Fluid Mechanics Concept Page - 4

Example

Problems on Pascal's law

Example: Figure shows a hydraulic press with the larger piston of diameter 35 cm at a height of 1.5 m relative to the smaller piston of diameter 10 cm.The mass on the smaller piston is 20 kg. What is the force exerted on the load by the larger piston. The density of oil in the press is 750 kg/m3${}^3$, (Take g =9.8m/s2$=9.8m/s^2$).

Solution:
F1A1=20×9.8π(.1)24=24968.15N${\displaystyle \frac{F_1}{A_1}}={\displaystyle \frac{20\times 9.8}{{\displaystyle \frac{\pi (.1{)}^2}{4}}}}=24968.15N$
F2A2=F2π(.35)24=10.4F2N${\displaystyle \frac{F_2}{A_2}}={\displaystyle \frac{F_2}{{\displaystyle \frac{\pi (.35{)}^2}{4}}}}=10.4F_{2}N$As the fluid is still, thus pressure at point A must be equal to pressure at point B.PA=F1A1+ρgh=24968.15+ρgh$P_A={\displaystyle \frac{F_1}{A_1}}+\rho gh=24968.15+\rho gh$; PB=F2A2+ρg(h+1.5)=10.4F2+ρg(h+1.5)$P_B={\displaystyle \frac{F_2}{A_2}}+\rho g(h+1.5)=10.4F_2+\rho g(h+1.5)$Now PA=PB⟹24968.15=ρg(1.5)+10.4F2$P_A=P_B⟹24968.15=\rho g(1.5)+10.4F_2$24968.15=750(9.8)(1.5)+10.4F2$24968.15=750(9.8)(1.5)+10.4F_2$⟹F2≈1.3×103N$⟹F_2\approx 1.3\times {10}^{3}N$

Example

Rise of liquid column in capillary tube in different geometries

Example:
A capillary tube when immersed vertically in a liquid rises to 3 cm. If the tube is held immersed in the liquid at an angle of 60o${}^o$ with the vertical, then find the length of the liquid column along the tube.Solution:
h=2Tcosθrρg$h={\displaystyle \frac{2Tcos\theta }{r\rho g}}$
h∝cosθ$h\propto cos\theta$
∴$\therefore$ h is double ⇒h=6cm$\Rightarrow h=6cm$.

Definition

Pressure Gauge

Pressure measurement is the analysis of an applied force by a ffluid on a surface. Pressure is typically measured in units of force per unit of surface area. Many techniques have been developed for the measurement of pressure and vaccum. Instruments used to measure and display pressure in an integral unit are called pressure gauges or vacuum gauges

Example

Variation of Pressure with Height in a static fluid of constant density

Example: A solid cone of height H and base radius H/2 floats in a liquid of density ρ$\rho$. It is hanging from the ceiling with the help of a string. What is the force by the fluid on the curved surface of the cone?
(P0=$(P_0=$ atmospheric pressure)

Solution:Let the atmospheric pressure be assumed to be Po$P_o$
If we assume a small ring  at height h from top of the cone, of radius r and thickness dr,
From geometry, h = 2r
The pressure on this ring would be, P = Po+ρgh$P_o+\rho gh$
Force on this ring, F = PA
dF = (Po+ρgh)2πrdr$(P_o+\rho gh)2\pi rdr$
Integrating this force from radius r = 0 to r = H/2
F = ∫H/20(Po+ρg2r)2πrdr${\int }_0^{H/2}(P_o+\rho g2r)2\pi rdr$
F = πH2(Po4+ρgH3)$\pi H^2(\frac{P_o}{4}+\frac{\rho gH}{3})$

Example

Force due to hydrostatic pressure on boundaries containing fluids

Example:
Water stands at a depth D behind the vertical upstream face of a dam as
shown in the figure. What is the force exerted by water per unit width of the
dam? (Density of water is ρ$\rho$)

Solution:
Pressure at a depth x = ρgx$\rho gx$
Force dF$dF$ on a horizontal of strip of cross section area 1.dx$1.dx$
 ⇒dF=ρgxdx$\Rightarrow dF=\rho gxdx$
Net force per unit width = ∫D0dF==∫D0ρgxdx=ρgx22|D0=ρgD22${\int }_0^{D}dF=={\int }_0^D\rho gxdx={\displaystyle \frac{\rho g{x}^2}{2}}{|}_0^D={\displaystyle \frac{\rho g{D}^2}{2}}$

Example

Torque due to hydrostatic pressure on boundaries containing fluids

Example: A square gate of size 2 m x 2 m hinged at its midpoint O as shown. It is held in position by an unknown force F (Given that σ$\sigma$ is density of fluid). Find the torque exerted by the fluid in the upper half of the gate.

Solution:The gate will be in equilibrium, if the sum of clockwise moments is equal to the sum of anticlockwise moments taken about hinge O.
(i) The moment of required force F about O is clockwise.
(ii) The moment of force due to fluid in the upper half of the gate about O is clockwise.
(iii) The moment of force due to fluid in the lower half of the gate about O is anticlockwise.

