
Interpretation:
The factor which influence the crystallinity of the polymers are to be determined. Also, the development and role of crystallinity in PET and nylon is to be explained.
Concept Introduction:
Polymers are the large molecules which are made up of many small repeating units bonded together in linear, branched or cross-linked manner.
They are of two types: natural and synthetic. Natural polymers are those which occur naturally, and synthetic polymers are those which are synthesized or man-made.
Crystallization of polymers is the process where the molecular chains of the polymers tend to align themselves when temperature is reduced. Crystallization in polymers happens in segments and affects

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Chapter 16 Solutions
Essentials Of Materials Science And Engineering
- A modulating signal f(t) is bandlimited to 5.5 kHz is sampled at a rate of 15000 samples/sec. The samples are quantized into 1024 levels. Calculate transmission bandwidth if the following modulation types are used for signal transmission: 1-ASK, QAM 2-QPSK, 8-PSK 3-FSK, 8-FSK with Af = 20 kHzarrow_forwardQ10 The full-load copper loss on the H.V. side of 100KVA, 11000/317 V, single-phase transformer is 0.62 kw and on the L.V. side is 0.48 kW. i) Calculate R1, R2 in ohms ii) Find X1,X2,if the percentage equivalent reactance is 4%, and reactance is divided in same proportion as resistance. Ans, 27.30, 0.175), 0.00482 . (7.5)arrow_forwardFind the binary sequence, for the following Differential Manchester code.arrow_forward
- Q2- What are the parameters and loss that can be determined during open-circuit test of singlephase transformer. Draw the circuit diagram of open-circuit test and explain how can you calculate the Parameters and loss.arrow_forwardQ6- the open circuit and short circuit tests on a 10 KVA, 125/250 v, 50 Hz single phase transformer gave the following results: O.C. Test: 125 V,0.6 A, 50 W ( on L.V.) S.C. Test: 20 V, 40 A, 177.78 W (on H.V. side) Calculate: i) Copper losses on half load ii) Full load efficiency at 0.8 leading p.f. iii) Half load efficiency at 0.8 leading p.f. iv) Regulation at full load at 0.9 leading p.f. Ans: 44.445 W, 97.23%, 97.69%, -1.8015%arrow_forwardQ3-A two-winding transformer has a primary winding with 208 turns and a secondary winding with 6 turns. The primary winding is connected to a 4160V system. What is the secondary voltage at no load? What is the current in the primary winding with a 50-amp load connected to the secondary winding? What is the apparent power flowing in the primary and secondary circuits? Ans. 120 V, 1.44 A, 6000 VAarrow_forward
- Q12- A three phase transformer 3300/400 V,has D/Y connected and working on 50Hz. The line current on the primary side is 12A and secondary has a balanced load at 0.8 lagging p.f. Determine the i) Secondary phase voltage ii) Line current iii) Output power Ans. (230.95 V, 99.11 A, 54.94 kW)arrow_forwardQ1- A single phase transformer consumes 2 A on no load at p.f. 0.208 lagging. The turns ratio is 2/1 (step down). If the loads on the secondary is 25 A at a p.f. 0.866 lagging. Find the primary current and power factor.arrow_forwardQ5-The efficiency of a 200 KVA, single phase transformer is 98% when operating at full load 0.8 lagging p.f. the iron losses in the transformer is 2000 watt. Calculate the i) Full load copper losses ii) half load copper losses and efficiency at half load. Ans: 1265.306 watt, 97.186%arrow_forward
- Q7- A 5 KVA, 500/250 V,50 Hz, single phase transformer gave the following reading: O.C. Test: 250 V,2 A, 50 W (H.V. side open) S.C. Test: 25 V10 A, 60 W (L.V. side shorted) Determine: i) The efficiency on full load, 0.8 lagging p.f. ii) The voltage regulation on full load, 0.8 leading p.f. iii) Draw the equivalent circuit referred to primary and insert all the values it.arrow_forwardQ4- A single phase transformer has 350 primary and secondary 1050 turns. The primary is connected to 400 V,50 Hz a.c. supply. If the net cross sectional area of core is 50 cm2, calculate i) The maximum value of the flux density in the core. ii) The induced e.m.f in the secondary winding. Ans: 1.029 T, 1200Varrow_forwarddesign rectangular sections for the beam and loads, and p values shown. Beam weights are not included in the loads given. Show sketches of cross sections including bar sizes, arrangements, and spacing. Assume concrete weighs 23.5 kN/m'. fy= 420 MPa, and f’c= 21 MPa.Show the shear and moment diagrams as wellarrow_forward
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