air is modeled as an ideal gas. Determine (a) the total power required by both compressors, in kw. (b) the mass flow rate of the water, in kg/s. KNOWN: Separate streams of air and water flow through a compressor and heat exchanger arrangement. FIND: (a) The total power required by both compressors, in kw, and (b) the mass flow rate of the water, in kg/s. SCHEMATIC AND GIVEN DATA: Air p = bar T = 300 K m 0.6 kg/s W cva p 4 = 9 bar T 4 = 800 K 4 W cvb Compressor A Compressor B p = 3 bar T = 600 K 3 p 3 = p T 3 = 450 K 6 5 T 6 = 30 o C p 6 = p 5 Heat exchanger Water T 5 = 0 o C p 5 = bar ENGINEERING MODEL:. Control volumes at steady state enclose the compressors and heat exchanger.. For each control volume, heat transfer with the surroundings is negligible and kinetic and potential effects can be ignored. 3. The air is modeled as an ideal gas. ANALYSIS:

9 (a) A mass balance for the air flowing through compressor A, the heat exchanger, and compressor B gives m m m 3 m 4 = 0.6 kg/s The energy rate balance for compressor A 0 = Q W cva cva m [(h h ) + ½ (V V ) + g(z z )] simplifies to W cva m ( h h ) The specific enthalpies for air at state and state are obtained from Table A-: h = kj/kg and h = kj/kg. Substituting values and solving yield kg kj kj kw W cva = 84. kw s kg kg kj s Similarly for compressor B, W cvb m ( h3 h4 ) The specific enthalpies for air at state 3 and state 4 are obtained from Table A-: h 3 = kj/kg and h 4 = 8.95 kj/kg. Substituting values and solving yield kg kj kj kw W cvb =. kw s kg kg kj s The total power required by both compressors is W total W cva W cvb = ( 84. kw) + (. kw) = 406. kw The negative sign indicates power is to the compressors. (b) Since the air and water do not mix in the heat exchanger, the steady state mass balance reduces to m m 3 = 0.6 kg/s The steady state form of the energy rate balance m 5 m 6

10 simplifies to 0 Q W cv cv i V m i i h i gzi V m e e h e e 0 m h m 5h5 m 3h3 m 6h6 gze or substituting results from the mass balance Solving for the mass flow rate of water gives 0 m ( h h3 ) m 5( h5 h6 ) m( h h3 ) m 5 h6 h5 The specific enthalpies of water at state 5 and state 6 are obtained from Table A-: h 5 h f5 = kj/kg and h 6 h f6 = 5.79 kj/kg. Substituting values and solving yield kg kj kj s kg kg m 5 kj kj kg kg m 5 =.3 kg/s 3

11 4. An 8-ft 3 tank contains air at an initial temperature of 80 o F and initial pressure of 00 lbf/in. 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 lbf/in. Employing the ideal gas model, determine the final temperature, in o 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, in o F. SCHEMATIC AND GIVEN DATA: Initial State State Final State State Process Air V = 8 ft 3 T = 80 o F p = 00 lbf/in. t = 90 s m = 0.03 lb/s e Air V = V = 8 ft 3 p = 30 lbf/in. ENGINEERING MODEL:. The control volume is defined by the dashed line on the accompanying diagram.. Air can be modeled as an ideal gas. ANALYSIS: The ideal gas model can be applied to the final state, state, to determine the temperature of the air remaining in the tank. p V = m RT Solving for temperature yields T = p V m R Pressure and volume are known at state. The mass in the tank at state, m, equals the initial mass in the tank, m, less the mass that leaks from the tank. Since the mass flow rate, m e, is constant, the amount of mass that leaks from the tank is m e t = (0.03 lb/s)(90 s) =.7 lb The initial mass, m, is obtained using the ideal gas equation of state

12 m = pv RT The gas constant, R, is the universal gas constant divided by the molecular weight of air. Temperature must be expressed on an absolute scale, T = 80 o F = 540 o R. Substituting values and applying the appropriate conversion factor yield m = lbf ft in. ft lbf 545 lbmol R 540R lb 8.97 lbmol 44 in. ft = 4.0 lb Collecting results Substituting m to solve for T yields m = 4.0 lb.7 lb =.3 lb T = lbf ft in. 44 in. ft lbf ft lb lbmol R lb 8.97 lbmol = o R = 38.5 o F Note the need to convert the final temperature from o R to o F to provide the answer in the requested units.

Find: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k.

Find: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k. PROBLEM 6.25 Three m 3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kpa. The air receives 1546 kj of work from the paddle wheel. Assuming the ideal gas model,