ru MECHANECE = 1, The mass of air in a room which is 3m x 5 m x 20m is known to be 350 kg, Find its density. (ALM67KgInP = BATE kal? L617 gm? 1.716 kam? P- M1. _320kg Vo SxS xaos Pile e@Oer oe + # a e Scanned with CamScanneraaa ine hundr grams OF Water are mixed Wit grams OF al (cu.m.). What is \eresul fing mixtures, assuming tat = ‘two > fuids mix completely? A.0.96, 1.04 cm'/g —B.0.82, 1.22 cm/g CC. 0.63,_1,59 cm3/g —_D. 0.86, 1.20 cm'/g S6* Fru ee = Mnig 2 My +My fas fear Bes furntes. Vinex Vyt\e FOKG frie = Ekg + BS ko Fer Be tes “rhe Se "oon 1 kg gs == 3) mix” 7— ee ee ao xe (= yas) poe Fix 960.48 cries @ODaoam 4 & 1 ste Scanned with CamScanneraaa 8. What is thé resulting pressure when one pourid of air at 15 sia and'200°F is heated at constant volume to 800°F? M=V5 B. 52.1 psia €.364 psia D..15 psia Paddle wi Th. fh =» Ge piQy _ AU) Th Te Bvt4D—g99 +460 — cy "IR e@Oeaoamses _ ute Scanned with CamScannerWee -sar- = 2 9, The volume of a gas under standard atmospheric pressure 76 cm Hg is 200 in®. What is the volume when the "pressure is 80 cm Hg if the temperatures unchanged?T,=Ta B, 90 in? C. 110 in? D. 30.4 in? Ful hl Th Te (xeem Geen) Cao em tg) Cle) Scanned with CamScanner10 kPa? (A.90 kPa B. 80 kPa C. 100 kPa D. 10 kPa GAGE PRESSURE (Relative pressure) fe pressures abet of below the atmecphere % con be fheasured by pressure gayges % monometers. ATMOSPHERIC PRessuRE (ot) pressure @ any point on the Sarth's sorface from the Wegnt of dhe Cir above it fabe= faim 2 f 9 look fa = fain FIO Vacuum a a SlRe@Oeoem 4 & Ves 01 2c Mectynemy De Scanned with CamScanner26, Determine the vertical pressure in N/m? due to a column of water 85 m high. A.8.33 x 10° B. 8,33 WE) C.8.33x 104 0.8.33 108 @ Water: - Gspecitie. weight) = Suater = 9-9) x = “gh ~ (me Is YG a = Ga y far 83,800 —— 2 weeoOoams & i oo” Scanned with CamScanner3. The pressure 34 meters below the ocean is nearest to:| A. 204 kPa B. 222 kPa S344 Kea] D. 362 kPa fiw = 1030 - @ Ocean (seawater) f= Pgh= 1080 (2.81)(04) = ove oe) @ are Scanned with CamScanner27 the abst press the baton of the cans 120 KP, how deep the ater ths point? A. 1.90 m B.185m 1.68. m D.1.55 m Fos = free + fe f= 120 ka < ior aas kPa = 12-ca5 kPa [8.G38.x10" = (080)(9.8))h a) =e « Fis @Oeroes s&s Toe Scanned with CamScannerZ 5, If the pressure at a point in the ocean is 60 kPa, what is the pressure 27 meter below this point? A. 256.3 kPa B, 521.3 kPa C3328 kPa ,1854 kPa PR rh ] e f= Gokfa P= GokFa + ((o3s)(9:81)(a9) ain | 1 lL per Scanned with CamScanneraaa & A pressure gage 6 m above the bottom of the tank containing a liquid reads 90 kPaj another gage height 4 m reads. termine the specific weight of the liquid. B, 5.1 kN/m? C. 3.2 KN/m? D.8.5 kN/m? 7 Ream 3 = Po hee =p f= Yh | Dips joa kfa- soxfa = Y@) 7 Teo ew om 8 8 Taam PONS teajamey Scanned with CamScannerene Sat 38. Find the lies pressure sufe reading if manometer ig, 60 in}HG. Mercury has a relative density of 13.6. * 30,2 kPa © B.32.5 kPa C. 79.8 kPa D. 0.35 kPa = Pgh S:G= 186 = fy GR "9 ' B= ekimn\aanGcoue I) ccs Your Suarer SS SS ae Aneeren ndeea a aD Scanned with CamScannerWhat is the specific gravity of th: D.0.11 SG-f f= Pah > Sb =o re Spuso* 33 BE Reaver water 19" “ =) SB as : 369 . == 2 Gay — = seoOeaomsse [Am ee tye Scanned with CamScanneraas (0 + 0.5h/ where w is in|kN/m?