Chapter 1 Problems

Problems Properties of Fluid 1. A certain liquid has a unit weight of 56 kN/m3. a. Compute the mass density. b. Compute its specific volume c. Compute for its specific gravity. Solution Given: w = 56 kN/m3 𝑤

a. Ρ = 𝑔

=

5600 9.81

ρ = 5,708 kg/m3 b. Specific volume =

1 𝜌

=

1 5708

Specific volume = 0.000175 m3/kg c. Specific gravity: Sp.gr. = w/ww =

56 9.81

Sp.gr. = 5.71

2. An object has a specific weight of 2.23 kN/m3 Compute the ff: a. Mass density b. Mass if the volume is 0.001 m3 c. Specific volume Solution a. Mass density ρ = mass density ρ=

𝑤 𝑔

=

2230 9.810

ρ = 227.32 kg/m3

b. M = ρ Vol. M = 227.32(0.001) M = 0.23 kg c. Specific volume =

1 𝜌

=

1 227.32

Specific volume = 0.0044 m3/kg

3. A rigid container is partly filled with a liquid at 1520 kPa. The volume of the liquid is 1.232 liters. At a pressure of 3039 kPa, the volume of the liquid is 1.231 liters. a. Compute the average bulk modulus of elasticity of the liquid. b. Compute the coefficient of compressibility. c. If the liquid has a density of 1593 kg/m3, what is the speed of sound in the medium. Given: P2= 3039 kPa P1 = 1520 kPa V1 = 1.232 li V2 = 1.232 li Solution : a. Bulk modulus of elasticity: ∆𝑃

EB = - ∆𝑉/𝑉 =

(P2−P1) V2−V1 V1

(3039−1520)

EB = - 1.231−1.232 1.232

EB = 1.87 x 106 kPa EB = 1.87 GPa

b. Coeff. Of compressibility: β = β=

1 K 1

1.87

β = 0.5347 GPa-1

c. Velocity of sound: 𝐾

V = √𝜌 V=

√

1.87 x 109 1593

V = 1083 m/s V = 1.87 x 109 Pa

Surface Tension 1. Surface tension of small drop of water. σ=

Pd 4

σ = surface tension P = diff. In pressure within the droplet and the atmosphere d = diam. Of droplet 2. Capillary. h=

θ

2σ cos θ ρgr

h = ht. Of capillary rise σ = surface tension θ = wetting angle θ = 0o for water θ = 140o for mercury r = radius of tube ρ g = specific weight of liquid

h

2r

Problems 1. The radius of the tube as shown in the figure is 1mm. The surface tension of water at 20o C is equal to 0.0728. a. Compute the capillary rise in the tube in mm. b. Compute the total force due to surface. c. Compute the weight of water above the surface due to surface tension. Solution

r = 1mm = .001m T = 200 σ =0.0728 N/m a. Capillary rise in the tube in mm.

h=

h

2r

Given:

h=

θ

2σ cos θ ρgr 2(0.0728) cos 0o 1000(9.81)(0.001)

h = 0.0148m h = 14.8 mm b. Total force due to surface tension F = 2 πr σ cos θ F = 0.0728(2π)(0.001) cos 0o F = 4.57 x 10-4 N c. Weight of water W = w V = 9810 π (0.001)2(0.0148) W = 4.56 x 10-4 N