1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressure difference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B is methane（甲烷）, that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and that liquid A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and 6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure difference over the instrument In meters of water, (a) when the change in the level in the reservoirs is neglected, (b) when the change in the levels in the reservoirs is taken into account? What is the percent error in the answer to the part (a)? Solution：
pa=1000kg/m3 pc=815kg/m3 pb=0.77kg/m3 D/d=8 R=0.145m
When the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubes ?2?Dx?d2R (1) 44so
?d?x???R (2)
?D?and hydrostatic equilibrium gives following relationship
2p1?R?cg?p2?x?cg?R?Ag (3) so
p1?p2?x?cg?R(?A??c)g (4)
substituting the equation (2) for x into equation (4)
gives
?d?p1?p2???R?cg?R(?A??c)g (5)
?D?22（a）when the change in the level in the reservoirs is neglected,
?d?p1?p2???R?cg?R(?A??c)g?R(?A??c)g?0.145?1000?815??9.81?263Pa?D?
（b）when the change in the levels in the reservoirs is taken into account
?d?p1?p2???R?cg?R(?A??c)g?D??d????R?cg?R(?A??c)g?D??6.5?????0.145?815?9.81?0.145?1000?815??9.81?281.8Pa51??222
error=
281.8?263＝6.7％
281.81.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometers are R1=400mm，R2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R3=50mm. Try to calculate the pressure at point A and B.
Figure for problem 1.4
Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by ?g,?H2O,?Hg, respectively. The pressure at point A is given by hydrostatic equilibrium
pA??H2OR3g??HgR2g??g(R2?R3)g
?gis small and negligible in comparison with?Hgand ρH2O , equation above can be simplified
pA?pc=?H2OgR3??HggR2
=1000×9.81×0.05+13600×9.81×0.05
=7161N/m2
9.81×0.4=60527N/m pB?pD?pA??HggR1=7161+13600×
1.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat
pa of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank A and the exit of
drainpipe are all open to H D d air.
h
pa Figure for problem 1.5 A
Solution:
Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane:
p12u12p2u2 ?gz1???gz2??2?2Where p1=0, p2=0, and u1=0, simplification of the equation 2u2 Hg ? 1
2
The relationship between the velocity at outlet and velocity uo at throat can be derived by the continuity equation:
?u2??u?o
??d??? ?????D?2
?D?uo?u2?? 2
?d?Bernoulli equation is written between the throat and the station 2-2 2p0u0u2 ? ? 3
22 ?Combining equation 1,2,and 3 gives
u21h?g12?1000?9.812?9.81 Hg??＝＝＝442?D??10002.44?1?1.25??1 ?1?? ?d?Solving for H H=1.39m
1.6 A liquid with a constant density ρ kg/m3 is flowing at an unknown velocity V1 m/s through a horizontal pipe of cross-sectional area A1 m2 at a pressure p1 N/m2, and then it passes to a section of the pipe in which the area is reduced gradually to A2 m2 and the pressure is p2. Assuming no friction losses, calculate the velocities V1 and V2 if the pressure difference (p1 - p2) is measured. Solution:
In Fig1.6, the flow diagram is shown with pressure taps to measure p1 and p2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,
2
V2?V1A1 A2For the items in the Bernoulli equation , for a horizontal pipe, z1=z2=0
Then Bernoulli equation becomes, after substitutingV2?V1A1 for V2, A2A12V2V1p1A12p2 0???0??2?2?21Rearranging,
A12?V(2?1)A1 p1?p2?221V1＝p1?p22??A1???A???2????2???1???
Performing the same derivation but in terms of V2,
p1?p22??A2?1???A???1????2V2＝?????
1.7 A liquid whose coefficient of viscosity is μ flows below the critical velocity for
laminar flow in a circular pipe of diameter d and with mean velocity V. Show
?p32?Vthat the pressure loss in a length of pipe is .
Ld2Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.
Solution:
The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area RR11 V ? udA ? u 2 ? rdr 1
2??A0?R0