Fluid mechanics - Answer sheet 4 PDF

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Solutions to questions to practice from the ES3D6 Fluid Mechanics Module. Lecturer: Peter Thomas...

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FLUID MECHANICS FOR MECHANICAL ENGINEERS Model Answers: Example Sheet 4 (1) Consider the free-falling body: f1(s, W, g, T) = 0 (a) List physical quantities and their fundamental units s = length L, W = force ML/T2, g = acceleration L/T2, T = time T There are 4 physical quantities and 3 fundamental units, hence (4-3) or one πterm. (b) Quantities s, W and T contain the 3 fundamental dimensions M, L and T so we choose those quantities. (c) The π-term is expressed in terms of the quantities multiplied together: π1 = sa Wb Tc g substitute in fundamental units considering π-term is non-dimensional: M0 L0 T0 = La (ML/T2)b Tc LT-2 (d) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = b Consider L: 0 = a + b + 1 Consider T: 0 = -2b + c - 2

Solving the above equation gives: a = -1, b = 0, c = 2. (e) Substitute into the π-term equation: π1 = s-1 W0 T2 g Rearranging and substituting π1 = 1/K to give an equation for s: s = K g T2 (2) Consider the pump with: f1(P, W, Q, H) = 0 (a) List quantities and the units P = Energy/time ML2/T3, W = density*gravity M/L2/T2, Q = L3T-1, Head = L

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There are 4 physical quantities and 3 fundamental units, hence (4-3) or one πterm. (b) Quantities W, Q and H contain the 3 fundamental dimensions M, L and T so we choose those quantities. (c) The π-term is expressed in terms of the quantities multiplied together: π1 = Wa Qb Hc P substitute in fundamental units considering π-term is non-dimensional: M0 L0 T0 = (ML-2T-2)a (L3T-1)b Lc ML2T-3 (d) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = a + 1 Consider L: 0 = -2a + 3b + c + 2 Consider T: 0 = -2a – b -3

Solving the above equation gives: a = -1, b = -1, c = -1. (e) Substitute into the π-term equation: π1 = W-1 Q-1 H-1 P Rearranging and substituting π1 = 1/K to give an equation for P: P=KWQH

(3) Consider the drag force on a body: f1(D, ρ, µ, U, L) = 0 (a) List quantities and the units D = Force ML/T2, ρ = density M/L3, µ = M/L/T, U = velocity L/T, L = length L There are 4 physical quantities and 3 fundamental units, hence (5-3) or two πterm such that: f 1(π1, π2) = 0 (b) Choose L, U and ρ as the repeating quantities that contain the 3 fundamental dimensions M, L and T. (c) The first π-term is expressed in terms of the L, U and ρ quantities multiplied together with one of the spare quantities, in this case D:

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π1 = La Ub ρc D substitute in fundamental units considering π-term is non-dimensional: M0 L0 T0 = (L)a (LT-1)b (ML-3)c MLT-2 (d) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = c + 1 Consider L: 0 = a + b – 3c + 1 Consider T: 0 = -b - 2

Solving the above equation gives: a = -2, b = -2, c = -1. (e) Substitute into the π-term equation: π1 = L-2 U-2 ρ-1 D (f) The second π-term is expressed in terms of the L, U and ρ quantities multiplied together with the other spare quantities, in this case µ: π2 = La Ub ρc µ substitute in fundamental units considering π-term is non-dimensional: 0 0 0 a -1 b -3 c -1 -1 M L T = (L) (LT ) (ML ) ML T

(g) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = c + 1 Consider L: 0 = a + b – 3c - 1 Consider T: 0 = -b - 1

Solving the above equation gives: a = -1, b = -1, c = -1. (h) Substitute into the π-term equation: π2 = L-1 U-1 ρ-1 µ (i) Giving: f 1(L-2 U-2 ρ-1 D, L-1 U-1 ρ-1 µ) = 0 or

D = (L2U2ρ) f2(LUρ/µ)

which can be written as D = (2K RE) (ρL2U2/2) where RE is the Reynolds number and K is a constant. Recognising that L2 is an area then the final equation can be written as:

