Solution Manual for Fundamentals of Momentum, Heat and Mass Transfer, 6th Edition

Chapter 2 End of Chapter Problem Solutions 2.1 Assume Ideal Gas Behavior = = For T = a + by = 530 – 24 y/n = = ln = ln With P = 10.6 PSIA, Po = 30.1 in Hg h = 9192 ft. 2.2 (a) =0 on tank (1) At H20 Level in Tank: P=Patm+ wg(h-y) From (1) & (2): h-y=1.275 ft. (2) (3) For Isothermal Compression of Air PatmVtank=P(Vair) P= Patm (4) Combing (1) & (4): y=0.12 ft. and h=1.395 ft. (b) For Top of Tank Flush with H20 Level =0 P=Patm+ At H20 Level in Tank: P=Patm+ wg(3-y) Combining Equations: F= 196(3-y)- 250 For Isothermal Compression of Air: (As in Part (a)) 3-y = 2.8 ft. F = 196(2.8) – 250 = 293.6 LBf 2.3 When New Force on Tank = 0 Wt. = Buoyant Force = 250 Lbf Vw Displaced = 250/ wg = 4.01 ft3 Assuming Isothermal Compression PatmA(3ft.)= P(4.01 ft3) = (Patm+ gy)(4.01) y=45.88 ft. Top is at Level: yor at 44.6 ft. Below Surface 2.4 = = o = = 1ln(1- ) = 300,000 ln(1-0.0462) = 14190 Psi Density Ratio: = = 1.0484 so P= 1.0484 2.5 Buoyant Force: FB= V= For constant volume: F varies inversely with T 2.6 Sea H20: S.G.=1.025 At Depth y=185m Pg= 1.025 wgy = 1.025(1000)(9.81)(185) = 1.86×106 Pa = 1.86 MPa 2.7 r = Measured from Earth’s Surface R = Radius of Earth = g= go P-Patm= At Center of Earth: r = R PCtr-Patm= Since PCtr PCtr Patm = = 176×109 Pa = 176 MPa 2.8 = g = g P-Patm= g(+h) = (1050)(9.81)(11034) = 113.7 MPa 1122 Atmospheres 2.9 As in Previous Problem P-Patm= gh For P-Patm= 101.33 kPa h=101.33/ g for H20: h = Sea H20: h = Hg: h = = 10.33m = 10.08m = 0.80m 2.10 5 4 3 2 1 P1=Patm+ Hg g(12’’) P1=P2 P2=P3+ K g(5’’) P3=P4 P4=PA+ w g(2’’) P4=P5 Patm+ Hg g(12’’)= PA ’’ PA= Patm+ K g(5’’) g[(13.6)(12)-2-0.75(5)]=Patm+5.81 PA = 5.81 PSIG 2.11 Force Balance on Liquid Column: A=Area of Tube -3A + 14.7A – ghA = 0 h= = 26.6 in. 2.12 A B D C PA=PB- o g(10 ft.) PC=PB+ w g(5 ft.) PD=PC- Hg g(1 ft.) PA-PD= PA-PAtm= Hg g(1)- w g(5)- o g(10) w g(13.6 x1-5-0.8 x 10 x 1) = 37.4 LBf/ft 2 2.13 P3=PA-d1g w = PB+( Hg g)x(d3+d4sin45) PA-PB = = 245 LBf/ft2= 1.70 Psi 3 2.14 3 P3 = PA – wgd1 P3= PB – wg(d1+d2+d3) + Hg gd2 Equating: PA-PB= Hg gd2- wg(d2+d3) = 2 wg[(13.6)(1/12)-7.3/12] = 32.8 LBf/ft = 0.227 Psi 2.15 I PI=PA+ wg(10”) PII=PB+ wg(4’’)+ Hg g(10”) PI=PII PA-PB = wg[-6+13.6(10)] = 56.3 psi II 2.16 Pressure Gradient is in direction of – & isobars are perpendicular to ( – ) ( – ) String will assume the ( – ) direction & Balloon will move forward. 2.17 At Rest: P= o Accelerating: P= Equating: ya= Level goes down. = (g+a)ya which yo 2.18 F = P6.6A – PatmA = h( r2) h=2m r=0.3m F=5546N yC.P.= + Ibb/A For a circle: Ibb= r4/4 yC.P.