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systems analysis and design

The Analysis And Design Of Linear Circuits 7th Edition Roland E Thomas, Albert J Rosa, Gregory J Toussaint - Solutions

=+a new total pressure characteristic for a speed of 700 rpm for the good selection zone. (b) Con-struct a new shaft power curve for 700 rpm. (c) Sketch the total pressure and power character-istics for both speeds on the same graph with a system characteristic for 11,000 cfm (800 rpm).
=+(d) What are the total pressure, volume flow rate, and shaft power at 700 rpm?
=+12-4.The fan of Fig. 12-8 is rated at standard sea-level density. Suppose it is to be used in Denver, CO, where the elevation is 5280 ft. (a) Construct total pressure and shaft power characteristics for the selection zone for the new condition. (b) Compute the percent change in power between sea
=+12-5.The fan of Fig. 12-10 is rated at sea level. Suppose it is to be used in Albuquerque, NM, where the elevation is 1618 m. (a) Construct the total pressure characteristic for 800 rpm. (b) Com-pute the percent change in power from sea level to 1618 m for 900 rpm and a volume flow rate of 150
=+12-6.Comment on the desirability of using the fan described in Fig. 12-8 to circulate (a) 15,000 cfm at about 1.0 in. wg total pressure, (b) 10,000 cfm at about 1.8 in. wg total pressure, (c) 4000 cfm at 2.1 in. wg total pressure.
=+12-7.Would the fan shown in Fig. 12-9 be suitable for use in a system requiring (a) 30,000 cfm al 5.0 in. wg total pressure? (b) 5000 cfm at 1.5 in. wg total pressure? (c) 15,000 cfm at 4.0 in.wg total pressure? Explain.
=+12-8 A duct system has been designed for 150 m3/min at 400 Pa total pressure. Would the fan shown in Fig. 12-10 be suitable for this application? Explain, and estimate the total efficiency, fan speed, and shaft power.
=+12.9.A small system requires 0,88 in. wg (0.22 kPa) total pressure at a flow rate of 1420 cfin(0.67 m3/s). Select a suitable fan using the data of Tahle 12-1. (a) Sketch the system and fan characteristics, showing the operating point.
=+(b) What are the fan speed and power?
=+12-10. A duct system has been designed to have 3.0 in. wg total pressure loss. It is necessary to use an elbow at the fan outlet that causes a system effect factor of 0.3 in. wg. The fan inlet also has an elbow with a system effect factor of 0.20 in. wg. (a) What total pressure should the fan for
=+The fan shown in Fig. 12-10 is operating in a system at 900 rpm with a flow rate of 150 m2/min.
=+12-11.The system was designed to circulate 170 m /min. Assuming that the duct pressure losses were accurately calculated, estimate the system effect factor for the fan.A SWSI, backward-curved blade fan discharges air into a 12 x 16 in. rectangular duct at the
=+12-12.rate of 4000 cfm. An elbow located 30 in. from the fan outlet turns up. Estimate the system effect for the elbow.
=+The fan of Problem 12-12 also has an inlet duct and elbow as shown in Fig. 12-14c. The diam-
=+12-13.eter is 14 in ., the duct length is 28 in ., and the turning radius is 10.5 in. Estimate the system effect.
=+12-14.A backward-curved, centrifugal fan discharges 10.000 cfm (4.7 m3/s) through a 20 × 20 in.(0.5 x 0.5 m) duct into a plenum. The duct length is 10 in. (0.25 m). Find the lost pressure (sys-tem effect) for this case. Assume sea-level pressure.A forward-curved blade fan with a 12 × 12 in.
=+12-15 2500 cfm. What length of outlet duct is required to prevent any system effect if an elbow in position A is to be used?
=+12-16.The fan of Problem 12-15 has an inlet duct configured as shown in Fig. 12-14d with an R/H ratio of 1.0 and H of 12 in. About how long must the duct be so that the system effect will not exceed 0.16 in. wg?
=+12-17.The fan shown in Fig. 12-9 was selected for a constant-volume system requiring 15,000 cfm and a total pressure of 4.5 in. wg. The operating speed was expected to be about 900 rpm.