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D = (2K RE) (ρAU2/2) (4) Consider the wave-making device: f1(W, X, f, d, ρ, g) = 0 (a) List quantities and the units W = power ML2/T3, X = length L, f = frequency 1/T, d = length L, ρ = density M/L3, g = acceleration L/T2 There are 6 physical quantities and 3 fundamental units, hence (6-3) or three π-terms such that: f 1(π1, π2, π3) = 0 (b) Choose d, ρ and g as the variables that contain the 3 fundamental dimensions M, L and T. (c) The first π-term is expressed as: π1 = da ρb gc W substitute in fundamental units considering π-term is non-dimensional: 0 0 0 a -3 b -2 c 2 -3 M L T = L (ML ) (LT ) ML T

(d) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = b + 1 Consider L: 0 = a - 3b + c +2 Consider T: 0 = -2c -3

Solving the above equation gives: a = -7/2, b = -1, c = -3/2. (e) Substitute into the π-term equation: π1 = d-7/2 ρ-1 g-3/2 W = W/(ρg3/2d7/2) (f) The second π-term is expressed as: π2 = da ρb gc X substitute in fundamental units considering π-term is non-dimensional: 0 0 0 a -3 b -2 c M L T = L (ML ) (LT ) L

(g) Equate powers of each of the fundamental units (i)

Consider M: 0 = b

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(ii) (iii)

Consider L: 0 = a - 3b + c +1 Consider T: 0 = -2c

Solving the above equation gives: a = -1, b = 0, c = 0. (h) Substitute into the π-term equation: π2 = d-1 ρ0 g0 X = X/d (i) The third π-term is expressed as: π3 = da ρb gc f substitute in fundamental units considering π-term is non-dimensional: 0 0 0 a -3 b -2 c -1 M L T = L (ML ) (LT ) T

(j) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = b Consider L: 0 = a - 3b + c Consider T: 0 = -2c – 1

Solving the above equation gives: a = 1/2, b = 0, c = -1/2. (k) Substitute into the π-term equation: π3 = d1/2 ρ0 g-1/2 f = f√(d/g) (l) Giving: f 1(W/(ρg3/2d7/2), X/d, f√(d/g)) = 0 W/(ρg3/2d7/2) = f2(X/d, Fr)

or

where Fr is the Froude number (5) Consider a smooth disc: f1(D, ρ, µ, N, T) = 0 (a) List quantities and the units D = length L, ρ = density M/L3, µ = viscosity M/L/T, N = rads/sec 1/T, T = torque ML2/T2 There are 5 physical quantities and 3 fundamental units, hence (5-3) or two πterms such that: f 1(π1, π2) = 0

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(b) Choose D, ρ and N as the variables that contain the 3 fundamental dimensions M, L and T. (c) The first π-term is expressed as: π1 = Da ρb Nc µ substitute in fundamental units considering π-term is non-dimensional: M0 L0 T0 = La (ML-3)b (T-1)c ML-1T-1 (d) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = b + 1 Consider L: 0 = a – 3b – 1 Consider T: 0 = -c – 1

Solving the above equation gives: a = -2, b = -1, c = -1. (e) Substitute into the π-term equation: π1 = D-2 ρ-1 N-1 µ = µ/(ρND2) (f) The second π-term is expressed as: π2 = Da ρb Nc T substitute in fundamental units considering π-term is non-dimensional: M0 L0 T0 = La (ML-3)b (T-1)c ML2T-2 (g) Equate powers of each of the fundamental units (i) (ii) (iii)

Consider M: 0 = b + 1 Consider L: 0 = a – 3b + 2 Consider T: 0 = -c – 2

Solving the above equation gives: a = -5, b = -1, c = -2. (h) Substitute into the π-term equation: π2 = D-5 ρ-1 N-2 T = T/(ρN2D5) (i) Giving: f 1(µ/(ρND2), T/(ρN2D5)) = 0 or

T/(ρN2D5)= f2(ρND2/µ)

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