Assignments CE 312 Fluid Mechanics (Fall 21) Assignment Set 2 - s Due: Wednesday October 6; 1: m Question A (7 marks) Consider the situation in the Figure where you see a gate laced under an angle θ=1 o that searates layers of water (ρ w =1 kg/m 3 ) and oil (ρ o =92 kg/m 3 ). Both layers have a height =1.2 m. The vessel is W=1 m wide in the direction erendicular to the lane of view. Determine the net force (in x and y direction) exerted by the two liquids on the gate. Gravity (g) oints in the negative y direction. Due to the density difference of oil and water there is a ressure difference over the gate. This = g ρ ρ y. The force on a small ressure difference is a function of y only: ( )( ) ortion of the late with height (since the late is under an angle the length of this ortion is ) has an x and a y-comonent: cos θ df = x cosθ W W dfy sinθ W tanθ W cosθ = = cosθ =. The total force we find by integration: 1 1 F = g y W = Wg y y = W g 2 2 ( ρ ρ )( ) ( ρ ρ ) ( ρ ρ ) x w o w o w o 2 2 1 2 Fy = g ( ρw ρo )( y) W tanθ = W g ( ρw ρo ) tanθ 2 Substituting the numbers gives F x =564 N, F y = 99.4 N. w o
Question B (4 marks) Estimate the and absolute ressure in the gas tank if the liquid in the manometer has a secific gravity of 1.93, and h=39 cm. Neglect the density of the gas in your estimates. This is retty much the same as the examle in the Lecture Notes 2. The absolute ressure in the gas is g = + hgρl ; the ressure is the ressure in the gas relative to the osheric ressure: g, = hgρl. With ρ l =1.93 1 3 kg/m 3 and =1 5 Pa we get g, =7.38 1 3 Pa, and g =1.74 1 5 Pa.
Question C (4 marks) A tree trunk is floating in the river. It can be considered a solid wooden cylinder having a circular cross section with diameter D=6 cm, and length L=1 m. The density of the wood is ρ wd =96 kg/m 3, the density of the water is ρ wtr =1 kg/m 3. Can a m=8 kg erson stand on the floating trunk without getting her/his feet wet? The maximum buoyancy force is reached if the trunk is fully submerged. Then π 2 π 2 π 2 Fbuo = ρwtr g D L. For the answer to be yes ρwtr g D L ρwd g D L + mg which 4 4 4 π 2 imlies m ( ρwtr ρwd ) D L = 113 kg which is true. 4
Question D (8 marks) An oen-ended can, L=3 cm long is originally full of air at =1 bar and 2 o C. The can is now immersed in water (ρ=1 kg/m 3 ) as shown in the figure. Assuming that the air in the can stays at 2 o C, and that the air in the can behaves as an ideal gas, what will be the height h with which the water rises in the can? (neglect surface tension effects) We can determine the ressure in the submerged can in two ways. Equating these rovides an equation in the unknown h. In the first lace, since the air in the can is an ideal gas with constant temerature, the roduct of volume and ressure stays constant when submerging the L can: LA = subm ( L h) A subm =. ere A is the cross-sectional area of the L h ( ) can. In the second lace, the ressure in the can is equal to the hydrostatic ressure at the air- = + ρg h. water interface in the can: ( ) And thus ( L h) subm L = + ρg h ( ). If we multily both sides by ( L h) both sides by we get L L h ( h)( L h), and divide ρg = +. This is a quadratic equation in h. 2 Writing it in the standard ax + bx + c = form gives ρg 2 ρg ρg ρg h 1+ + L h + L = (always a good exercise to check if the units of ρg -1 all terms are the same; they are all meters). Inserting numbers: =.98 m, ρg ρ g 1+ + L = 1.3234 (dimensionless), ( ) 2 ρg L =.882 m. As a result: 1.3234 ± 1.3234 4.98.882 h =. The lus sign in this equation gives an answer 2.98 that does not make sense (h would be bigger than L); the minus sign answer is h=6.2 cm which does make sense.
Question E (7 marks) Marine biologists with an interest in forms of life in the very dee sea want to know the ressure at =11 km below sea level. They base their estimate on the equation = g ρw and come u with =1.78 1 8 Pa=178 bar (they took g=9.8 m/s 2 and ρ w =1 kg/m 3 ). An engineer comes by and says that under these extreme circumstances they may want to consider the comressibility of water, i.e. the fact that the density of water increases if the ressure increases. She gives the biologists the following equation that relates water density to ressure: ρ = ρ ( 1+ β ) with ρ =1 kg/m 3, and β =5 1-1 Pa -1 and then walks away. Please hel the biologists and determine the ressure at =11 km taking into account the comressibility of water. Since the density is not considered constant anymore we have to revert to the more fundamental fluid statics equation from the lecture notes: = ρg where (different from the z lecture notes) there is no minus sign because we chose to let the z-coordinate oint down instead of u. For in this equation we can take the ressure. If we also realize that the ressure is only a function of z we can write u the following ordinary differential d with = if z= (at the sea surface the dz absolute ressure is osheric). We can solve the ODE by searation of variables: d = gρdz. If we integrate both sides of the equation from the surface to we get 1+ β equation (ODE): = g ρ ( 1+ β ) ( ) ( ) 1 ln 1 β ρ β β ρ ( + ) = g so that ( ) = ( g 1) only a little bit higher than the estimate based on constant density. 1 e =1.18 1 8 Pa. This is β