=+(a) When the system was started, measurements were made and a flow rate of 10,000 cfm and total pressure of 2.0 in. wg were observed. Explain the probable cause of the discrepancy. (b)At another time, after the problem of part (a) was corrected, a flow rate of 12,000 cfm and a total pressure of
=+12-18. Refer to Fig. 12-19 and compute the percent difference in shaft power between flow conditions 1 and 2.
=+12-19. A variable-volume system using a fan as shown in Fig. 12-19 will operate at about 15,000 cfm a majority of the time. (a) Estimate the power saved, in kW-hr for one day, as compared with a constant-volume system operating at point 1. (b) Would it be possible to operate at the min-imum flow
=+12-20.The fan of Fig. 12-18 is operating as shown in a VAV system. Compute the percent decrease in shaft power between flow conditions 1 and 2.
=+12-21.Assume that the fan and VAV system shown in Fig. 12-18 operate at an average capacity of 25,000 cfm over a given 24-hour period. (a) Estimate the power savings as compared with a constant-volume system operating at point 1. (b) Estimate the power saving as compared with
=+a VAV system with no fan control: that is, where point I will move along the full open characteristic.
=+12-22.Estimate the lost pressure in 50 ft (15 m) of 12 × 10 in. (30 x 25 cm) metal duct with an air-flow rate of 2000 cfm (0.94 m3/s). The duct is lined with I in. of the type A liner shown in Fig.12-23
=+12-23. Assuming the duct of Problem 12-22 is operating with standard sea-level air, estimate the lost pressure for air at the same temperature but at an altitude of 5000 ft (1525 m).
=+12-24.A circular metal duct 20 ft (6 m) in length has an abrupt contraction at the inlet and an abrupt expansion at the exit. Both have an area ratio of 0.6. The duct has a diameter of 10 in. (25
=+12-25.Compare the lost pressure for a bellmouth entrance (r/D = 0.06, Table 12-10B) and an abrupt entrance (0 = 180 degrees, Table 12-10A) for a duct velocity of (a) 1000 ft/min (5 m/s) and
=+(b) 4000 ft/min (20 m/s). (c) Compare the results.
=+12-26. Compute the lost pressure for a 14 in. (350 mm) pleated, 90-degree elbow with a volume flow rate of 1200 cfm (0.6 m3/s) of standard air.
=+12-27.Compute the lost pressure for a 16 × 16 in. (400 x 400 mm) 90-degree mitered elbow with a volume flow rate of 2500 cfm (1.2 m3/s) of standard air (a) with single-thickness vanes, design 3. and (b) without vanes.
=+12-28 Compute the lost pressure for a diverging wye fitting with a 45-degree branch. The flow rate in the 12 in. (30 cm) upstream section is 800 cfm (0.38 m3/s), and the flow rate in the 6 in. (15 cm)branch is 250 cfm (0.12 m3/s). The downstream section has a diameter of 10 in. (25 cm).
=+12-29.Compute the lost pressure for a diverging tee fitting. Use the data of Problem 12-28.
=+12-30.What is the lost pressure for an 18 x 18 in. (46 × 46 cm) duct discharging into a large plenum?The flow rate is 4500 cfm (2.1 m /s), and the duct expansion ratio A /A, is 6.0. Table
=+12-9B applies to this situation. (a) Assume an abrupt entrance. (b) Assume a 20-degree transi-tion exists at the entrance to the plenum.
=+12-31.Compute the lost pressure for a converging fitting. The flow rate in the 8 in. (20 cm) upstream section is 500 cfm (0.24 m3/s), and the flow rate in the 8 in. (20 cm) branch is 500 cfm(0.24 m /s). The downstream section has a diameter of 12 in. (25 cm). Assume (a) a 45-degree wye and (b) a
=+12-32.Compute the loss in total pressure for each run of the duct system shown in Fig. 12-35. The ducts are of round cross section. Turns and fittings are as shown. Use the loss coefficient and the equivalent length approaches (Table 12-14), and compare the answers. (a) Use English units. (b) Use
=+12-33.Refer to Problem 12-25, and compute the equivalent lengths for the two different entrances assuming a duct diameter of 12 in. (30 cm). Compute the lost pressure using the equivalent lengths.
=+12-34.Refer to Problem 12-26, and compute the equivalent lengths for the elbow. Compute the lost pressure using the equivalent lengths.
