explain please Consider a jet of water issuing out at the bottom of the large ta
ID: 1859910 • Letter: E
Question
explain please
Consider a jet of water issuing out at the bottom of the large tank as shown below. The jet impinges vertically on a cylindrical block of wood floating on water. The block diameter is 20 cm and height is 15 cm. With the impinging jet, the block is just submerged in the water (density 1000 kg/m3 i.e. the top surface of wood block is at the water level). Assuming quasi-steady flow (i.e. assume that fluid velocity at the fluid surface is negligible) and neglect the variation of the jet velocity from the jet origin till it hits the wooden block. Also neglect the weight of the 'jet' fluid on the wooden block. Find the density of the wooden block. Clearly indicate all important steps, all conservation laws used, the CV used, and box answer. [Hint: First find the velocity of the jet coming out of the tank, and then perform a force balance around the wooden block. Consider weight of the wooden block as well as buoyancy force on the wooden block.].Explanation / Answer
Velocity while coming out is given by expresson:
Pressure at bottom of tank = 1/2*p_w*V^2 = p_o*g*h_o + p_w*g*h_w
Putting values, 1/2*1000*V^2 = 800*9.81*0.5 + 1000*9.81*0.7
V = 4.646 m/s
Force on block due to jet = Momentum change of impingng jet = (p_w*A*V)*V = (1000*0.0003*4.646)*4.646 = 6.4746 N
Buoyant force on block = p_w*volume of block*g = 1000*(3.14/4*0.2^2 *0.15)*9.81 = 46.2051 N
Weight of block = p*Volume of block*g = p*(3.14/4*0.2^2 *0.15)*9.81 = p*0.0462 N
Balancing vertical direction forces on the block, p*0.0462 + 6.4746 = 46.2051
Solving this, density of block p = 859.9 kg/m^3
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.