Moment of force F (unknown) about O is F x 1 clockwise.
Moment of the force exerted by fluid above O is given by
τ1=∫1oσgy(2dy)(1−y)${\displaystyle {\tau }_1={\int }_o^1\sigma gy(2dy)(1-y)}$
[where σgy$\sigma gy$ is the pressure of the fluid of depth y. Here, 2dy is the area of a layer of thickness dy at y. Also, (1 - y) is the moment - arm about O].

τ1=2σg∫10[ydy−y2dy]${\tau }_1=2\sigma g{\int }_0^1[ydy-y^{2}dy]$
=2σg[y22−y33]10=2σg(12−13)$=2\sigma g{[\frac{y^2}{2}-\frac{y^3}{3}]}_0^1=2\sigma g(\frac{1}{2}-\frac{1}{3})$
=2σg6=σg3$=\frac{2\sigma g}{6}=\frac{\sigma g}{3}$ clockwise

Similarly, the moment due to the liquid in the lower half (i.e., below O) is
τ2=∫10σg(y+1)(2dy)(y)${\tau }_2={\int }_0^1\sigma g(y+1)(2dy)(y)$
= 2σg[y33+y22]10$2\sigma g{[\frac{y^3}{3}+\frac{y^2}{2}]}_0^1$
= 2σg[13+12]$2\sigma g[\frac{1}{3}+\frac{1}{2}]$
= 5σg3${\displaystyle \frac{5\sigma g}{3}}$ anticlockwise
∴τ+τ1=τ2$\therefore \tau +{\tau }_1={\tau }_2$
⇒τ+σg3=5σg3$\Rightarrow \tau +\frac{\sigma g}{3}=\frac{5\sigma g}{3}$
⇒τ=4σg3Nm$\Rightarrow \tau =\frac{4\sigma g}{3}Nm$

Example

Variation of Pressure with height in a fluid with varying density

Example: In a cylindrical container, open to the atmosphere from the top, a
liquid is filled upto 10 m depth. Density of the liquid varies with depth from the surface as ρ(h)=100+6h2$\rho (h)=100+6h^2$ where h$h$ is in meter and ρ$\rho$ is in kg/m3$kg/m^3$. What will be the pressure at the bottom of the container ? (atmospheric pressure = 105Pa${10}^{5}Pa$, g=10m/sec2$g=10m/se{c}^2$)

Solution: dp=ρgdh$dp=\rho gdh$
∫P=PP=P0=∫h=10h=0(100+6h2)gdh${\int }_{P=P_0}^{P=P}={\int }_{h=0}^{h=10}(100+6h^2)gdh$
Pf−P0=(1000h+20h3)h=10h=0$P_f-P_0={(1000h+20h^3)}_{h=0}^{h=10}$
⇒Pf=105+0.3×105=1.3×105Pa$\Rightarrow P_f={10}^5+0.3\times {10}^5=1.3\times {10}^5{P}_a$

Example

Variation of height of liquid in a tank with hole at the bottom

Example: A tank of cross-section A contains a liquid of density up to a height H.
there is a hole of cross-section area a, at the bottom of the tank. Find the variation in height of liquid and the time in which the liquid level is reduced to half?

Solution:
velocity  of  ejection  of  water  at  any  height  X  is  2gx−−−√$\sqrt{2gx}$
andAV2=aV1$andA{V}_2=a{V}_1$
acceleration a1=dv1dt=ddt2gx−−−√$a_1=\frac{d{v}_1}{dt}=\frac{d}{dt}\sqrt{2gx}$
=2g−−√2x1/2×dxdt$={\displaystyle \frac{\sqrt{2g}}{2x^{1/2}}}\times \frac{dx}{dt}$
butdxdt=V2$but\frac{dx}{dt}=V_2$
⇒a1=2g√2x1/2V2=2g√2x1/2(aA)V1$\Rightarrow a_1=\frac{\sqrt{2g}}{2x^{1/2}}{V}_2=\frac{\sqrt{2g}}{2x^{1/2}}(\frac{a}{A}{)}^{V_1}$
a1=g(aA)$a_1=g(\frac{a}{A})$
v−u=−a1t$v-u=-a_{1}t$
⇒2gH2−−−−√−2gH−−−−√=−g(aA)t$\Rightarrow \sqrt{2g\frac{H}{2}}-\sqrt{2gH}=-g(\frac{a}{A})t$
⇒t=Aa(2√−1)Hg−−√$\Rightarrow t=\frac{A}{a}(\sqrt{2}-1)\sqrt{\frac{H}{g}}$

Definition

Variation of pressure with height in a fluid

For an infinitesimally small change in height, change in pressure is given by
dP=ρgdh$dP=\rho gdh$

Note:
This equation is useful in situations where density and acceleration due to gravity are functions of height.

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