=+12-35. The duct system shown in Fig. 12-36 is one branch of a complete air-distribution system. The system is a perimeter type located below the floor. The diffuser boots turn up 90 degrees. Size the various sections of the system, using the equal-friction method and round pipe. A total pres-sure
=+The system shown in Fig. 12-37 is supplied air by a rooftop unit that develops 0.25 in. wg total
=+12-36.pressure external to the unit. The return air system requires 0.10 in. wg. The ducts are to be of round cross section, and the maximum velocity in the main run is 850 ft/min, whereas the branch velocities must not exceed 650 ft/min. (a) Size the ducts using the equal-friction method. Show
=+12-37. Design the duct system shown in Fig. 12-38 for circular ducts. The fan produces a total pres-sure of 0.70 in. wg at 1000 cfm. The lost pressure in the filter, furnace, and evaporator is 0.35 in. wg. The remaining total pressure should be divided between the supply and return with 65
=+12-38.Design the duct system shown in Fig. 12-39 using the balanced-capacity method. Circular ducts are to be used and installed below a concrete slab. The total pressure available at the plenum is 0.18 in. wg, and each diffuser has a loss in total pressure of 0.025 in. wg. Use the
=+12-39. Refer to Fig. 12-34 and rework Example 12-14 assuming that the volume flow rate for each ter-minal is increased by 20 percent. Also, assume the use of round pipe throughout. The increased capacity requires a static pressure of 0.75 in. wg (187 Pa) at each takeoff. (a) Size the system using
=+(b) Specify the required fan total pressure and capacity,
=+12-40. Solve Problem 12-39 using SI units. Specify fan characteristics.
=+12-41.Refer to Fig. 12-40, and construct the energy grade line (total pressure versus length) for the system shown. The change in total pressure in in. wg is shown for each part of the system. What total pressure must each fan produce?
=+12-42.Refer to Fig. 12-40, and assume that a return fan does not exist. Construct the energy grade lines, and give the fan total pressure.
=+12-43.Refer to Fig. 12-40, and assume that the supply fan is moved just upstream of the coil section.Construct the energy grade lines, and give the total pressure for the fans.
=+12-44.Refer to Problem 12-43, and assume that the return fan does not exist. Construct the energy grade lines, and give the fan total pressure.
=+12-45 The makeup air duct for a low-velocity system must be sized to handle a maximum of 2000 cfm(0.94 m3/s). The length of the duet is 40 ft (12.2 m), and it has two mitered elbows (Table 12-8B, Design 3). The air inlet grille has a lost pressure of 0.25 in. wg (62 Pa) at 2000 cfm (0.94 m3/s),
=+pressure between outdoors and the mixing box. (b) If the mixing box pressure is maintained constant at the value computed above, which will be below atmospheric pressure, what preis sure loss must the adjusted damper induce at a minimum flow rate of 1000 cfm (0.47 m/m))
=+(c) What is the effective loss coefficient for the damper of (b)?Refer to Problem 12-36 and Fig. 12-37. Assume that the plenum is replaced by a draw-through
=+12-46.air-handling unit (AHU). The filters have a pressure loss of 0.10 in. wg. the coil has a preware loss of 0.5 in. wg, and the AHU has a miscellaneous casing loss of 0.05 in. wg. The fan can produce 0.90 in. wg total pressure at the design flow. The return system still requires 0.10 in. wg.
=+13-1. Use Eqs. 13-14 and 13-2 to show that the units of the mass-transfer coefficient h, are mass of dry air per unit area and time. Further show that c ., in Eq. 13-13 must be on the basis of unit mass of dry air.
=+13-2.It is necessary to estimate the rate at which water is removed from an airstream by a cooling coil. The mass-transfer coefficient is not known, but the coil was tested at the same Reynolds number under sensible heat-transfer conditions, and the heat-transfer coefficient was 10 Btu/(hr-ft2-F)
=+13-3.Estimate the mass-transfer coefficient for moist air flowing normal to a 1 in. (25 mm) diame-ter tube (Nu = 0.615 Re047). The air velocity is 100 ft/min (0.5 m/s). Assume standard air.
=+13-4.The Nusselt number for flow in a tube is given by Eq. 13-4, where the constants. C .,a, and b are 0.023, 0.8, and 0.3, respectively. Estimate a mass-transfer coefficient for moist air flowing in a 12 in. (30 cm) tube at the rate of 600 cfm (0.3 m3/s). Assume standard air.
