FLOW AND PRESSURE MEASUREMENT 89
For pressures below 192 kN/m^2 :GD
CDA 0
v 2√
√
√
√ 2 P 1 v 1(
k
k 1)[
1
(
P 2
P 1
)k 1 /k]
(equation 6.26)Substituting for 1/v 1 DP 1 /Pavaandv 2 Dvagives:GD
CDA 0
va√
√
√
√ 2 Pava(
k
k 1)[
1
(
P 2
P 1
)k 1 /k]and: G^2 DCDA 0 /va^22 Pava[k/k 1 ][1P 2 /P 1 k^1 /k]
DCDA 0 /va^22 Pavað 3 .5[1P 2 /P 1 ^0.^286 ]WhenP 1 D192 kN/m^2 ,G 192 D 0. 55 G 350 ,P 2 , atmospheric pressure, 101.3 kN/m^2 and:
0. 55 G 350 ^2 DCDA 0 /va^22 Pavað 3 .5[1 101. 3 / 192 ^0.^286 ]When the final pressureP 1 is reached, the flowrate is 0. 25 G 350.
∴ 0. 25 G 350 ^2 DCDA 0 /va^22 Pavað 3. 5 1 101. 3 /P 1 ^0.^286
Dividing these two equations gives:
(
0. 55
0. 25
) 2
D
1 101. 3 / 192 ^0.^286
1 101. 3 /P 1 ^0.^286
and: P 1 D 102 .3kN/m^2
PROBLEM 6.15
Water discharges from the bottom outlet of an open tank 1.5 m by 1 m in cross-section.
The outlet is equivalent to an orifice 40 mm diameter with a coefficient of discharge of
0.6. The water level in the tank is regulated by a float valve on the feed supply which
shuts off completely when the height of water above the bottom of the tank is 1 m and
which gives a flowrate which is directly proportional to the distance of the water surface
below this maximum level. When the depth of water in the tank is 0.5 m the inflow and
outflow are directly balanced. As a result of a short interruption in the supply, the water
level in the tank falls to 0.25 m above the bottom but is then restored again. How long
will it take the level to rise to 0.45 m above the bottom?
Solution
The mass flowrateGis related to the headhfor the flow through an orifice when the
area of the orifice is small in comparison with the area of the pipe by:
GDCDA 0 √
2 gh (equation 6.21)