In the above equation, is a functio have to use the error-trial method S the velocity of coal gas is 20m/s, then we can use the formula to calculate the inside diameter of the pipe 1000 =0.421m 3600×z、 choose pipe size o426×6mm, the inside diameter of pipe d=426-2×6-414mm。 The practical velocity in the 10000 206m/s 3600×2×04142 dhy0414×20.6×0.75 R =426×10 0015×10-3 E =0.2/414=0.00048 According to the value of Re and - and check fig 1-24 in the book we can find=0.018 nput the values of u and hinto equation a 0.62 20.62 009844+152-+15×2-=7464810 >∑b it illustrations that the (426 6mm is available 26. Water at 20C passes through a steel pipe whose inside diameter is 300mm. In order to measure the flow rate of water, the arrangement as shown in the figure is adopted. There is a branch pipe( ( 60x3. 5mm) which is parallel with a 2m main pipe. The total length including all local frictional loss is 10m. There is a rotary meter fixed in the branch pipe. From the reading of the meter, we know that the flow rate of water is 2. 72m/h. Try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficients of the main pipe and the ranch pipe are 0.018 and 0.03, respectively Solution: This problem is attributed to the parallel pipeline, take subscript I as main pipe, and subscript 2 as branch pipe. The friction loss for parallel pipelines is ∑hn=∑b V=Vs+vs? The energy loss in the branch pipe is
In the above equation,λis a function of u. So we have to use the error-trial method. Suppose the velocity of coal gas is 20m/s，then we can use the formula to calculate the inside diameter of the pipe d 0.421m 20 4 3600 1000 = = choose pipe size of φ426×6mm , the inside diameter of pipe d=426-2×6=414mm。 The practical velocity in the pipe is 0.2 / 414 0.00048 4.26 10 0.015 10 0.414 20.6 0.75 20.6 / 0.414 4 3600 10000 5 3 2 = = = = = = = − d du R u m s e According to the value of Re and d ，and check fig.1-24 in the book，we can findλ=0.018， Input the values of u and λinto equation a 2 2 1 40 20.6 20.6 0.018( 15) 1.5 744.6 / 810 / 0.414 2 2 f J kg J kg p h + + = it illustrations that the φ426×6mm is available. 26. Water at 20℃ passes through a steel pipe whose inside diameter is 300mm. In order to measure the flow rate of water, the arrangement as shown in the figure is adopted. There is a branch pipe(Φ603.5mm) which is parallel with a 2m main pipe. The total length including all local frictional loss is 10m.There is a rotary meter fixed in the branch pipe. From the reading of the meter, we know that the flow rate of water is 2.72m/h. Try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficients of the main pipe and the branch pipe are 0.018 and 0.03, respectively. Solution：This problem is attributed to the parallel pipeline，take subscript1 as main pipe, and subscript 2 as branch pipe. The friction loss for parallel pipelines is 1 2 1 2 s S S f f V V V h h = + = The energy loss in the branch pipe is
∑h2=2 In the equation 1=0.03 l2+∑l2=10m d2=0.053 2.7 0.343m/s 3600×-×0.053 ut the data into equation c 2=003×-100.3432 ∑h 0.333/kg 0.0532 The energy loss in the main pipe is ∑hn=∑ 1u=0 333 0.333×0.3 0.018×2≈236mls The water discharge of main pipe is Vn=3600×2×0.32×236=601m3/h Total water discharge is V=601+272≈603.7m3/h 27. Water supply system is shown in the figure. The water surface of the elevated tank keeps constant. Water is discharged form branch pipes BC and BD, respectively. The distances between the water surface of tank and the exits of two branch pipes are both llm. The inside diameters of pipeline AB, BC and BD are 38mm, 32mm and 26mm, respectively. And the length of pipeline AB, BC and BD are 58m, 12. 5m and 14m, respectively. The frictional coefficient \of pipeline AB and BC are both 0.03. Try to calculate a) When the valve of branch pipe BD is closed, what's the max water discharge of branch pipe BC(in m/h)? b) When all the valve are wide open, what is the water discharge of the two branch
