Op Amp problems from Irwin and Nelms Circuits textbook, with solutions.Full description
Solved problems on different structural engineering topicsFull description
magnetic circuits
Solved problemsDescription complète
Full description
applied linear algebra
Mass Transfer Solved problemsFull description
Solved Problems: 5-1 A 2.8 liter four cylinder square engine (bore=stroke) with two intake valves per cylinder is designed to have a maximum speed of 7500 rpm. Intake temperature is 333K. Calculate: a. intake valve area (cm 2, in2) b. diameter of intake valve (cm, in.) c. valve lift (cm, in)
Vd = (2.8 L)/4 = 0.7 L = 0.0007 m 3 B = S; Vd = (π/4)B2S = (π/4)S3 = 0.0007 m3 S = 0.0962 m. = 9.62 cm = B Upmax = 2SN = (2 stroke/ rev)(0.0962 m/s)(7500/ 60rev/s) 60rev/s) = 24.1 m/s
Ai = CB2
where C is a constant = 1.3
.//
Ai = (1.3)(0.0962m)2 Ai = 0.000792 m2
Ai = 7.92 cm 2 = 1.23 in 2
b. For each valve: Ai = (πdv2)/4 = 7.92 cm2 dv =
. =
dv = 2.25 cm = 0.886 in.
c. lmax = dv/4 = (2.25 cm)/4
lmax = 0.56 cm = 5.6 mm = 0.22 in
5-2 Two engine options are to be offered in a new automobile model. Engine A is naturally aspirated with a compression ratio of 10.5:1 and cylinder inlet conditions of 60°C and 96kPa. Engine B is supercharged with aftercooling and has cylinder inlet conditions of 80°C and 130 kPa. To avoid knock problems, it is desirable to have the airfuel temperature at the start of combustion in engine B to be the same as in engine A. Calculate: (a) Temperature at start of combustion in engine A, using air-standard Otto cycle analysis. [0C] (b) Compression ratio of engine B which would give the same temperature at the start of combustion. (c) Temperature reduction in the aftercooler of engine B if the compressor has an isentropic efficiency of 82% and inlet conditions are the same as in engine A. [0C] Solution:
a. Using Fig. 3-5 and equation 3-4 in the book: T2 = T1 (rc)k-1 = (333 K)(10.5)0.35 = 758K = 485 C ° b. Using equation 3-4 758K = 353K(r c)0.35 , rc= 8.88 c. Using Figure 5-19 and equation 5-15, with k =1.4
5-3 A six-cylinder,3.6-literSI engine is designed to have a maximum speed of 6000RPM. At this speed the volumetric efficiency of the engine is 0.92. The engine will be equipped with a two-barrel carburetor, one barrel for low speeds and both barrels for high speed. Gasoline density can be considered to be 750kg/m3.
=0. 9 4 =0.74
Calculate: 1. Throat diameter for the carburetor (assume discharge coefficient 2. Fuel capillary tube diameters (assume discharge coefficient
5-6. A 2A-liter, four-cylinder engine is equipped with multipoint port fuel injection, having one injector per cylinder. The injectors are constant-flow devices, so the fuel flow rate into the engine is controlled by injection pulse duration. Maximum power is obtained at WOT when injection duration is continuous. At this condition, engine speed is 5800 RPM with stoichiometric gasoline and an inlet pressure of 101 kPa. At idle condition, the engine speed is 600 RPM with stoichiometric gasoline and an inlet pressure of 30 kPa. Volumetric efficiency can be considered 95% at all conditions. Calculate: (a) Fuel flow rate through an injector. [kg/sec] (b) Injection pulse duration in seconds at idle conditions. (c) Injection pulse duration in degrees of engine rotation at idle conditions.
Solution: For one cylinder Vd = 2.4 L / 4 = 0.6L = 0.0006 m 3 a.) use air flow rate (one cylinder) ma = = 1.1181(0.0006)(5800/60 1.1181(0.0006)(5800/60 rev/s)(0.95)/(2) = 0.03254 kg/sec ma = ma/AF = 0.03254 kg/s(14.6) = 0.00223 kg/s b.) ma = [(1.181)(30/101)(0.0006)(600/6 [(1.181)(30/101)(0.0006)(600/60)]/2 0)]/2 = 0.001052 kg/s mf = = (0.001052 kg/s)/14.6 = 0.0000721 kg/s mf(idle) = (0.0000721 kg/s)(2 rev/cycle)(600/60 rev/cycle)(600/60 rev/s) = 0.0000144 kg/ cycle time of injection t = (0.0000144kg) / (0.00223 kg/s) = 0.0065 sec
/
c.) rotational speed @ idle (600/60 rev/s)(360°/rev) = 3600°/sec Time of injection = (3600°/sec)(0.0065 (3600°/sec)(0.0065 sec) = 23.4