=+Use Fig. 13-9 and Table 13-2 to select a suitable tower(s) for the conditions of Problem 13-13.
=+13-17.Repeat Example 13-3, changing the entering water temperature to 105 F. Compare the transfer
=+13-18.units with those of Table 13-1. How will this affect the tower dimensions?Repeat Example 13-3, changing the entering air wet bulb temperature to 79 F. Compare the
=+13-19.transfer units with those of Table 13-1. How will this affect the tower dimensions?The condensers for a centrifugal chiller plant require 200 gpm with water entering at 85 F and
=+13-20 leaving at 100 F: the outdoor ambient air wet bulb temperature is 76 F. (a) Select a suitable cooling tower using Fig. 13-9 and Table 13-2, and (b) compute the cooling range, approach.and the tower capacity.
=+Suppose the cooling tower of Fig. 13-7 is to be used to produce chilled water during the cool
=+13-21.winter months when the air wet bulb temperature is low. If the cooling load on the coils is 250,000 Btu/hr (73 kW) and the warm water temperature is 70 F (21 C), what wet bulb tem-perature is required to satisfy the coil load?Consider the cooling tower performance data of Fig. 13-7. (a) At
=+13-22.operate in Albuquerque, NM, to reject 500,000 Btu/hr (147 kW) with 2.5 gpm ton [0.045 L/(s-kW)]?
=+ (b) At what condition would the tower operate for the design wet bulb for Charleston.SC?
=+A refrigeration plant is rated at 1,200,000 Btu/hr (352 kW) with a cooling range of 10 F (5.6 C)
=+13-23.and cold water temperature of 80 F (27 C). What is the maximum wet bulb temperature allow able for (a) 240 gpm (113 L/s) and (b) 320 gpm (189 L/s)?A model G cooling tower (Table 13-2) is to be used to cool water from 97 F (36 C) to 85 F
=+13-24.(29 C). (a) What is the maximum wet bulb temperature allowable? (b) Suppose the cooling range is increased to 15 F (8 C). What is the maximum wet bulb temperature allowable?
=+13-5. Estimate the rate at which water is evaporated from a 1000 acre lake on an August day when the dry bulb and wet bulb temperatures are 100 and 75 F. respectively (43,560 ft2 = I acres Assume a heat-transfer coefficient / of 5 Btu/(hr-ft2-F) between the moist air and the lake sur face and a
=+13-6.The sensible heat-transfer coefficient for a dry surface has been determined to be 9 Bin(hr-ft -- F) [50 W/(m2-C)| at a certain Reynolds number. Estimate the total heat transfer to the surface per unit area at a location where the wall temperature is 50 F (10 C) and the state of the moist
=+13-7.In order to determine the latent cooling load produced, estimate the rate at which water is evan-orated from an indoor swimming pool. The pool area is maintained at 75 F (24 C) dry bulb and
=+63 F (17 C) wet bulb while the pool water has a temperature of 80 F (27 C). The pool has dimensions of 300 × 150 ft (100 × 50 m), Assume a natural convection condition between the pool water and air of 1.5 Btu/(hr-ft2-F) [8.5 W/(m2-C)].
=+13-8.A housekeeper hangs a wet blanket out to dry in a high wind. The blanket weighs 4 lb dry and 16 lb wet and has dimensions of 7 ft by 8 ft. Assume outdoor conditions of 90 F db and 50 per-cent relative humidity and that the blanket is at a temperature of 90 F. Estimate the time required for
=+13-9. Redesign the air washer in Example 13-1 assuming counterflow of the air and water.
=+13-10. Determine the final state of the air, the cross-sectional area, and the height of a counterflow spray dehumidifier that operates as follows:(n=60 F (16 C)"(=95 F (35 C)*n =50 F (10 ℃)/whl = 82 F (28 C)G _= 1200 Ibm/(hr-ft2) [1.63 kg/(s-m3)]Q_ = 5000 cfm (2.36 m2/s)Q = 55 gpm (3.5 × 10-3
=+13-11. Solve Problem 13-10 assuming parallel flow of the air and water spray.