2 2 2 2 2 2 2 2 u d l l h e f + = In the equation 2 = 0.03 u m s d l l e m 0.343 / 0.053 4 3600 2.72 0.053 10 2 2 2 2 2 = = = + = input the data into equation c hf 0.333J / k g 2 0.343 0.053 10 0.03 2 2 = = The energy loss in the main pipe is 0.333 2 2 1 1 1 1 = 2 = 1 = u d l hf hf So u 2.36m / s 0.018 2 0.333 0.3 2 1 = = The water discharge of main pipe is Vh 0.3 2.36 601m / h 4 3600 2 3 1 = = Total water discharge is Vh 601 2.72 603.7m / h 3 = + 27.Water supply system is shown in the figure. The water surface of the elevated tank keeps constant. Water is discharged form branch pipes BC and BD, respectively. The distances between the water surface of tank and the exits of two branch pipes are both 11m.The inside diameters of pipeline AB, BC and BD are 38mm, 32mm and 26mm, respectively. .And the length of pipeline AB, BC and BD are 58m, 12.5m and 14m, respectively. The frictional coefficient λof pipeline AB and BC are both 0.03. Try to calculate : a) When the valve of branch pipe BD is closed, what’s the max water discharge of branch pipe BC(in m³/h)? b) When all the valve are wide open, what is the water discharge of the two branch
pipes(in m/h)? The roughness of branch pipe BD can be considered as 0. 15m. The density of water is 1000kg/h, and the viscosity of it is ICp Solution: (1)When the valve of branch pipe BD is closed, the max water discharge of branch pipe BC will be found as follows Set up the bernoulli equation between the surface l--l and the section C--C, and take section C--C as the referring plane,so 8ZI P ∑ in the equation Z=llm Zc=0 PI=ps ke+2b=981×1109(a) ∑h=∑b,B+∑hBC (b) h,A=2 (03×58 +0.5)=4=23.528(C 0.038 hBC=(003x、12.5、1BC=586ug ∑ (d) 0.0322 ∵uABd lAB C Input equation e into equation c ∑hAB=2315×052=113812 (f) input equation f, d into equation b ∑h=1158n2+586012c=1744c =lg, input the value of∑h into the equation a unc=2. 45m/s
pipes (in m³/h)? The roughness of branch pipe BD can be considered as 0.15m.The density of water is 1000kg/h, and the viscosity of it is 1Cp. Solution： (1)When the valve of branch pipe BD is closed, the max water discharge of branch pipe BC will be found as follows Set up the Bernoulli equation between the surface 1--1 ’ and the section C--C ’ , and take section C--C’ as the referring plane，so + + = + + + f c c c h u p gZ u p gZ 2 2 2 1 2 1 1 in the equation Z1 =11m ZC = 0 c p p u = 1 1 0 hence, 9.81 11 107.9 2 2 + f = = c h u （a） hf =hf ,AB +hf ,BC （b） 23.15 (C) 2 0.5) 0.038 58 (0.03 2 ( ) 2 2 2 , AB AB AB C e f AB u u u d l l h = + = + + = 2 2 , 5.86 2 ) 0.032 12.5 (0.03 BC BC f BC u u h = = （d） 2 4 2 2 ) 38 32 ( ( ) AB BC bc AB BC AB u u u d d u = = （e） input equation e into equation c， = = 2 2 hf ,AB 23.15 0.5uBC 11.58uBC （f） input equation f,d into equation b 2 2 2 hf =11.58uBC + 5.86uBC =17.44uBC input the value of into the equation c BC f v u h a = ， ， u BC =2.45m/s
V=3600×0.0322×245=7.1m/h 2 )when all valves are wide open, the water discharge of the two branch pipes According to the frictional equation of parallel pipelines, we ∑h,BC=gzD P The exits of two branch pipes are in the same level, we can simplify the equation, Choose the outboard of the two additional pipes as the lower reaches f, BD (a) ∑b=(-d 12.5,l (003 +1)-C=662c 2 0.0322 ∑ h, rp=(n 0.0262(26922+0.5B input the value of∑hc∑ hEReinto eqution a, 6.36BC=(269.2+0.5)uBD the relation of the flow rate of the main duct and the additional pipe is BC+VaD d BuB =d BcuBc +d Bpu Bl 0.0382l42=0.032lBc+0026uBD rearrange the above equation 708lxc+0.469 Set up the bernoulli equation between the surface l--l and the section C--C, and take section C--C as the referring plane, we have gz, +1+P1 ∑ sInce ZI=lIn Zc=0 l1≈0 0 PI=Pc simplify it ∑h=∑bAB+∑h=1079 ∑h,A=23152,∑h,BC=6362k
