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.