> }rd his in meters. Determine the 7, ‘The weight densty of a mud is given pressure, in kPa, at depth of Sm. Ao eRe 1B. 56.25 kPa C.62.5 kPa D. 78.54 kPa wa “3 a4 = Reh Ss -oH © fy (Sa = (‘orana Scanned with CamScanner10. A two-meter square plane surface is immersed vertically below the water surface. The immersion is such that the ‘two edges of the square are horizontal. If the top of the square is 1 meter below the water surface, what is the total water pressure exerted on the plane surface? A. 43,93 kN B, 52.46 KN C.64.76 IN y Dees watts Suk © cmmersed [adbmergs =: Fe 9A aerh =(481n0" A )ancon) (i +2) oe eoniwe@Oeam sd & Scanned with CamScanner11. Find the total water pressure on a vertical circular gate, 2 meters in diameter, with its top 3.5 meters below the water surface. a A387 KN | B. 107.9 KN C. 169.5 kN D. 186.5 kN water Sur face - _¢ a teqtrid of Ze — ple =O = Fah unde = redkus A Pe(eanat ts) (Fea? (asr'e.) ES Scanned with CamScanner13, Ablock of wood requires a force of 40 to Keep immersed in water and afore of 100N to kep itimmesed | in glycerd(sp.gr. = 1.3). Find the weight and sp. gr of the wood. AO] B.06 _ oR ‘0.08 | Fas yV SG ,F wee 2 yw N Fea Sur + 2heteal = 0 @ Water’ ater Sign oF the tid Eee SVEPAWS Quine et tre Pedy (4-81x18') Vygooo ~ 4d-W*0 | — 9. 81x10" Vion ~ Wyeao = 40 0) w te © Glycerin’ Neda (Sit) | C-BY(8-81.415" ) Vroom ~ 160 ~ Wyrgon =P [ATS 3\yoap ~ Wwe = (50 | 100 fs. Veooo = 22 rt? © yea * GON : S600" Scanned with CamScanner._29, A stone weighs 105 Ibs in air and 83 Ib in water. Find the specific gravity of the stone. A298 8.035 Tay D:2.21 Derived = ; SG = Ware = lob = 47 : Wate - Wwater Jos - 33 © _ Th CHO G Oo s S cla Scanned with CamScanner12. An iceberg having specific gravity o0Sie floating on salt water of so 103, Ifthe volume of the ice above the water surface is 1000 m’, what is the total volume of the ice? A. 8523 m? B, 7862 m? D, 6325 m? DERLVED : Veseweresp = SG coy Vera SG.roux0 Scanned with CamScannera S al B an OA "19, An otfce has a Gide of Sege of Ol6land 2 coe conta of 3, Deteie he coset for the orifice. Co c Cy B. 0.99 C.0.97 D. 0.96 OCC = = 0-62 = Cy Co.63) ST aovlwne@Ooam sw [ee OPA RE Mas douy A Bw ces Scanned with CamScannerCi 2-o-Sar- ooo . aT S ~ 21, The theoretical velocity of flow through an orifice 3m below the surface of water in a tall tank is: A, 8.63 ms B, 9.85 m/s C5ums LOD TRE Torricetti's Ein : V=Ca\agh S coeff. of discharge VE Ha x4. s193 Scanned with CamScannerc -Wir-o-sar- ° oe __35, Determine the velocity of the fluid in a tank at the exit, given that surface hi = 2m and hy = 5 cm, A.8.2mjsec f BGDiHIseC I] C53 mjsec D, 20 m/sec V=Cyfagh \F antein(a a5) i) a 7c LWaT a See aah oat Scanned with CamScanneraaa . 