Question

A vertical storage tank with a hemispherical bottom and
a cylindrical shell of 2 m. inside diameter and 4 m. high
is filled with water. A sharp-edge orifice 150 mm
diameter with C = 0.62 is located at the lowest point.
26. Determine the time to empty the cylindrical part of
the tank through the orifice.
A. 134.04 sec C. 152.11 sec
B. 145.6 sec D. 160 sec
27. Determine the time to empty the hemispherical
tank only through the orifice.
A. 60.41 sec C. 75.11 sec
B. 53.61 sec D. 48.11 sec
28. Determine the time to empty the whole tank
through the orifice.
A. 213.61 sec C. 220.41 sec
B. 187.65 sec D. 194.45 sec

Answer 1

Solution - Giren - R> zim. * 2m=D dia oforifice Cat 150mm = 0.15 m. (a=0.62. 4m im orifice of dia = 150mm o orifice area =a= TTX0.15 soli 4 mr. Time required to empty cylindrical part =? Consider small strip of volume below Eda 7,=4+1=5m. -1m Let Ŕ be the radius of tank. dv=TR. dz {as we knowth at dv dt as here tank is being emptied sign comes) dv ve » Qdt

1 dt -TRY dz (t-o) = -TIR Qidt = -1Ř. dzi relo cirty through orifice J2qz -:Q = Ca a vegz where a = area of sufice (da. Vagz.at - TRY - de integrate both sides s dt = S 122) az Ca a 120 zi as emptying & adag starts ? varie from ti to IZź to empty Cylindrical portion ny > t = - TRY 42 [522-J2;] Cd. a J2q ca arag. Vz -IR O t - TRY zh+1782 JEI Zz x [2 vz] 21 (-a dag 2TR Cd-a. Jag Now, substitute values;

w.kit R=1m; (a = 0.62 a= Tx0.15 mr. g=9.81 m/sea Z,= 5m! Zz=im. it= 2ft xiv 0.62x + x 0.15 x/2x9.81 4 => t = 8. 0-62X0-15 x 19.62 [15-ri *[15-] . t= 160.032 sec option D silta 160 see . Answer. I Time required to empty cylindrical portion. Now, after emptying cylindrical portum Consider small volume as below dz. zi 1 z. 91 Sz,= im ? Z₂zom ry From figure R² = (R-z trn brr R²-(R-3)?

we know that do Q.at - dv here. – dv=-A.dz - Tar, dz. = T[R_(R-Z] dz =- [R² _ [2 +2% -2R2)] dz =-(-_Z +2Rz) az Far= T[="_282) dz Q = Cd a 29Z and Ca a vzgz . dt = π (z -2R2) az ſz at = 22-2R2) dz cd-a 29 at ( 2 1 2_2R. 2 ) dz cd-a /2g integrate on both sides. » Stat Ca.a.v2g z > [t] ? 22 Ca a V2g T 23/2_2RzKdz Z 0 512 36 2 R. Z W? zi

For this case ZER=in 2. = om 2252 L-03- GR. Z 3 ( a. Izg t- -|(2022 48.64)- Ca a 129 ( 2XR31_UR-p? 3 T Ga Jeg [04 aeth 4 esh) .(o-(-1485/9) T t= Ca a vzg 14R/ 15 Cd a reg here; R=1m, Ca=0.62 as I x 0.15 m ( 148 59 148156 15 0-62***0-15"x32x9.89 4 • Answer 60.41 see col t= loption A by time to empty hemisphe- rical portion. Total time required to empty whole tants=T= 1604 60.41 T=220.41sec Answer (option'e'