Problem 4 232 in your text page 71 For the setup shown in Fi

Problem 4 (2.32 in your text, page 71). For the setup shown in Figure P2.32, calculate the manometer reading H. 16 kPa 30 cm Oil S = 0.92 Water 20 cm 40 kPa Hg Figure P2.32

Solution

At the bottom of H, pressure on both arms of the manometer must be equal.

On LHS we have 40kPa pressure in the bulb, 20 cm column of water.

On RHS we have H cms of mercury column, 30 cm of oil column and 16kPa pressure in the bulb. S=0.92 is taken as specific gravity /relative density of the oil.

Density of water d_w = 1000kg/m³

Density of Hg        d_h   = 13,534 kg/m³

Density of oil         d_o   = 1000*0.92   = 920 kg/m³

Pressure of any liquid column = hdg, h-height of the column, d- density, g= 9.81m/s²

Equating the pressures on LHS and RHS we have

40e+3 + 0.20*1000*9.81 = H*13534*9.81+0.30*920*9.81+16e+3

Solving the above equation we have H = 0.1752 m

                                                                       = 17.52 cm