Question 1 |

The pressure in a pipe at X is to be measured by an open manometer as shown in figure. Fluid A is oil with a specific gravity of 0.8 and Fluid B is mercury with a specific gravity of 13.6. The absolute pressure at X is \mathrm{kN} / \mathrm{m}^{2} (round off to one decimal place).

[Assume density of water as 1000 \mathrm{~kg} / \mathrm{m}^{3} and acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2} and atmospheric pressure as \left.101.3 \mathrm{kN} / \mathrm{m}^{2}\right]

[Assume density of water as 1000 \mathrm{~kg} / \mathrm{m}^{3} and acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2} and atmospheric pressure as \left.101.3 \mathrm{kN} / \mathrm{m}^{2}\right]

115.2 | |

253.6 | |

254.3 | |

140.5 |

Question 1 Explanation:

Equating pressure at \mathrm{A}-\mathrm{A}^{\prime}

\begin{aligned} & P_{a t m}+\rho_{B} \gamma_{w} \times \frac{25}{100}=P_{x}-\rho_{A} \gamma_{w} \times \frac{75}{100} \\ & \therefore \quad P_{x}=P_{a t m}+\rho_{B} \gamma_{w} \times 0.25+\rho_{A} \gamma_{w} 0.75 \end{aligned}

\begin{gathered} =101.3+13.6 \times 9.81 \times 0.25+0.8 \times 9.81 \times 0.75 \\ =140.54 \mathrm{kN} / \mathrm{m}^{2} \end{gathered}

Question 2 |

A three-fluid system (immiscible) is connected to a vacuum pump. The specific gravity values of the fluids (S1, S2) are given in the figure.

The gauge pressure value (in kN/m^{2}, up to two decimal places) of p_{1} is ______

The gauge pressure value (in kN/m^{2}, up to two decimal places) of p_{1} is ______

-8.73 | |

-4.78 | |

-2.54 | |

0 |

Question 2 Explanation:

Taking P_{1} is in gauge pressure.

\begin{aligned} P_{A}=& P_{1}+\left(0.88 \times 10^{3}\right) \cdot(9.81)(0.5) \\ &+\left(0.95 \times 10^{3}\right)(9.81)(1) \\ \left(10^{3}\right)(9.81)(0.5)=& P_{1}+\left(0.88 \times 10^{3}\right) \cdot(9.81)(0.5) \\ &+\left(0.95 \times 10^{3}\right)(9.81)(1) \\ P_{1}=&-8.73 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Question 3 |

A closed tank contains 0.5 m thick layer of mercury (specific gravity = 13.6) at the bottom. A 2.0 m thick layer of water lies above the mercury layer. A 3.0 m thick layer of oil (specific gravity = 0.6) lies above the water layer. The space above the oil layer contains air under pressure. The gauge pressure at the bottom of the tank is 196.2 kN/m^{2}. The density of water is 1000 kg/m^{3} and the acceleration due to gravity is 9.81 m/s^{2}. The value of pressure in the air space is

92.214 kN/m^{2} | |

95.644 kN/m^{2} | |

98.922 kN/m^{2} | |

99.321 kN/m^{2} |

Question 3 Explanation:

\begin{aligned} &P_{a i t} \text { is in gauge pressure. }\\ &\begin{array}{l} P_{a i r}+\left(0.6 \times 10^{3}\right)(9.81)(3)+\left(10^{3}\right)(9.81)(2) \\ \quad+\left(13.6 \times 10^{3}\right)(9.81)(0.5)=196.2 \times 10^{3} \\ P_{a i r}=92.214 \mathrm{kN} / \mathrm{m}^{2} \end{array} \end{aligned}

Question 4 |

Bernoulli's equation is applicable for

viscous and compressible fluid flow | |

inviscid and compressible fluid flow | |

inviscid and incompressible fluid flow | |

viscous and incompressible fluid flow |

Question 5 |

The figure shows a U-tube having a 5 mmx5 mm square cross-section filled with mercury (specific gravity = 13.6) up to a height of 20 cm in each limb (open to the atmosphere).

If 5 cm^{3} of water is added to the right limb, the new height (in cm, up to two decimal places) of mercury in the LEFT limb will be_____

If 5 cm^{3} of water is added to the right limb, the new height (in cm, up to two decimal places) of mercury in the LEFT limb will be_____

1.73 | |

19.26 | |

20.74 | |

2.73 |

Question 5 Explanation:

\begin{aligned} P_{A} &=P_{B} \\ \left(13.6 \times 10^{3}\right) g \times(2 x) &=\left(10^{3}\right) g(20) \\ x &=\frac{10}{3.6}=0.735 \mathrm{cm} \end{aligned}

So, the new height (in cm to two decimal place of

mercury in the left limb will be

=20+0.74=20.74 \mathrm{cm}

There are 5 questions to complete.