Fluid Mechanics and Hydraulic Machinery Miscellaneous
Direction: Consider a linear programming problem with two variable and two constraints. The objective function is maximize x1 + x2. The corner points of the feasible region are (0, 0), (0,2) (2, 0) and (4/3, 4/3)
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The ratio PA − PB . (1/2)ρu0²
(where pA and pB are the pressures at section A and B, respectively, and ρ is the density of the fluid), is
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Applying Bernoulli’s equation at sections A and B, we get
PA + u0² + ZA = PA + Vm² + ZB ρg 2g ρg 2g PA + u0² = PB + Vm² ρg 2g ρg 2g or PA + PB = Vm² + u0² Pg ρg 2g 2g
Now, the ratioPA − PB/(1/2)ρ0² = ρVm² − ρu0² 2 2 1 ρu0² 2 = Vm² − u0² = Vm ² − 1 u0² u0 Substituting the Value of Vm = 1 u0 1 − (δ/H)
we getPA − PB = 1 − 1 (1/2)ρ0² [1 − (δ/H)]² Correct Option: B
Applying Bernoulli’s equation at sections A and B, we get
PA + u0² + ZA = PA + Vm² + ZB ρg 2g ρg 2g PA + u0² = PB + Vm² ρg 2g ρg 2g or PA + PB = Vm² + u0² Pg ρg 2g 2g
Now, the ratioPA − PB/(1/2)ρ0² = ρVm² − ρu0² 2 2 1 ρu0² 2 = Vm² − u0² = Vm ² − 1 u0² u0 Substituting the Value of Vm = 1 u0 1 − (δ/H)
we getPA − PB = 1 − 1 (1/2)ρ0² [1 − (δ/H)]²
- The dimension of surface tension is
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Surface tension (σ)
σ = F = N = kg − m = MT−2 L m s2 × m Correct Option: D
Surface tension (σ)
σ = F = N = kg − m = MT−2 L m s2 × m
- If P is the gauge pressure within a spherical droplet, then gauge pressure within a bubble of the same fluid and of same size will be
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P = Gauge pressure within a spherical droplet
Pdroplet = 4σ for droplet d Pbubble = 8σ for bubble d
Pbubble = 2 PdropletCorrect Option: D
P = Gauge pressure within a spherical droplet
Pdroplet = 4σ for droplet d Pbubble = 8σ for bubble d
Pbubble = 2 Pdroplet
- The pressure gauges G1 and G2 installed on the system show pressure of ρG1 = 5.00 bar and ρG2 = 1.00 bar. The value of unknown pressure P is [Atmospheric pressure = 1.01 bar]
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P = PG1 + PG2 + Patm
P = 5 + 1 + 1.01 = 7.01 barCorrect Option: D
P = PG1 + PG2 + Patm
P = 5 + 1 + 1.01 = 7.01 bar
- Oil in a hydraulic cylinder is compressed from an initial volume 2 m3 to 1.96 m3. If the pressure of oil in the cylinder changes from 40 MPa to 80 MPa during compression, the bulk modulus of elasticity of oil is
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Bulk modules of elasticity, 1 dh K = −dh dv/V1 K = −V1 −dp dv K = −2 × 40 −0.04
K = 2000 MPaCorrect Option: B
Bulk modules of elasticity, 1 dh K = −dh dv/V1 K = −V1 −dp dv K = −2 × 40 −0.04
K = 2000 MPa