39. The static head corresponding to a flow velocity of 10m/sec is: A. 6.10 m B.5.09m | C. 1.67 m D. 8.36 V=\(agh = lo= [axtaien ES 2S Str Cangas s Tees ela Scanned with CamScanneroas 37. Calculate the discharge in ters per second through 2 Sin diameter orice under a head of 7.6¢F water. Assume “coefficient of discharge of 0.65 ADByrse BIST cn Mpie Yun? Perea GQ =fy velocity a Re E(sinx fen toy ) = 0.0103 ne Ve calagh = 0.65| 3(q.aG¢) = 7818 2 Qe(oone "(4 gia 2)= gins Wx Ss Ss wr SSS SSS SS = ctr @Deoomss Vom vo) Maree oe Scanned with CamScanneroae 40, A water tank with diameter 5 mis filled with water. If the water level is 24 mabove the nozzle and the nozzle is 10 ft above the ground, how long will it take the water to hit the ground? A1sec B15 sec pis! D.2.18 sec desn aum 3 ay ———— he \ot +at" SEIS * =senome us Tiare UNDE renv ney Aa Scanned with CamScannera-o-Sar- 7 Caney RTT quant a We wed ay ottermine , 14. Reynolds number may be calculated from the type of flaw pation: A. diameter, density and absolute viscosity \. bominde Flow Jeritiot ley B. diameter, velocity, and surface tension aturbulent Flew diameter, velocity, and al Viscosity D. characteristic length, mass flow rate per unit area, and absolute viscosity Re= VD . VOR, OFD Ay dd Rul, Vrmean velocity D> pipe diameter Aig? dynamic —— Pa. s Nr kinemete " me fe Ra density ae Plow rea ele @Oaomas & Tee on Scanned with CamScannerLE o Sar- aopeae _ 22. Water having kinematic viscosity v = My x 104 m’Js flows in a 100-mm diameter pipe at a velocity of 4.5 m/s. The Reynold’s Number is: B 258, aa C. 387,4 D. 298,750 h R Streams ea — a ———— Kia between Coriti cnt) 9 Qob]- 3999 Se fee ASOD I Ie @ OOOO 4 w Tone ©) are rey a= Scanned with CamScanneroe rs ———S ss - __ 30. Ethylene glycol at 60°C, with a velocity of 4 cmys enters a 2.5 cm ID tube, At 60°C the viscosity is 4.75 x 10 Ly Js, determine the Reynolds Number, omere A200 - B20) C. 2000 D. 1200 i. 7 G Px aoe) (asc x) 4.5 x1 e MY, se@OeRom sh & Vase olan Scanned with CamScannerLWie-sar- an . 31. Air at atmospheric pressure and with a mean velocity of 0.5 m/s, flows inside a square section duct of side b = 2.5 cm. If the air temperature is 350°K, determine the Reynolds Number (at 350°K viscosity = 20.76 x 10* m’/s). A206 8.302 C.108 "De Rez qo = (ostiiem . 2 UK Gone STERN oo H1e COA sw Ts 0] ARE Putney A Ow Oed & oO Scanned with CamScanner2 Water is flowing in a pipe with radius of 12 inches and a velocity of 11 m/sec. Tet cea ‘$€¢. What is the Reynolds Number? A264 8.5930" ¢.5800 0. 9189 Re= WP iy en Ag Re (0h Yaniaion Bey ooe aS, (ioral ig) Scanned with CamScannerWEo-Ser- ° =? » 18. atthe of at pasng tough pe with dane of 20 mm and speed of 0.5 mysec? ALM fO'mys B2SLx10%m)s “MSTA MIS 0.187x10*m's Qztv= Lee) Ch aris eer mas # Teal CPW wenn "Aa da Scanned with CamScanner__ 20, Determine the mean velocity of an engine al at OC fling at 0.01 kg/s through a square duct 1 cm in cross section. At 60°C density is 864 kg/m’, == B.0.1517 ms C. 0.2263 m/s D. 0.2630 m/s ae, a "Gan tea a wi]. D W (3° Scanned with CamScanner33. Water is flowing through a cast iron pipe at the rate of 3500 GPM. The inside diameter of pipe is 8 in. Find the flow velocity. ny 13. SAU in == B.8.35misee | cane . 