Question

Question 3 (a) A river flows with a flowrate of Q (mº/s). Develop an expression for calculating available hydraulic power at a site on the river when there is a significant fall, h (m), in the river bed. [6] (b) Consider a rectangular stream of width 2.3 m. The average velocity of the stream is 0.8 m/s and the depth of water is 0.4 m. If there is a sharp fall of 2 m at a location along the stream, find the available hydraulic power (kW) at this location. If a turbine having an efficiency n=60% is installed at the site, what electrical power can be generated? [Note: assume water density p = 1000 kg/m>] [4] (c) For the truss in Figure Q3, in which all members are 1.5 m in length, you are required to: (i) Calculate the reaction forces at the supports and sketch on diagram. (ii) Calculate forces in all members of the truss and indicate these on diagram. (iii) Calculate the strain in all members of the truss, if the cross-sectional area and Young's modulus are 250 mm² and 205 kN/mm², respectively. All members are 1.5 m in length. [10] 12 KN 7.5 kN Figure Q3

Answer 1

or where and h. 3 mo Sol: (06) Crivewi flowrate Q(ms) Fall in river bed = h (m) •Power available at the site is given by p=pgesh KW] Ang = wish kW) Ang w = Unit weight of water in (KN/m3) g= acceleration due to gravity (m/z). p = mass density of water (Kg/m3). Griven : Auerage relo locity = oid m/s. width of Stream Depth of water = 0.4m. fall, h=hm in Discharge, od= Area of flowx Average velocity (2.3x014) Xo18 = 0.736 mes. Power available at this location = pq sh = 1000 x 9.81x0.736 x 2 - 14440.32 W = 14.40 kw! Ang → Electrical that can be generated is given aby Pe = a (pgnus) = m x 14.40 kW x 14.40 = le si 2 Si Pn powe 6o. Pe Anl 8.64 kW 100 CS Scanned with CamScanner

(6) is Caso 12 KN 1.5/2 2015 LV D in EMA (V8x3) =O giver B –(341919 M Vo 20 Ev=o, gives 2 1.5 75 kW 16 = 60° 609 68 С. +(7.5x1.299) -t.5m k 1.5m o VB = -0.2475 KN VA=12.2475k tano T.5/2 or x = 1.5 tan 60 VATVB =12 2=1.299m i VA-0.2475=12. VA A = 12.2475 KN (i) D Forces in Members: FOA A: Joint LV-0 in FA simbo° = 12.2475 KN. A FCA ů Fox = 14.14 KN I (Compressive) VA=12.247 Ell=0, FCA FCA = 14.14 Cosbo° 2) FcA=7,07 KN Joint B: EV=Og gives (tensile FEB Simboo -0.2475=0. FEB ü l FEB = 0.2857 kN (tensile) EH=Og gives es FCB FCB Fes los 60° TFCB = 0.1428 KN (Compresive) 60° С. FDA Costo E 68 C 0.2475 KN CS Scanned with CamScanner

B A C FAC =0 ما B Adding @ and 12120 Joint C: {=0, we have E Foc For los 60° + FEG los 600 - Fac-Fcb =0 FEC 68 3660 (Foc + Fec) – 14.4256 kN -- FCB Ev=0g we have FEc sinboo - Foc sim 60° FEC - Foc=0 2FEC = 14.4256 FEC = 7.2128 kN (tensile) in Foc = 7.2120 KN (Compressive) Joint D: {H =Og giver FED FAD Ces 60° - Foc les bo-FED=0 FAO Foc AD - Foc FED Costo° (14.14 - 7.2128)Cos 60" = FED FED = 3.4636 KN (Compressive) (ii) Elongation of any member, &l= fL SL 오 L P=ancial force A = cross-sectional area i IEAD 14.14 = 0.00027 2500 do 5 Similarly strains in other members are calculated- E 600 A C AE SL = longitudinal Strain. I Oy AE 오 AE น e = = E = Young's modulus? CS Scanned with CamScanner