VBC 0.032 2.45 7.1m / h 4 3600 2 3 = = 2)when all valves are wide open，the water discharge of the two branch pipes According to the frictional equation of parallel pipelines，we + + + = + + + f BD D D f BC D c c c h u p h gZ u p gZ , 2 , 2 2 2 The exits of two branch pipes are in the same level, we can simplify the equation, Choose the outboard of the two additional pipes as the lower reaches hf ,BC =hf ,BD （a） 2 D 2 D , D 2 2 2 , (269.2 0.5) 2 1) 0.026 14 ( 6.36 2 1) 0.032 12.5 (0.03 2 ( ) B B f B BC BC BC BC e BC e f BC u u h u u u d l l h = + = + + = + = + = input the value of h , h into eqution a f,BC f,BD ， 2 2 6.36 (269.2 0.5) uBC = + uBD （b） the relation of the flow rate of the main duct and the additional pipe is AB BC BD AB AB BC BC BD BD AB BC BD u u u d u d u d u V V V 2 2 2 2 2 2 0.038 = 0.032 + 0.026 = + = + rearrange the above equation u AB uBC 469uBD = 0.708 + 0. （c） Set up the Bernoulli equation between the surface 1--1 ’ and the section C--C ’ , and take section C--C’ as the referring plane，we have + + = + + + f c c c h u p gZ u p gZ 2 2 2 1 2 1 1 since Z1 =11m ZC = 0 p pc u = = 1 1 0 uc 0 simplify it hf =hf ,AB +hf ,BC =107.9 = = 2 , 2 hf ,AB 23.15uAB , hf BC 6.36uBC
therefore23.15u2+6,36u2=1079 in equation b and c UB, MBc, ugD and hare indeterminates and nis the function of uBD or-trail method can be used to find the velocity 可采用试差法求解。设uBD=1.45m/s,则 R d4p0.026×145×100037700 1000 E =0.15/26=0.0058 d based on the value of Re and s we get A=0.034, input the value of and u Bp into equation b 6.3612=(269.2×0034+0.5)1452 1.79m/s nput the value of u to equation c, l4=0.708×1.79+0.469×1.45=1.95m/s input the value of uBC, U into equation d 23.15×1.952+6.36×1.792=1084 the result is consistent with the assumption (108.4=107.9), so uBp is excepted, and the water discharge of the two pipes are =3600×z×00322×179=518m3/h 00262×145=277m3/h 28. A standard orifice meter is fixed in the pipe(38x2 5mm). The diameter of the orifice meter is 16. 4mm. The toluene at 20 C flows in the pipe. The pressure difference between both sides of orifice is measured by a"U-shape pressure meter, and the indicating liquid is mercury. The connected tube is full of toluene. The reading of the gauge is 600mm. Try to calculate the flow rate he toluene(in kg/h)
therefore 23.15 6.36 107.9 2 2 u AB + u BC = （d） in equation b and c ， , , are indeterminates and is the function of AB BC BD BD u u u u and the error-trail method can be used to find the velocity 可采用试差法求解。设uBD =1.45m/s,则 = = du Re 37700 1000 1 0.026 1.45 1000 = = 0.15 / 26 = 0.0058 d based on the value of Re and d we get λ=0.034，input the value of λand uBD into equation b， we have 2 2 6.36u BC = (269.2 0.034 + 0.5)1.45 u BC =1.79m/s input the value of uBD，u BC into equation c， u AB =0.708×1.79+0.469×1.45=1.95m/s input the value of u BC ，u AB into equation d 23.15×1.95²+6.36×1.79²=108.4 the result is consistent with the assumption （108.4≈107.9），so u BD is excepted，and the water discharge of the two pipes are V m h V m h BD BC 0.026 1.45 2.77 / 4 3600 0.032 1.79 5.18 / 4 3600 2 3 2 3 = = = = 28．A standard orifice meter is fixed in the pipe (Φ382.5mm).The diameter of the orifice meter is 16.4mm.The toluene at 20℃ flows in the pipe. The pressure difference between both sides of orifice is measured by a “U”-shape pressure meter, and the indicating liquid is mercury. The connected tube is full of toluene. The reading of the gauge is 600mm. Try to calculate the flow rate of the toluene (in kg/h)