105 m/sec airerthy 3) fae avo Ce =? Gres ee @ (2S) Ge san | Ic sine ey) ReOeacasu Scanned with CamScanner16. What is the expected head loss per mile of closed circular pipe (17-in inside diameter, friction factor of 0.03) * when 3300 gal/min of water flow under pressure? B. 0.007 ft C. 3,580 ft D. 64 ft = a Q =v head Lose (hp) a = s ir ong ue) =68)(yix 2BH) (ask? Tie EP 222 Se ier WEE aGat) ye acess | ccvclw@ Daag a & Tse na Scanned with CamScanneree oat = Pease /Puter ~ ru@hnina ego Tits od dro tush Ratoe wads REET: Velocty of 1 ms. The pipe is 50 m long and 150 mm in diameter. Find the head lost du to fiction, AQTImY 8.045 PIC OSM LLM gg Kuater hy~ $k .¥e fe Reet OEM ae") (8) D 39 pat — \ page o p.0i(s0) ——= hee Te5 30a.) OSe6) Scanned with CamScanneraA8 36. Determine the friction head lass for fully developed laminar flow of ethylene glycol at 40°C (p = 1101 ka/m? through a 5 cm diameter, 50.m long pipe, if friction factor is 0.242 and a flowrate of 0.1 kg/s. Asam > [Bam > 634m * D.532m wm = __—S arm : ——— qm oi a he Genie _ 0.0468)" 7 TFs = Toi =n RG Ne Wo \ = 0-0463 "Io a ee Fle @ Oo om 4 & se Scanned with CamScanner3 —— ons 7 24, at omer se of nt ttn pp tl be used to cay 490 opm vith st of ed of 1056 ft per mile? Assume f = 0.019. “Inia A. 625 mm | BS7emm | C479 mm D. 352 mm Q= dag gat, HP (im (3): 0.2836" {5 hn = Fagg” sagt @Ds hpstoseft, im 3.99"). cmt = 3.dam pe m Su aa “ne RAM = im: (sem) = Ieoan = hee aL LV? > daaooreo) [CRF Sg Gs) a(4-81 ) opin @Oeoae 4 w Scanned with CamScanner25, The reciprocal of Bulk Modulus of any fuidis called. A.volumestrain B,volume stress © C.compressiility') . shape elasiaty g= | Scanned with CamScanner, 15. The sum of the ire head, elevation head, and velocity head remains constant, this is known as: C. Torrecell’s Theorem + V*= Jash B. Boyle's law > P= Fao D. Archimedes’ Principle > Pvayan uy Pressure hed v2 CP Z aKa Canc teat — ————, Velocity, elevation herd corIR OC ORaa 6 w Tee Seen) Scanned with CamScanner2-9-Sar > —— = 4 Aturbineis ated at 650 hp then the flow of water throug tis 0.85 msec Assuming an efeny of 8%, What is the head acting on the turbine? A58m B.66m 160m (D.6m pe fe Pig =? QYh oO. = sds tHow oo ——— ac aonls @Moro 8 w so 1 ae ma Scanned with CamScanner2 aaa = = - —9-u67 17. Water at a gauge pressure of 3.8 atm at street level flows it to an office building at a speed of a8 ays through a pipe 5.0 cm in diameter. The pies taper down to 2.6 cm in diameter by the top floor, 20 m above. Calculate the flow velocity and tge gauge pressure in such a pipe on the top floor. Assume no branch pie and ignore viscosity. C. 4.44 m/s, 2.34 x105 Pa B. m/s, 8.2x105 Pa D. 3.33 m/s, 3.24 x105 Pa Fy, = 3.8aimx lol-garxefa _ 39s 0a8Ta Tam a a 4 || \yn 0-G0"ls Bay Vo = Be Vs ds= 8:2 ~ 9.0S5m ss dy= aG ,~ 002m — To" \s CONTLNULTY EQUATION! (occa? Ve(oveit re) fg= 3.2atn eo cle @Oamama 4 w Paso 0 Ge. Scanned with CamScanner2 fi bhi bah! fa + GV + Paha Ly Ss potenticl Energy ee Kinetic Energy, per unit vilun per unit volume Riz das,oas fa \= 0-60 “Is f= lobo _k9 ao ° “= \a72.59 "Is he= aom I secoult@ODoagms & Vaso 0 | © 2c Hae Scanned with CamScanner2 P rh? poh fa + aes + Paha = ? 4 Potential Energy —— = ae Kinetic Erergy > Ar unit valvin per unit volume ——— c= P= 3as,oa¢ fo : = Y= O-Go “Ie z : PP = lobo _k9 m? an —— 2 eee \a= 9.59 "s he = am = I TO ETE TST seo vlR CORR Hw Toss 0) Bac im A Gm ae & oe O Scanned with CamScanner