=+13-12. A parallel-flow air washer is to be used as an evaporative cooler. Air will enter at 100 F (38 C)dry bulb and 62 F (17 C) wet bulb and leave at 75 F (24 C) dry bulb. Other operating condi-tions are as follows:(n =80 F(27C)G _= 1000 1bm/(hr-ft2) [1.36 kg/(m2-s)]he = 700 Btu/(hr-ftª-F) [13
=+13-13.A counterflow cooling tower cools water from 104 to 85 F when the outside air has a wet bulb temperature of 76 F. The water flow rate is 2000 gpm and the air-flow rate is 210,000 cfm. Cal-culate the transfer units for the tower.
=+13-14.A counterflow cooling tower cools water from 44 to 30 C. The outdoor air has a wet bulb tem-perature of 22 C. Water flows at the rate of 0.32 m3/s and the water-to-air mass-flow ratio is 1.0. Estimate the transfer units for the tower.
=+13-15.Estimate the tower dimensions for Problem 13-13. The air mass velocity may be assumed to be 1800 lbma/(hr-ft2), and the overall mass-transfer coefficient per unit volume Ua _, is about 125 1bm/(hr-ft)).
=+13-16.Estimate the tower dimensions for Problem 13-14. Assume a mass velocity for the air of 2.7 kg/(s-m2).
Find the voltage gain vO/vS and current gain iO/ix in Figure P4–1 for r¼5 kV. VS 100 ww ix 500 ww io + 400 + rix 2 kvo + | FIGURE P4-1
Find the voltage gain vO/v1 and the current gain iO/iS in Figure P4–2. For iS¼5 mA, find the power supplied by the input current source and the power delivered to the 2-kV load resistor. w + + 100 100i1 is V1 100 2 vo 2 FIGURE P4-2
Find the voltage gain vO/vS and current gain iO/ix in Figure P4–3 for g¼2 103 S. For vS¼5 V, find the power supplied by the input voltage source and the power delivered to the 2-kV load resistor. VS 1 www + 3kNvx g.vx Vx 500 ww 10 FIGURE P4-3 Vo
(a) Find the voltage gain vO/vS and current gain iO/ix in Figure P4–4.(b) Validate your answers by simulating the circuit in OrCAD. VS 1 + 2.2 W FIGURE P4-4 io 6.3 www 50ix 10 kvo
Find the voltage gain vO/vS in Figure P4–5. Vs + | + + Vx 50Vx 1 Vo + FIGURE P4-5
Find an expression for the current gain iO/iS in Figure P4–6.Hint: Apply KCL at node A. Rs (A) is VS +1 w RE BIE w Rc IE FIGURE P4-6
(a) Find the voltage vO in Figure P4–7.(b) Validate your answer by simulating the circuit in OrCAD. 5V +1 10-3vx 1 www + 1 www 1.5 . 4.7 + Vo FIGURE P4-7
(a) Find an expression for the gain iO/vS in Figure P4–8 in terms of RX.(b) Select a value for RX so that the gain is 0.227. VS +1 1 ww Rx w Vx +1 FIGURE P4-8 io 2.2 103 Vx
Find an expression for the voltage gain vO/vS in Figure P4–9. VS +1 Rs w- - Vx+ gvx Rovo FIGURE P4-9
(a) Find an expression for the voltage gain vO/vS in Figure P4–10.(b) Let RS¼10 kV, RL¼10 kV and m¼100. Find the voltage gain vO/vS as a function of RF. What is the voltage gain when RF is an open circuit, a short circuit, and for RF¼100 V?(c) Simulate the circuit in OrCADby varying RF from
Select g in the circuit of Figure P4–11 so that the output voltage is 10 V. 1 WWW. 2.2 1 mV Vx vo gvx FIGURE P4-11 +
Design a dependent source circuit that has a closed-loop voltage gain of 10 using a VCVS with a m of 100. The source circuit is a voltage source vS in series with a 1-kV resistor, and the load is a 3.3-kV resistor. (Hint: See Figure P4–10.)
Find the Thevenin equivalent circuit that the load RL sees in Figure P4–13. Repeat the problem with RF replaced by an open circuit. Rs RF Rp w w VS Vx + w FIGURE P4-13 VT, RT RL
Find the Thevenin equivalent circuit that the load RL sees in Figure P4–14. VS is Rs ww Rp | +1 1 +1 ris RL Thvenin circuit FIGURE P4-14
Find RIN in Figure P4–15 is[ ww vs + R + r.is RIN FIGURE P4-15

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