Mechanical Properties Of Fluids 2 L-2 Mechanical Properties of Fluids
→ \rightarrow → → \rightarrow → Mechanical Properties of Fluids → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Pressure due to Liquid Column
P = P 0 + ρ g h P=P_0+\rho g h P = P 0 + ρ g h
P 0 P_0 P 0
ρ g h \rho g h ρ g h
→ \rightarrow → → \rightarrow → Pressure due to Liquid Column → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Problem
The surface of water in a Storage tank is 20 m {m} m
Given: density of water = 1 × 10 3 k g / m 3 = ρ waler 1 \times 10^3 {kg} / {m}^3=\rho_{\text {waler }} 1 × 1 0 3 k g / m 3 = ρ waler
Δ P = ρ g h = ( 1 × 10 3 k g / m 3 ) ( 98 m s − 2 ) ( 20 m ) \Delta P = \rho g h = (1 \times 10^{3} kg/m^3)(98 m s^{-2}) (20 m) Δ P = ρ g h = ( 1 × 1 0 3 k g / m 3 ) ( 98 m s − 2 ) ( 20 m )
= 1.96 × 10 5 N / m 2 1.96 \times 10^{5} N/m^2 1.96 × 1 0 5 N / m 2
= 1.96 × 10 5 P a = 1.96 \times 10^5 Pa = 1.96 × 1 0 5 P a
Mechanical Properties of Fluids → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Problem
What is the difference in blood pressure between top of the head and the bottom of the feet of 1.60 m tall person standing vertically?
Density of blood is 1060 k g / m 3 kg/m^3 k g / m 3
Pressure difference in given by
Δ P = ρ g h \Delta P = \rho g h Δ P = ρ g h
Pressure due to Liquid Column → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Problem
When you run up a tall hill or run down the hii quickly the ears ' pop'. Suppose this did not happen, What would be the force in the eardrum of area 0.5 c m 2 0.5 {cm}^2 0.5 c m 2 1000 m 1000 {m} 1000 m P a i r = 1.29 k g / m 3 P_{air} =1.29 {kg} / {m}^3 P ai r = 1.29 k g / m 3
P = h ρ \rho ρ 1.29 k g / m 3 1.29 kg / m^3 1.29 k g / m 3 9.8 m s − 2 9.8 m s^{-2} 9.8 m s − 2
F =P A = (12642 N / m 2 N/m^2 N / m 2 × ( 0.5 × 10 − 4 m 2 ) \times (0.5 \times 10^{-4} m^2) × ( 0.5 × 1 0 − 4 m 2 )
F = 6.32 N
Problem → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Problem
Determine the variation in pressure in the earth's atmosphere as a function of the height y y y
assuming ' g g g ( P a t m = 1.013 × 10 5 N / m 3 ) (P_{atm} = 1.013 \times 10^{5} N/m^3) ( P a t m = 1.013 × 1 0 5 N / m 3 )
ρ ρ 0 = P P 0 ⋅ \frac{\rho}{\rho_0} = \frac{P}{P_0} \quad \cdot ρ 0 ρ = P 0 P ⋅
d P d y = − ρ g ⋅ \frac{dP}{dy} = - \rho g \quad \cdot d y d P = − ρ g ⋅
ρ = P P 0 × ρ 0 ⇒ d p d y = − P P 0 ρ 0 g ⋅ \rho = \frac{P}{P_0} \times{\rho_0} \Rightarrow \frac{dp}{dy} = -\frac{P}{P_0} \rho_0 g \quad \cdot ρ = P 0 P × ρ 0 ⇒ d y d p = − P 0 P ρ 0 g ⋅
∫ P 0 P = − ∫ 0 y ρ 0 g P 0 d y \int_{P_0}^P = - \int_0^y \frac{\rho_0 g}{P_0} dy ∫ P 0 P = − ∫ 0 y P 0 ρ 0 g d y ⇒ l n P P 0 = ρ 0 g P 0 y \Rightarrow ln \frac{P}{P_0} = \frac{\rho_0 g}{P_0} y ⇒ l n P 0 P = P 0 ρ 0 g y
Problem → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Problem
