A Nuclear Reactor :: science

A Nuclear Reactor The term Nuclear Reactor means an interaction between two or more Nuclei, Nuclear Particles, or Radiation, possibly causing transformation of the nuclear type; includes, for example, fission, capture, elastic container. Reactor means the core and its immediate container. Nuclear Reactors are used to produce electricity . The numbers of Nuclear Reactor plants have grown sufficiently . Electricity is being generated in a number of ways, it can be generated by using Thermal Power. It can be employed by using two basic systems a Steam Supply System and an Electricity Generating System these two systems are related to each other. The Steam Supply System produces steam from boiling water by the burning of coals and the Electricity Generating System produces electricity by steam turning turbines. The Nuclear power plants of this century depend on a particular type of Nuclear Reaction, Fission (The splitting of a heavy nucleus like the uranium atom to form two lighter "fission ! fragments" a s well as less massive particles as the Neutrons). In the Nuclear Reactors this splitting is induced by the interaction of a neutron with a fissionable nucleus. Under suitable conditions, a "chain" reaction of fission in which events may be sustained. The energy released from the fission reactions provide heat, part of which is ultimately converted into electricity. In the present day Nuclear power plants, this heat is removed from the Nuclear fuel by water that is pumped past rods containing fuel. The basic feature of the nuclear reactor is the release of a large amount of energy from each fission event that occurs in the nuclear reactors core. On the average, a fission event releases about 200 million electron volts of energy. a typical chemical reaction, on the other hand releases about one electron volt. The difference, roughly a factor of 100 million electron volts. The complete fission of one pound of uranium would release roughly the same amount of energy as the combination o f 6000 barrels of oil or 1000 tons of high quality oil. The reactor cooling fluid serves a dual purpose. Its most urgent function is to remove from the core the heat that results when the energy released from the Nuclear reactions is transformed by the collisions into the random nuclear motion. An associated function is to transfer this heat into an outside core, typically for the production of electricity. The designer provides for a nuclear core in a container through which a cooling fluid is pumped.

Assignment Solution 01

North South University ETE 321 – Spring 2010 Instructor: Nahid Rahman Assignment #1 Total Marks: 100 Worth: 7. 5% 1. Consider the sinusoidally modulated DSB LC signal shown below. The carrier DSB-LC frequency is ? c and the message signal frequency is ? m. (a) Determine the modulation index m. Solution: Amax = 25 Amin = 5 ? 25 ? 5 = = 0. 67 + 25 + 5 (b) Write an expression for the modulated signal ? (t). Solution: 1 1 ) = (25 ? 5) = 10 = ( ? 2 2 1 1 ) = (25 + 5) = 15 = ( + 2 2 = + cos = cos + ( ) cos Assignment 1 Sol Page: 1 of 12 = 15 cos + 10 cos cos (c) Derive time domain expressions for the upper and lower sidebands.Solution: = 15 cos + 10 cos cos = 15 cos + 5 cos( + ) + 5 cos( ? ) Upper sideband: 5 cos( + ) ) Lower sideband: 5 cos( ? (d) Determine the total average power of the modulated signal , the carrier power and the two sidebands. Solution: Power of carrier signal = (15 cos )2 = + ? (15)2 2 cos cos = (cos( + ) + cos( ? )) 2 1 = 112. 5 W 2 (5)2 2 (5)2 Power of upper sideband = (5 cos( Power of lower sideband = (5 cos( ))2 = = 12. 5 W = 12. 5 W Power of modulated signal = 137. 5 W (e) Assuming that the message signal is a voltage signal, calculate the PEP (Peak Envelop Power) across a 100? load. Solution: PEP = 2 ))2 = eed to obtain the RMS value by dividing the peak by v2. (f) Determine the modulation efficiency ?. Solution: 12. 