Fluid Mechanics and Hydraulic Machinery Miscellaneous
Direction: Consider a linear programming problem with two variable and two constraints. The objective function is maximize x_{1} + x_{2}. The corner points of the feasible region are (0, 0), (0,2) (2, 0) and (4/3, 4/3)

The ratio P_{A} − P_{B} . (1/2)ρu_{0}²
(where p_{A} and p_{B} 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
P_{A} + u_{0}² + Z_{A} = P_{A} + V_{m}² + Z_{B} ρg 2g ρg 2g P_{A} + u_{0}² = P_{B} + V_{m}² ρg 2g ρg 2g or P_{A} + P_{B} = V_{m}² + u_{0}² P_{g} ρg 2g 2g
Now, the ratioP_{A} − P_{B}/(1/2)ρ_{0}² = ρV_{m}² − ρu_{0}² 2 2 1 ρu_{0}² 2 = V_{m}² − u_{0}² = V_{m} ² − 1 u_{0}² u_{0} Substituting the Value of V_{m} = 1 u_{0} 1 − (δ/H)
we getP_{A} − P_{B} = 1 − 1 (1/2)ρ_{0}² [1 − (δ/H)]² Correct Option: B
Applying Bernoulli’s equation at sections A and B, we get
P_{A} + u_{0}² + Z_{A} = P_{A} + V_{m}² + Z_{B} ρg 2g ρg 2g P_{A} + u_{0}² = P_{B} + V_{m}² ρg 2g ρg 2g or P_{A} + P_{B} = V_{m}² + u_{0}² P_{g} ρg 2g 2g
Now, the ratioP_{A} − P_{B}/(1/2)ρ_{0}² = ρV_{m}² − ρu_{0}² 2 2 1 ρu_{0}² 2 = V_{m}² − u_{0}² = V_{m} ² − 1 u_{0}² u_{0} Substituting the Value of V_{m} = 1 u_{0} 1 − (δ/H)
we getP_{A} − P_{B} = 1 − 1 (1/2)ρ_{0}² [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 s^{2} × m Correct Option: D
Surface tension (σ)
σ = F = N = kg − m = MT^{−2} L m s^{2} × 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
P_{droplet} = 4σ for droplet d P_{bubble} = 8σ for bubble d
P_{bubble} = 2 P_{droplet}Correct Option: D
P = Gauge pressure within a spherical droplet
P_{droplet} = 4σ for droplet d P_{bubble} = 8σ for bubble d
P_{bubble} = 2 P_{droplet}
 The pressure gauges G_{1 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 = P_{G1} + P_{G2} + P_{atm}
P = 5 + 1 + 1.01 = 7.01 barCorrect Option: D
P = P_{G1} + P_{G2} + P_{atm}
P = 5 + 1 + 1.01 = 7.01 bar
 Oil in a hydraulic cylinder is compressed from an initial volume 2 m^{3} to 1.96 m^{3}. 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/V_{1} K = −V_{1} −dp dv K = −2 × 40 −0.04
K = 2000 MPaCorrect Option: B
Bulk modules of elasticity, 1 dh K = −dh dv/V_{1} K = −V_{1} −dp dv K = −2 × 40 −0.04
K = 2000 MPa