the physical property of gas is almost the same as that of air at 50. As shown in the figure, a U-shape differential pressure meter whose indicating liquid is water is fixed in the pipeline befare the gas entering the blower. The reading of the meter is 30mm. The inside diameters of both the delivering pipe and the exiting pipe are 250mm. The equivalent length of the pipe line, the fitting, and the valve is 50m(The resistance of entering and leaving the tower are not included ), the vertical distance between the air exit and the entrance of the blower is 20m. Suppose the pressure drop of the gas through the filler layer in the tower is 200mm H2O, the roughness of the pipe wall a can betaken as 0. 15m and the atmospheric pressure can be considered as latm, Try to calculate the effective power of the blower Solution: choose the joint of the differential manometer at the entrance of the air blast as section I'and medial wall of the anti-aircraft pipe as section 2- ' the Bernoulli equation between the two sections g21+++ Pi+w ∑h z1=0 In this equation P,=30x 9.798=294N/m(the gague pressure p2=o( the gague pressure) 3600 =20.4m/s 3600×2×0252 Because the pressure of the air does not change a lot inside the system, the density of air can be calculated under the conditions of latm and 50C 29 27 =1.094kg/m 224273+50 input the data into the equation and we will get W=∑h+981×20 294 1094 ∑h-2 h +∑ 1+∑l at latm and 50C, the viscosity of air H=1.96x10-N.s/
the physical property of gas is almost the same as that of air at 50℃. As shown in the figure, a “U”-shape differential pressure meter whose indicating liquid is water is fixed in the pipeline before the gas entering the blower. The reading of the meter is 30mm.The inside diameters of both the delivering pipe and the exiting pipe are 250mm.The equivalent length of the pipe line, the fitting, and the valve is 50m (The resistance of entering and leaving the tower are not included), the vertical distance between the air exit and the entrance of the blower is 20m. Suppose the pressure drop of the gas through the filler layer in the tower is 200mm H2O, the roughness of the pipe wall ε can be taken as 0.15m and the atmospheric pressure can be considered as 1atm, Try to calculate the effective power of the blower. Solution： choose the joint of the differential manometer at the entrance of the air blast as section 1——1’and medial wall of the anti-aircraft pipe as section 2——2’, the Bernoulli equation between the two sections . + + + e = + + +hf u p W gZ u p gZ 2 2 2 2 1 2 1 1 2 2 In this equation Z m Z 20 0 2 1 = = u u m s p p N m 20.4 / 0.25 4 3600 3600 0(the gague pressure) 30 9.798 294 / (the gague pressure） 2 1 2 2 2 1 = = = = = = Because the pressure of the air does not change a lot inside the system, the density of air can be calculated under the conditions of 1atm and 50℃， 3 1.094 / 273 50 273 22.4 29 = kg m + = input the data into the equation and we will get = + − = − 72.5 1.094 294 We hf 9.81 20 hf hf =h f管 + hf填 f管 h = 2 2 u d l l e （ + 进塔 + 出塔） + at 1atm and 50℃, the viscosity of air µ=1.96× 5 2 10 N s / m −
R=2=025×204×10942.85×105 E_0.15 =0.0006 heck the fig 1-24 and find 2=0.019 ∑h=(019+1+0.5 20.42 2=1103J/kg h=200×919=1791J/kg 1094 ∑h=1103+1791=2894 The effective work of the blower W=2894-72.5≈2822J/kg The mass flow rate of the liquid v=3600×1.094/3600=1.094kg/s The effective power of the blower N。=Ww,=2822×1.094=3087W≈3.09%W 24. As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe(inside diameter is 100mm) is connected to the bottom of the reservoir. There is gate valve fixed in the pipiline, and there is a"U-shape differential meter whose indicating liquid is water fixed in some position 15m away from the entrance of the pipeline, one part of the meter is connected to the pipe, and the other is connected to air. The connected tube of the differential pressure meter is full of water. The length of the straight pipe between the pressure measurement point and the exit of the pipeline is 20m a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is pened partially, r=400mm, h=1400mm. The friction coefficient n is 0.025, and the local resistance coefficient of the pipeline entrance is 0.5. What is the flow rate(in m3/h)? b) When the gate valve is wide open, what is the static pressure in the pressure measurement point (in gauge pressure, N/m2). It is given that le/d=15, and the friction coefficient 2 is still 0.025
