UNIT-2: PRESSURE AND ITS MEASUREMENTS Q 1 State and prove Pascal’s Law (July 2013, 2014, jan 2015, July 2015) Ans:Consider a small vertical cylinder of fluid in equilibrium, where positive z is pointing Suppose the origin z = 0 is set at the free surface of the fluid. Then the vertically upward. pressure variation at a depth = below the free surface is governed by z -h ( p  p) A  W  pA 0 pA  gA z  - p  gz .(1) Eq or dp - -  g dz p (as z → Therefore, the hydrostatic pressure increases Linearly with depth at the rate of the specific weight = of the fluid.  ρg Homogeneous fluid: is constant ρ By simply integrating the above equation-1: ∫ ∫ - dp  g dz -gz    C is constant of integration C When = 0 (on the free surface), (the atmospheric pressure). z p= C = p0= Hence, p - g z p0   Pressure given by this equation is called ABSOLUTE PRESSURE, i.e., measured above perfect vacuum. However, for engineering purposes, it is more convenient to measure the pressure a
UNIT-2: PRESSURE AND ITS MEASUREMENTS Q 1 State and prove Pascal’s Law (July 2013, 2014, jan 2015, July 2015) Ans:Consider a small vertical cylinder of fluid in equilibrium, where positive z is pointing vertically upward. Suppose the origin z = 0 is set at the free surface of the fluid. Then the pressure variation at a depth z = -h below the free surface is governed by ( p  p) A  W  pA pA  gA z  0 p  - gz dp dz  - g or - Eq .(1) (as z → Therefore, the hydrostatic pressure increases Linearly with depth at the rate of the specific weight  = ρg of the fluid. Homogeneous fluid: ρ is constant By simply integrating the above equation-1: ∫ dp  - ∫  g dz p  -gz  C C is constant of integration When z = 0 (on the free surface), p= C = p0= (the atmospheric pressure). Hence, p  - g z  p0 Pressure given by this equation is called ABSOLUTE PRESSURE, i.e., measured above perfect vacuum. However, for engineering purposes, it is more convenient to measure the pressure a
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