Q1 Give answers to all of the following questions (5 marks each):

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1 FLUID MECHANICS First Year Exam Solutions 03 Q Give answers to all of the following questions (5 marks each): (a) A cylinder of m in diameter is made with material of relative density 0.5. It is moored in fresh water by one end and the water level is m above the middle of the cylinder. Find the tension of the mooring cable. When the cylinder is submerged to the middle, the tension is zero (σ = 0.5). Additional buoyancy due to further submerging is compensated by the cable tension: T = ρ g π d 4 l = 000 kg/m3 9.8 m/s π m m 7700 N 4 (b) A Pitot-static tube is place in an air flow (ρ =.3kg/m 3 ). A connected manometer shows pressure difference 0mm of water. Whats is velocity of the flow? Using Bernoulli s equation for a stagnation point we can write: P + ρ a U = P 0 U = P/ρ a. The pressure reading is in mm of water, that is P = ρ w g h. Then U = ρ w g h ρ a and U = (000/.3) 9.8 m/s m = 7.4 m/s (c) A mm space between two parallel plates is filled with viscous fluid. One plate is moving with velocity m/s. Find the mean velocity of the flow between the plates and the flow rate if plates width is 0cm. The velocity profile is linear, and the mean velocity is the average between velocities on the walls: U = 0.5 m/s. The flow rate is Q = a b U = 0 3 m 0. m 0.5 m/s = 0 4 m 3 /s (d) Water (µ = 0 3 Pa s) flows through a pipe of 5mm in diameter with mean velocity 0.m/s. Can the formula f = 6/Re be applied to calculate the friction factor for this flow?

2 Reynolds number of the flow: Re = ρ V d µ = 000 kg/m3 0. m/s m 0 3 P a s = 000. Re < 000 flow is laminar and we can use formula f = 6/Re for calculating the friction coefficient. (e) A lock gate is m high and 3m wide. It can rotate around a vertical hinge at one side. Calculate the maximal force applied to the gate by water, and maximal moment about the hinge. Maximal possible depth is h = m, the maximal pressure is P = ρ g h, for the triangular distribution the total load is F = P A = ρ g h b = 000 kg/m3 9.8 m/s m 3 m = N. The load is applied in the middle of the width, therefore the moment about the vertical axis is: M = F b/ = N.5 m = 8890 N m (f) Calculate the minimal head of the pump required to pump m 3 of water per second to the height 0m trough a pipe with cross section area 0.m. Assume constant friction coefficient f = 0.0 and neglect all losses in fittings. The pump head is the gravity head plus losses, which can be found by Darcy s equation and are minimal when the pipe is vertical: H = 0 m H = Z + h = Z + 4 f L d Q g A 0 m ( m 3 /s) 9.8 m/s (0. m ) = m Q A tank 900mm square in plain is filled with water and is drained via a 60mm diameter pipe, 6m in length. The pipe has two bends along its length with head loss coefficients k = 0.6 each, and for the entry of the pipe k = 0.5. The outlet of the pipe is 30mm below the level of the base of the tank (Fig. ). The value of a friction factor f = can be assumed constant.. What is the depth h of water in the tank when the rate of flow through the pipe is 7 liters per second? [7 marks]

3 . How fast the pressure at the bottom of the tank changes for this value of h? [6 marks] 3. What is the value of the flow rate when the tank is half empty? [5 marks] 4. Justify application of steady-flow equations to this problem. [ marks] A. Z 30 mm h K = 0.5 K = 0.6 a K = 0.6 Fig.. Steady flow energy equation along : Z Q a g = ( 4 f L d + K Σ ) Q a g, where a is the pipe cross section area, K Σ = i K i is the sum of all loss coefficients, and Z(t) is the the height of the water surface in the tank above the outlet at time t. For Z(t) we have ( Z = + K Σ + 4 f L ) Q d a g, and when Q = 7 l/s the value of Z is: Z = ( m 0.06m ) 7 6 m6 0 s m π s m =.843m, 4 and the corresponding depth h = =.63m.. The flow rate can be expressed via the rate of change of water level as: Q = A dh dt, 3

4 where A is the area of the water surface in the tank. Bottom pressure is P = ρ g h and it changes with the rate dp dt = ρ g dh dt = ρ g Q A m3 kg = 000 m 9.8m s = 84.8 P a 3 s 0.9 m s 3. Z Q Z = Q Z Q = Q Z Q Z 4. a A Z Z =.63/ = 0.56; Q = 7 l/s 0.56 = 5.4 l/s Q3 A liquid with σ =.5 flows from a 50mm diameter pipe A through an abrupt enlargement into a 00mm diameter pipe B, the two pipes being coaxial and horizontal. At some distance downstream of the enlargement in pipe B is a Pitot tube which is facing upstream and connected to one limb of a vertical U-tube manometer containing mercury of relative density The other limb of the manometer is connected to a static pressure hole in the wall of pipe A. Neglecting frictional effects at the pipe walls, calculate the mass flow rate of the liquid when the difference in manometer levels h = 46mm. A head loss coefficient at an abrupt enlargement from area A to area A can be calculated by the formula k = ( A /A ). [0 marks] A B h Fig. 4

5 Manometer shows difference between static pressure at A and stagnation pressure at B. Condition of fluid equilibrium in the manometer gives: P A + σ ρ g h = P B + ρ U B + ρ g h, where ρ is the density of the liquid, and σ is mercury density relative to the density of the liquid: σ = σ Hg /σ = 3.56/.5. The pressure difference is: Head loss in a sudden expansion: where the heads are: P A P B = ρ U B H A H B = U A g + ( σ ) ρ g h. ( A ) A = ( U A U B ), A B g H A = P A ρ g + U A g and H B = P B ρ g + U B g. Therefore P A P B ρ g + U A U B g = ( U A U B ) g and substituting the value of the pressure difference we have:, ( σ ) h = ( U A U B ) g U A g. After rearrangement and substitution of U A = Q/A A and U B = Q/A B we obtain: ( Q ) = ( σ A ) g h, B A B A A and finally Q = (σ ) g h A A A B A A A B ; Q = ( ) 9.8m s 0.046m π 0. 4 m 4 4 = m3 s. Mass flow rate: ṁ = ρ Q = kg m m3 s = 0.65kg s. 5

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