5 5 2.85 10 1.96 10 0.25 20.4 1.094 = = = − du Re 0.0006 250 0.15 = = d check the fig.1——24 and find λ=0.019 h = + + = J k g f 1103 / 2 20.4 1 0.5) 0.25 50 (0.019 2 管 h J k g f 1791 / 1.094 200 9.798 = 填 = hf =1103 +1791= 2894 The effective work of the blower: We = 2894 − 72.5 2822J / kg The mass flow rate of the liquid w kg s s = 36001.094 / 3600 =1.094 / The effective power of the blower Ne =Wews = 28221.094 = 3087W 3.09kW 24. As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe(inside diameter is 100mm) is connected to the bottom of the reservoir.There is a gate valve fixed in the pipiline, and there is a “U”-shape differential meter whose indicating liquid is water fixed in some position 15m away from the entrance of the pipeline, one part of the meter is connected to the pipe, and the other is connected to air. The connected tube of the differential pressure meter is full of water. The length of the straight pipe between the pressure measurement point and the exit of the pipeline is 20m. a) When the gate valve is closed, R=600mm，h=1500mm; when the gate valve is opened partially, R=400mm，h=1400mm.The friction coefficient λ is 0.025, and the local resistance coefficient of the pipeline entrance is 0.5. What is the flow rate (in m³/h)? b) When the gate valve is wide open, what is the static pressure in the pressure measurement point (in gauge pressure, N/m²). It is given that le/d≈15, and the friction coefficient λ is still 0.025
Solution: (1)When the gate valve is opened partially, the water discharge Set up Bernoulli equation between the surface of reservoir 1-1 and the section of pressure point 2-2,and take the center of section 2-2 as the referring plane, then Z PI ∑h ,1-2 (a) In the equation p,=0(the gauge pressure) P2=PhgR- Throgh=13600×981×04-1000×9.81×14=39630N/m ul, z1=0 We can get the value of Z2 when the gate valve is wide open. When the gate valve is fully closed, the water is standstill. We can get the following based on the hydrostatic equation PHzog(Z1+ h)=puer where h=1. 5m data into equation d z1=1350×06-15=66m 1000 (+5)=(0.025×+0.5)=2.13 input the above data into equation a 981×6.66= 239630+2.3u 21000 the velocity is u3 13m/s he flow rate of water is n=3600×2d2=360×2×0.12x33=885m3/h 2)the pressure of the point where pressure is measured when the gate valve is wide-open Set up Bernoulli equation between the surface of reservoir 1-l and the section of the inner side of the pipeline 2-2,, and take the center of the pipe as the referring plane, then ZI g since Z1=6.66m
Solution：(1) When the gate valve is opened partially, the water discharge is Set up Bernoulli equation between the surface of reservoir 1—1 ’ and the section of pressure point 2—2’，and take the center of section 2—2 ’ as the referring plane, then + + = + + + ,1 2 2 2 2 2 1 2 1 1 2 2 hf — u p gZ u p gZ （a） In the equation p1 = 0 (the gauge pressure) 2 2 2 p = Hg gR − H O gh =136009.81 0.4 −10009.811.4 = 39630N / m 0 0 1 1 = Z u We can get the value of Z2 when the gate valve is wide open. When the gate valve is fully closed, the water is standstill. We can get the following based on the hydrostatic equation. H 2O g(Z1 + h) = Hg gR （b） where h=1.5m R=0.6m Input the known data into equation d 2 2 2 ,1_ 2 1 2.13 2 0.5) 0.1 15 (0.025 2 ( ) 1.5 6.66 1000 13600 0.6 u u u d l h Z m f = + c = + = − = = input the above data into equation a 9.81×6.66= 2 2 2.13 1000 39630 2 u u + + the velocity is u=3.13m/s the flow rate of water is Vh d u 0.12 3.13 88.5m / h 4 3600 4 3600 2 3 = = = 2） the pressure of the point where pressure is measured when the gate valve is wide-open. Set up Bernoulli equation between the surface of reservoir 1—1 ’ and the section of the inner side of the pipeline 2—2’，and take the center of the pipe as the referring plane, then + + = + + + ,1 3 2 2 2 2 1 2 1 1 2 2 hf — u p gZ u p gZ （c） since Z1 = 6.66m
