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Elenna [48]
2 years ago
15

A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with specific gravity of 1.7. The ma

nometer reading is 320 mm. What is the pressure difference
Engineering
2 answers:
ipn [44]2 years ago
5 0

Answer :pressure difference is Δp=5,32pa

Explanation : the expression for pressure difference at the top of the manometer can be given as

P1-P2=Rg(ρm-ρs)...... 1

For gases, ρs is negligibly small when compared to ρm . So substitute zero for in equation (1) and re-write the final expression

Δp=Rg(ρm)

Here,

Δp is the differential pressure applied to manometer in Pa or psi

R is the difference in the levels of the two interfaces in meters 320mm

To metre =0.320m

g is the acceleration of gravity

ρm is the manometer liquid density in 9.81m/s²

Specific gravity(sg) of liquid 1.7

Density of water is 997 kg/m³

Sg=density of substance/density of water

1.7=density of substance/997

Density of substance=1.7*997

=1694.9kg/m³

Hence solving for pressure difference

Δp=Rg(ρm)

=0.320*9.81*1694.9

Δp=5,32pa

MArishka [77]2 years ago
3 0

Answer:

5320.6 Pascal

Explanation:

Manometer is a pressure measuring device use to measure gas pressure .

Pressure difference in Manometer is a function of density,gravity and the height difference of the liquid.

Pressure difference = density x acceleration due to gravity x difference in height of liquid

Density of liquid = specific gravity of object x density of water.

Density of water = 997 kg/m^3

Specific gravity of liquid = 1.7

Density of liquid = 997 x 1.7 =1694.9kg/m^3

g= 9.81 m/s^2

h =320mm = 0.320m

Pressure difference = 1694.9 x 9.81 x 0.320 = 5320.6 Pascal

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Answer:

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T = \frac{a\cdot{t}}{r^2} = \frac{(1.3\cdot10^{-7})(3600)}{0.045^2}= 0.231> 0.2

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2 years ago
The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate
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Answer:

a) Mechanical efficiency (\varepsilon)=63.15%  b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

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For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and \rho the density of the water:

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