= 2m + = 2.011m 2.19 Height of H20 column above differential element= h-4+y For (a) – Rectangular gate- dA = 4dy dFw=[ wg(h-4+y)+Patm]dA dFA=[Patm+(6Psig)(144)]dA = =0 [ g(h-4+y)-864](4dy)=0 h=15.18 ft. For (b): dA=(4-y)dy [ g(h-4+y)-864](4-y)dy=0 h=15.85 ft. 2.20 Per unit depth: Fy|up = wg =0 r2/2 {buoyancy} Fy|down = g r2+ 2 wg(r – r2/4) Equating: = g r2+ = w / = w 2 wgr (1- /4) = 0.432 w = 432 kg/m3 2.21 a) To lift block from bottom F = {wt. of concrete}+{wt. of H20} = cgV+[ wg(22.75’)+Patm]A = (150)g(3x3x0.5)+[62.4g(22.75)+14.7(144)]x(3×3) = 675+31828 = 32503 lbf b) To maintain block in free position F = {wt. of concrete}-{Buoyant force of H20)} = 675- wgV = 675-[62.4g(3x3x0.5)] = 675 – 281 = 394 lbf 2.22 Distance z measured along gate surface from bottom =500(15)- g(h-zsin60)dz=0 g =7500 g =7500 (62.4)g =7500 h3 = h=8.15 ft. = 541 2.23 Using spherical coordinates for a ring At y=constant dA=2 r2sin d P= g[h-rcos +rcos ] dFy=dFcos Fy = (h-rcos +rcos )(2 r2sin =2 gr2 (sin Let: c=2 = c[ d + =c d ] +r )(1- d ) gr2 sin = c[(h- d ) ] )- (0- )] Now for Fy=0 sin = =[1- 0=(h- )( ]1/2 & r=d/2 )+ ( ) = h- + Giving h= = = 1/2 For d=0.6m h= = 1/2 = 1/2 2.24 0 y 12 ft H2O, A 10 ft Mud, B PA-Patm= wg(12)=24g PB-Patm=24g+40g=64g Between 0 & A: P-Patm= wgy Between A & B: P= wg(12)+ mg(y-12) Per unit depth: F= = dy (192)+ (50) =18790 lbf Force Location: Fxy= = = dy+ (576+2040)+ =288,400 ft. LBf = = 15.35 ft. (2973-2040) )dy 2.25 Force on gate= A = (1000)(9.81)(12) (2)2 = 369.8 kN yC.P.= = 0.0208m (Below axis B) = =0 P(1) = (369.8×103)(0.0208) = 7.70 kN 2.26 Fc Fw 10 m Fw = A= (1000)(9.81)(4)(10)(1) = 392 kN ycp is 2/3 distance form water line to A ~ 6.66m down form H20 line ~ 3.33m up from A = Fc(9)=392(3.33) Fc = 145.2 kN 2.27 Width = 100m H20@27 =997 kg/m3 F= A = (997)(9.81)(64)x(160)(100) = 10.016 x 109 N = 10.02×103 MN For a free H20 surface ycp= (128m) = 85.3m {below H20 surface} =106.7 m {measured along dam surface} 2.28 Spherical Float Upward forces F + FBuoyant Downward forces WT W= gV= R3 ) Fb= g gVz= R3)z z= fraction submerged F= z= R3)- g R3 ) 2.29 = G is center of mass of solid = 2[1/2(L/2)(0.1L)(L) ( )-(0.9L)(L)(L) (0.05L )]+M {Part of original submerged volume is now out of H20} (Part that was originally out is now submerged} M= L4 [ + 0.045] = L4 (0.02833) = 0.00556 L4 2.30 Air Butyl Alcohol Benzene Water 17 feet 19 feet 34 feet 25 feet (a) the pressure at the butyl alcohol/benzene interface (b) the pressure at the benzene/water interface (c) the pressure at the bottom of the tank 2.31 For Blood: For Mercury: So, Solve for the height of the blood, Eliminate the gravity terms, We take 120 mm Hg as the height here, and 120 mm = 0.12 m, and the density of mercury is 845 = : 2.32 Freon 12 ft 2 ft Nitrogen 4 ft 7 ft Benzene 2.5 ft 4 ft 7.5 ft 10 ft 5 ft 8 ft 11 ft 5 ft 10 ft 7 ft 7.5 ft Water 5.8 ft Mercury Glycerin