5 + 12. 5 = = 18. 18% 137. 5 Amax is the peak value of the modulated signal. To calculate the DC power, we = ( )2 v2 = (25 )2 v2 100 = 3. 125 W Assignment 1 Sol Page: 2 of 12 2. A DSB-SC modulated signal can be generated by multiplying the message signal with a periodic pulse generator and passing the resultant signal through a band-pass filter. = 2 cos 200 + cos 600 ( )= 1 2 + 2 ? (? 1) ? 1 cos ( =1 2 ? 1 (2 ? 1)) (a) Find the DSB-SC signal component in V(t). Solution: Input to the BPF: = ? 1 1 2 ? (? ) = ( ){ + cos ( (2 ? 1))} =1 2 ? 1 2 1 2 2 ? 1 2 1 = ( ) { + cos + †¢ cos 3 + †¢ cos 5 + other terms} 2 3 5 1 2 2 2 ( )+ ( ) cos ( ) cos 3 ( ) cos 5 = ? + + other terms 2 3 5 Output of the BPF: 2 = cos 2 = 2 cos 200 + cos 600 cos (b) Specify the unwanted components in V(t) that need to be removed by a BPF of suitable design. Solution: 1 2 2 ( ), ( ) cos 3 ( ) cos 5 , ,other terms 2 3 5 (c) Assume the carrier frequency is 500 Hz. Sketch the spectral density of the resulting DSB-SC waveform. Solution: = 2 cos 200 + cos 600 =2 ? 200 + 2 + 200 + ? 600 + + 600 Assignment 1 Sol Page: 3 of 12 = 2 = 1 cos ? = 2 1 = 2 ? 00 rad = 1000 rad See plot below. (d) In the sketch for Part (c), specify lower and upper sidebands. + + 1 2 + 1 2 = 1 + 1 Assignment 1 Sol Page: 4 of 12 3. Let f(t) be a real signal. The transmitter transmits the following modulated signal = cos + sin Where is the Hilbert transform of f(t). (a) Explain that the modulated signal is a lower sideband SSB signal using an example of = cos . Solution: Note that there was an error in the question. The frequency of f(t) should be ? m inst ead of ? c. Any students with a reasonable attempt to this question will be awarded full marks. However, the solution below refers to the corrected problem. cos = sin cos + sin sin = cos = cos ? Since, cos ? = cos cos + sin sin ? = + ? + ? + For ? > 0, the impulse function is located to the left of the carrier frequency. For ? < 0, the impulse function is located to the right of –? c. Therefore, the modulating function produces lower sideband signals. (b) Determine the frequency of the modulated signal. Solution: From the expression of , the frequency of the modulated signal is ? . Assignment 1 Sol Page: 5 of 12 4. An SSB signal is generated by modulating an fc = 1 MHz carrier by the message signal = 2 cos 2000? t + cos 4000? t . The amplitude of the carrier signal is Ac = 1. a) Determine the Hilbert transform of f(t). Solution: = 2 cos 2000? t + cos 4000? t ? ? = 2 cos 2000? t ? + cos 4000? t ? 2 2 = 2 sin 2000? t + sin 4000? t (b) Determine the time domain expression of the lower SSB and upper SSB signals. Solution: ?  ± t = cos ? sin ? 2 sin 2k? t + sin 4k? t sin = 2 cos 2k? t + cos 4k? t cos = 2 cos 2k? t + cos 4k? t cos ? 2 sin 2k? t + sin 4k? t sin = 2 cos 2k? t cos + cos 4k? t cos ? sin 4k? t sin ? 2 sin 2k? t sin = 2 cos 2k? t cos ? sin 2k? t sin + cos 4k? t cos ? sin 4k? t sin  ± 2000 + cos  ± 4000 = 2 cos (c) Sketch the magnitude spectrum of the lower SSB. Solution: ? t = 2 cos ? 2000 + cos ? 000 ? ? = 2 + ? 2000 + 2 ? + 2000 + + ? 4000 + ? + 4000 A 2? ? -? c -? c+4000? -? c+2000? ?c-4000? ?c-2000? ?c ? (d) The coherent detection of the lower SSB signal consists of multiplying the received modulated signal by cos followed by a low pass filter. If the local (receiver) oscillator generates a phase error ? (i. e. the message signal is now multiplied by cos + , write the expression at the output of the low-pass filter and discuss how the phase error will affect the demodulated signal. Solution: Assignment 1 Sol Page: 6 of 12 Input of the L PF: = = = cos cos cos + + cos cos A cos B = sin A cos B = = os + sin + + cos sin + cos + cos A + B + cos A ? B sin A + B + sin A ? B cos 2 + + sin ? + sin 2 + sin ? A = ? sin A = cos + cos 2 + ? sin + sin 2 + Output of the LPF: = = cos ? sin 2 cos 2000? t + cos 4000? t cos ? 