1 Solutions to Chapter 1 problems Problem 1.1: This problem requires the reading of Table 1.1 on which the units and dimensions of important concepts and properties are given. Let us take one example. Let us verify the dimension of pressure in terms of M, L and T; it is given as [p] = [M L - 1 T - 2 ] in the table. Recall that the unit of pressure in SI is N/m 2 . Recall also that a unit of force of 1 Newton, N, is equal to 1 kg m/s 2 . Thus, the unit of pressure can also be expressed as kg m - 1 s - 2 and, hence, the dimension of pressure can also be expressed as follows: [M L - 1 T - 2 ]. This is what is given in the table. Problem 1.2: The answer, as given in the text, is one divided by the correct answer. Based on the formula given (which is correct), the unit of G must be [ G ] = [M - 1 T - 2 L 3 ]. The gravitational constant can be found in any physics text; it is G = 6 . 670 × 10 - 11 N m 2 kg - 2 or G = 6 . 670 × 10 - 11 kg - 1 m 2 s - 2 . Hence, the unit of G must be [ G ] = [M - 1 T - 2 L 3 ]. Problem 1.3: This problem is the dimensional analysis of a pendulum. The period of oscilla- tion is τ . The length mass of the object attached to the end of the pendulum is m . The length of the pendulum is l . The acceleration of gravity is g . Let us examine the expression τ = Cm a l b g c The dimensions of each parameter is as follow: [ τ ] = C [ m ] a [ l ] b [ g ] c [ T ] = C [ M ] a [ L ] b L T 2 c The exponents equate as follows: a = 0, b + c = 0 and 1 = - 2 c . Thus, c = - 1 / 2, b = 1 / 2 and a = 0. This means that τ = Cm 0 l 1 / 2 g - 1 / 2 = C q l/g, where C is a constant. This illustrates that τ is independent of the mass, m . Problem 1.4: We want to examine the power required to rotate a disk in a viscous fluid. We assume (based on the problem statement) that the power is related to the other properties as follows: P = f ( D, ω, ρ, ν ) The unit of power is N m s - 1 or kg m 2 s - 3 . Thus the dimension of power and its relationship with the dimensions of the other properties are as follows: [ P ] = C [ D ] a [ ω ] b [ ρ ] c [ ν ] d " ML 2 T 3 # = C [ L ] a 1 T b M L 3 c " L 2 T # d Equating the exponents we get: 1 = c , 2 = a - 3 c + d and - 3 = - b - 2 d . Let d be the undetermined parameter. Hence, c = 1, a = 5 - d and b = 3 - 2 d . Thus, P = CD 5 - d ω 3 - 2 d ρν d This can be rearranged as follows: P = ρω 3 D 5 ν Dω 2 d
or P = ρω 3 D 5 f ν Dω 2 or C P = P ρω 3 D 5 = f ν Dω 2 This is one of a number of forms that can be written for this relationship. Part (b) provides other nondimensional quantities that can be written from dimensional analysis. Of course, in this case, the power coefficient reduces to a function of only one independent parameter.
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- Fall '19
- Aerodynamics, Drag coefficient, Airfoil