P1=P2 几3=(2 h 1+l +)4 =1001+15052 4.8lu2 input the above data into equation c 9.81×6.66=2+481u2 the velocity is: u3.51 m/s Set up Bernoulli equation between the surface of reservoir 1-l and e . for the same situation of water level gZI P since Zi=6.66m z2=0 1 l2≈3.5lm/s p,=o(page pressure ∑hn2=(4元+5)2=02×1+05)27-=262J/kg input the above data into equation d 3512P2 981×6.66= +26.2 21000 p2 25. The coal gas in the gas tank is sent through a 40m steel horizontal pipe to an ordinary pressure equipment. The gauge pressure in the gas tank is not lower than 62mmH2O. There is
1 2 1 2 0 0 p p u Z = = 2 2 2 ,1_ 3 4.81u 2 15) 0.5] 0.1 35 [0.025( 2 ( ) = = + + + + = u u d l l h c e f input the above data into equation c， 9.81 2 2 4.81u 2 6.66 = + u the velocity is: u=3.51 m/s Set up Bernoulli equation between the surface of reservoir 1—1 ’ and 2——2’, for the same situation of water level + + = + + + ,1 2 2 2 2 2 1 2 1 1 2 2 hf — u p gZ u p gZ （d） since Z1 = 6.66m 2 1 2 1 0 0 3.51 / 0(page pressure Z u u m s p = = ） J k g u d l hf c 26.2 / 2 3.51 0.5) 0.1 15 (0.025 2 ( ) 2 2 ,1_ 2 = + = + = input the above data into equation d， 9.81×6.66= 26.2 2 1000 3.51 2 2 + + p the pressure is: p2 = 32970 25. The coal gas in the gas tank is sent through a 40m steel horizontal pipe to an ordinary pressure equipment. The gauge pressure in the gas tank is not lower than 62mmH2O.There is a
gate valve in the pipeline (/d=15), the local resistance coefficients of the entrance and the exit of the pipeline are 0.5 and 1, respectively. It is given that the average density of the coal gas is 0.75kg/m and the viscosity is 0.015cP under operating conditions. It is required that the flow rate of gas is 10000m/h. Try to choose a steel pipe which has a suitable diameter. The roughness of the pipe wall E is 0.2mm nominal size, mm 350 400 450 500 60 Outside diameter, mm 377426 478 530 630 6 6 6 6 thickness of the wall, mm 8 8 8 8 Solution: Set up the Bernoulli equation between the section l-l which is the surface of the entrance of the pipeline and the section of the outlet of the pipeline 2-2, and take the center of the pipe as the referring plane, then 8z1 B In the equation Z1=Z2=0 ≈0 P,=62 x9.798=6075N/m(the gauge pressure) P2=0 input the above data into the equation and simplify it 607.5 0.75 kg ∑h=(+)+(:+5) Because of /, /d=15,5=0.5,4.=1,so 15=810
gate valve in the pipeline (le/d≈15),the local resistance coefficients of the entrance and the exit of the pipeline are 0.5 and 1,respectively.It is given that the average density of the coal gas is 0.75kg/m³ and the viscosity is 0.015cP under operating conditions. It is required that the flow rate of gas is 10000m³/h.Try to choose a steel pipe which has a suitable diameter. The roughness of the pipe wall ε is 0.2mm. nominal size，mm 350 400 450 500 600 Outside diameter，mm 377 426 478 530 630 thickness of the wall，mm 6 6 6 6 6 8 8 8 8 8 —— —— —— 10 10 Solution： Set up the Bernoulli equation between the section 1—1 ’ which is the surface of the entrance of the pipeline and the section of the outlet of the pipeline 2—2’，and take the center of the pipe as the referring plane, then + + = + + +hf u p gZ u p gZ 2 2 2 2 1 2 1 1 2 2 In the equation Z1 = Z2 = 0 0 0 2 1 u u (the gauge pressure) input the above data into the equation and simplify it hf 810J / kg 0.75 607.5 = = 2 ( ) 2 ( ) 2 2 u u d l d l h c e e f = + + + Because of / 15, = 0.5, =1, e d c e l so 810 2 1.5 2 15 40 2 2 + + = u u d （ ） 0 62 9.798 607.5 / 2 2 1 = = = p p N m