A manometer is fitted to a pipe line containing water. the centre of the pipeline is at a height of 200mm from the free surface of manometric fluid in the right limb of the u-tube & the deflection of manometric fluid (specific gravity 15) is 80mm in the right hand limb of the u-tube. determine the pressure (N/m^2) of water in the pipe line.
Answers
Answer:
Case 1 - U-tube upright differential manometer connected at two points in a pipe at same level
The construction and arrangement of a manometer connected at two different points, A and B, of a pipe is shown in figure 7.1.
Fig. 7.1 U-tube upright differential manometer
Let,
ρ1 = density of liquid flowing in the pipeline
ρ2 = density of manometer liquid (assume mercury)
S = Specific gravity of liquid for which pressure has to be determined
S1 = Specific gravity of manometer liquid
hA be the pressure in terms of height of fluid in the pipe at point A
hB be the pressure in terms of height of fluid in the pipe at point B
h is the distance of mercury level in the right limb from the datum line XX�
h1 is the height of manometer liquid level in the right limb from the centre of pipe at point B.
Case 2 - U-tube upright differential manometer connected between two pipes at different levels and carrying different fluids
Fig. 7.2 Vertical differential manometer (pressure difference between two pipes)
Let,
S1 = Specific gravity of liquid in pipe A
S2 = Specific gravity of liquid in pipe B
S = Specific gravity of manometer liquid
hA be the pressure head in terms of height of fluid in the pipe at point A
hB be the pressure head in terms of height of fluid in the pipe at point B
h is the distance of mercury level in the right limb from the datum line XX�
h1 is the height of manometer liquid level in the left limb from the from the datum line XX�
h2 is the height of manometer liquid level in the right limb from the from the centre of pipe at point B.
7.2 U-tube Inverted Differential Manometer
In such types of manometers light fluids for e.g. oil is used as manometer fluid. In the previous derivation, the term (h�S) is added, but here in the left and right limb equations, it is necessary to subtract (h�S) term.
Fig. 7.3 Inverted differential manometer
Let,
S1 = Specific gravity of liquid in pipe A
S2 = Specific gravity of liquid in pipe B
S = Specific gravity of manometer liquid
hA be the pressure head in terms of height of fluid in the pipe at point A
hB be the pressure head in terms of height of fluid in the pipe at point B
h is the distance of manometer liquid level in the right limb from the datum line XX�
h1 is the height of manometer liquid level in the left limb from the from the datum line XX�
h2 is the height of manometer liquid level in the right limb from the from the centre of pipe at point B
7.4 Numerical
Q1. A U-tube monometer is used to measure the pressure of oil (specific gravity 0.85) flowing in a pipe line. It�s left end connected to pipe and right limb is open to the atmosphere. The centre of pipe is 100 mm below the level of mercury in the right limb. If the difference of mercury level in the two line is 160 mm, then determine the head and pressure.
Solution:
h = 13.6 � 160 � 10-3 � 0.85 � 60 � 10-3
= 2176 � 10-3 � 51.00 � 10-3
= (2176-51) � 10-3
= 2125 � 10-3
= 2.125 m
P = ρgh
= 1000 � 9.81 � 2.125
= 20846.25 N/m2
Q2. A U-tube monometer containing mercury was used to find the negative pressure in the pipe. The right limb was open to the atmosphere; find the vacuum pressure in pipe if the difference of mercury level in two pipes is 100 mm and height of water in the left limb from the centre of pipe was found to 40 mm.
Solution:
h = - (h1S1+ h2S2)
�� = - (40�1+100�13.6) � 10-3
�� = - (40+1360) � 10-3
�� = -1400 � 10-3
�� = -1.4 m
P = ρgh
�� = 13734 N/m2
Q3. A simple U-tube manometer containing mercury is connected to a pipe in which an oil of specific gravity 0.80 is flowing. The pressure in the pipe is vacuum. The other end of the manometer is open to atmosphere, find the vacuum pressure in pipe if the difference of mercury level in two limbs is 200 mm and height of oil in left end from the centre of pipe is 150 mm below.
Solution.
�h = h2S2- h1S1
= (200 � 13.6 � 150 � 0.8) � 10-3
= (2720-120) � 10-3
= 2600 � 10-3
= 2.6 m
P = ρgh
= 800 � 9.81 � 2.6
= 20404.8 N/m2
please mark as Brainliest
