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- 1. Basic quantity is defined as a quantity which cannot be derived from any physical quantities <ul><li>Derived quantity is defined as a quantity which </li></ul><ul><li>can be expressed in term of base quantity </li></ul>cd candela Luminous Intensity mol mole N Amount of substance A ampere I Electric current K kelvin T/ Temperature s second t Time kg kilogram m Mass m metre l Length Symbol SI Unit Symbol Quantity
- 2. d. 29 cm = ? in e. 12 mi h -1 = ? m s -1
- 3. Learning Outcome: At the end of this chapter, students should be able to: <ul><li>Use dimensional analysis to check homogeneity </li></ul><ul><li>and construct equation of physics. </li></ul>
- 4. Dimensional Analysis <ul><li>Dimension is defined as a technique or method which the physical quantity can be expressed in terms of combination of basic quantities . </li></ul><ul><li>It can be written as </li></ul><ul><li>[physical quantity or its symbol] </li></ul><ul><li>Table 1.5 shows the dimension of basic quantities. </li></ul>mole N [amount of substance] or [ N ] K [temperature] or [ T ] A A @ I [electric current] or [ I ] s T [time] or [ t ] m L [length] or [ l ] kg M [mass] or [ m ] Unit Symbol [Basic Quantity]
- 5. <ul><li>Dimension can be treated as algebraic quantities through the procedure called dimensional analysis. </li></ul><ul><li>The uses of dimensional analysis are </li></ul><ul><ul><li>to determine the unit of the physical quantity . </li></ul></ul><ul><ul><li>to determine whether a physical equation is dimensionally correct or not by using the principle of homogeneity . </li></ul></ul><ul><ul><li>to derive/construct a physical equation . </li></ul></ul><ul><li>Note: </li></ul><ul><ul><li>Dimension of dimensionless constant is 1 , </li></ul></ul><ul><ul><li>e.g. [2] = 1, [refractive index] = 1 </li></ul></ul><ul><ul><li>Dimensions cannot be added or subtracted. </li></ul></ul><ul><ul><li>The validity of an equation cannot determined by dimensional analysis. </li></ul></ul><ul><ul><li>The validity of an equation can only be determined by experiment. </li></ul></ul>Dimension on the L.H.S. = Dimension on the R.H.S
- 6. Determine a dimension and the S.I. unit for the following quantities: a. Velocity b. Acceleration c. Linear momentum d. Density e. Force Solution : a. The S.I. unit of velocity is m s 1 . Example 1.2 : or
- 7. b. Its unit is m s 2 . d. S.I. unit : kg m 3 . c. S.I. unit : kg m s 1 . e. S.I. unit : kg m s 2 .
- 8. Determine Whether the following expressions are dimensionally correct or not. a. where s , u , a and t represent the displacement, initial velocity, acceleration and the time of an object respectively. b. where t , u , v and g represent the time, initial velocity, final velocity and the gravitational acceleration respectively. c. where f , l and g represent the frequency of a simple pendulum , length of the simple pendulum and the gravitational acceleration respectively. Example 1.3 :
- 9. Solution : a. Dimension on the LHS : Dimension on the RHS : Dimension on the LHS = dimension on the RHS Hence the equation above is homogeneous or dimensionally correct. b. Dimension on the LHS : Dimension on the RHS : Thus Therefore the equation above is not homogeneous or dimensionally incorrect. and and
- 10. Solution : c. Dimension on the LHS : Dimension on the RHS : Therefore the equation above is homogeneous or dimensionally correct.
- 11. The period, T of a simple pendulum depends on its length l , acceleration due to gravity, g and mass, m . By using dimensional analysis, obtain an equation for period of the simple pendulum. Solution : Suppose that : Then where k , x , y and z are dimensionless constants. Example 1.4 : ………………… (1)
- 12. By equating the indices on the left and right sides of the equation, thus By substituting eq. (3) into eq. (2), thus Replace the value of x , y and z in eq. (1), therefore The value of k can be determined experimentally. ………………… (2) ………………… (3)
- 13. Determine the unit of in term of basic unit by using the equation below: where P i and P o are pressures of the air bubble and R is the radius of the bubble. Solution : Example 1.5 :
- 14. Since thus Therefore the unit of is kg s -2
- 15. <ul><li>Deduce the unit of (eta) in term of basic unit for the equation below: </li></ul><ul><li>where F is the force, A is the area, v is the change in velocity and l is the change in distance. </li></ul><ul><li>ANS. : kg m -1 s -1 </li></ul><ul><li>A sphere of radius r and density s falls in a liquid of density f . It achieved a terminal velocity v T given by the following expression: </li></ul><ul><li>where k is a constant and g is acceleration due to gravity. Determine the dimension of k . </li></ul><ul><li>ANS. : M L -1 T -1 </li></ul>Exercise 1.1 :
- 16. <ul><li>a. What is meant by homogeneity of a physical equation? </li></ul><ul><li>b. The escape velocity, v for a tomahawk missile which escape the gravitational attraction of the earth is depend on the radius of the earth, r and the acceleration due to gravity, g . By using dimensional analysis, obtain an expression for escape velocity, v . </li></ul><ul><li>ANS. : </li></ul><ul><li>Show that the equation below is dimensionally correct. </li></ul><ul><li>Where R is the inside radius of the tube, L is its length, P 1 -P 2 is the pressure difference between the ends, is the coefficient of viscosity ( N s m -2 ) and Q is the volume rate of flow ( m 3 s -1 ). </li></ul>Exercise 1.1 :
- 17. <ul><li>Acceleration is related to velocity and time by the following expression </li></ul><ul><li>Determine the x and y values if the expression is dimensionally consistent. </li></ul><ul><li>ANS. : x= 1; y= 1 </li></ul><ul><li>Bernoulli’s equation relating pressure P and velocity v of a fluid moving in a horizontal plane is given as </li></ul><ul><li>where is the density of the fluid and k is a constant. Determine the dimension of the constant k and its unit in terms of basic units. </li></ul><ul><li>ANS. : M L 1 T 2 ; kg m 1 s 2 </li></ul>Exercise 1.1 :

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