P = P 0 e − ( ρ 0 g P 0 ) y P=P_0 e^{-(\frac{\rho_0 g}{P_0}) y} P = P 0 e − ( P 0 ρ 0 g ) y
ρ 0 g P 0 \frac{\rho_0 g}{P_0} P 0 ρ 0 g ( 1.29 k g / m 3 ) × ( 98 m s − 2 ) ( 1.013 × 10 5 N / m 2 \frac{(1.29 {kg} / {m}^3) \times(98 {ms}^{-2})}{(1.013 \times 10^5 {N} / {m}^2} ( 1.013 × 1 0 5 N / m 2 ( 1.29 k g / m 3 ) × ( 98 m s − 2 ) 1.25 × 10 − 4 m − 1 1.25 \times 10^{-4} m^{-1} 1.25 × 1 0 − 4 m − 1
P = P 0 2 P = \frac{P_0}{2} P = 2 P 0
P = P 0 e ( 1.25 × 10 − 4 m − 1 ) y P = P_0 e^{(1.25 \times 10^{-4} m^{-1})} y P = P 0 e ( 1.25 × 1 0 − 4 m − 1 ) y
y = l n 2 1.25 × 10 − 4 m − 1 y = \frac{ln 2}{1.25 \times 10^{-4} m^{-1}} y = 1.25 × 1 0 − 4 m − 1 l n 2
= 0.693 ( 1.25 × 10 − 4 m − 1 ) m \frac{0.693}{(1.25 \times 10^{-4} m^{-1})} m ( 1.25 × 1 0 − 4 m − 1 ) 0.693 m
= 5550 m = 18,000 ft
Problem → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Atmospheric Pressure and Gauge Pressure
Atmospheric Pressure: P 0 = 1.013 × 10 5 N / m 2 P_0=1.013 \times 10^5 {N} / {m}^2 P 0 = 1.013 × 1 0 5 N / m 2
1 b a r = 1 × 10 5 N / m 2 1 bar = 1 \times 10^5 {N} / {m}^2 1 ba r = 1 × 1 0 5 N / m 2
Tire gauge
P = P a t m + P gauge pressure P=P_{atm}+P_{\text{gauge pressure}} P = P a t m + P gauge pressure
If a tire gauge measure a pressure of 200 kPa then the partial pressure is 200 kPa + 100 kPa.
Problem → \rightarrow → → \rightarrow → Atmospheric Pressure and Gauge Pressure → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Measurement of Pressure
Problem → \rightarrow → → \rightarrow → Measurement of Pressure → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Different Units of Pressure
1 a t m = 1.013 × 10 5 N / m 2 1 atm = 1.013 \times 10^5 {N} / {m}^2 1 a t m = 1.013 × 1 0 5 N / m 2 1.013 × 10 5 P a = 101.3 k P a 1.013 \times 10^5 {Pa}=101.3 {kPa} 1.013 × 1 0 5 P a = 101.3 k P a
1 b a r = 1 × 10 5 N / m 2 1 bar = 1 \times 10^5 {N} / {m}^2 1 ba r = 1 × 1 0 5 N / m 2
=1.013 bar = 76 c m of H g 1.013 \text { bar }=76 {cm} \text { of } {Hg} 1.013 bar = 76 c m of H g
= 760 mm of Hg = 760 torr
= 1.03 × 10 4 \times 10^4 × 1 0 4 H 2 O {H_2 O} H 2 O 4 0 C 4^0 C 4 0 C
Blood pressure 120/80, systolic pressure is 120 mm Hg and diastolic pressure is 80 mm Hg.
Atmospheric Pressure and Gauge Pressure → \rightarrow → → \rightarrow → Different Units of Pressure → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Mercury Barometer
Measurement of Pressure → \rightarrow → → \rightarrow → Mercury Barometer → \rightarrow → → \rightarrow →
Mechanical Properties Of Fluids 2 L-2 Thank You
Different Units of Pressure → \rightarrow → → \rightarrow → Thank You → \rightarrow → → \rightarrow →
Resume presentation
Mechanical Properties Of Fluids 2 L-2 Mechanical Properties of Fluids $\rightarrow$ $\rightarrow$ Mechanical Properties of Fluids $\rightarrow$ Pressure due to Liquid Column $\rightarrow$ Problem