1 2 sin 2000? t + sin 4000? t sin 2 = cos 2000? t cos + cos 4000? t cos 1 ? sin 2000? t sin ? sin 4000? t sin 2 1 cos 4000? t cos ? sin 4000? t sin 2 = cos 2000? t cos ? sin 2000? t sin + cos A cos B ? sin A sin B = cos A + B = cos 2000? t + ? + cos 4000? t + ? Assignment 1 Sol Page: 7 of 12 5. A given DSB-LC transmitter develops an unmodulated power output of 1 KW across a 50-ohm resistive load.When a sinusoidal test tone with a peak amplitude of 5. 0 V is applied to the input of the modulator, it is found that the spectral line for each sideband in the magnitude spectrum for the output is 40% of the carrier line. Determine the following quantities in the output signal: (a) The modulation index. Solution: = + c os cos cos + cos cos = 1 1 = cos + cos ? + cos + 2 2 When a sinusoidal test tone with a peak amplitude of 5. 0 V is applied to the input of the modulator, it is found that the spectral line for each sideband in the magnitude spectrum for the output is 40% of the carrier line. 1 : = 0. 40 2 = : = 0. 0 (b) The peak amplitude of the lower sideband. Solution: A given DSB-LC transmitter develops an unmodulated power output of 1 KW across a 50-ohm resistive load. = /v2 = 1000 Am is the amplitude (or â€Å"peak†) of the modulated signal. We need to use the rms value when calculating DC power. = 1000 2 = 10 = 316. 27 Peak amplitude of the sideband = = 158. 11 (c) The ratio of total sideband power to carrier power. Solution: Total power of the sidebands = = Carrier Power = Ratio = : = . cos : = cos . + = . 0. 8 ? . = = 0. 32 + cos + (d) The total power of output. Solution: Assignment 1 Sol Page: 8 of 12 Total Power = + = 33kW e) The total average power in the output if the peak amplit ude of the modulation sinusoid is reduced to 4. 0 V. Solution: Changing the modulation sinusoid peak amplitude will affect the modulation index. 4 = 5 4 = ? 0. 8 = 0. 64 5 Ratio of total sideband power to carrier power = . : = . 0. 64 = 0. 2048 Total Power = + = 30. 12kW Assignment 1 Sol Page: 9 of 12 6. Suppose that a message signal f(x) has bandwidth B Hz. If f(x) is modulated by one of the modulation schemes DSB-SC or SSB or VSB, then for demodulation, the receiver must generate a (local) carrier in phase and frequency synchronous with the incoming carrier. This is referred to as synchronous or coherent demodulation. ) (a) Draw a block diagram for the demodulator. Solution: (b) Assume that there is a frequency error in the local carrier (the phase is correctly estimated). Give the expression of the Fourier transform of the output of the demodulator for the case of DSB-SC modulation, sketch the spectrum of the output signal, and compare it with the spectrum of the original signal ( you may assume an arbitrary shape of F (w)). Solution: = cos ? cos +? = = = = = 1 2 ? = 2 ? ? ? ? ? cos cos ? cos ? os ? +? cos +? +? cos 2 +? + cos 2 + ? + ? 2 ? +? Without frequency error: Assignment 1 Sol Page: 10 of 12 With frequency error: (c) Repeat (b) for the case of SSB-SC modulation (you may do so by choosing either upper SSB or lower SSB). Solution: + sin = cos Input to the LPF: ? = cos = = = = ? ? ? ? +? + cos cos + ? cos cos cos cos ? 1 2 ? ? sin +? +? cos + + cos 2 ? +? sin sin +? +? cos cos +? +? + = 1 2 sin ? cos ? + +? + 1 2 ? ? sin 2 sin ? ? 2 ? = ? ? 2 ? ? ? 2 + + ? 2 ? +? Assignment 1 Sol Page: 11 of 12 With frequency error: d) Suppose that you are an engineer who responds to design a modulation system for a coarse environment in which it is difficult to generate a local carrier in frequency synchronous with the incoming carrier during some period of transmission. Which modulation system would you like to recommend, DSB-SC or SSB? Justify your answer. Solutio n: For DSB-SC, we notice a distortion in frequency. For SSB, we only observe a frequency shift. Therefore, it would be better to use SSB for a coarse environment. Assignment 1 Sol Page: 12 of 12

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