Z1=022=14m P1=0p2=9.807×104N/m ∑h+∑ha2=2n2+10 from the data above, we get W=9,81×14+807×10 +12.5×22=2854J/k 1000 he pump's effective power is N=W,=2854×7.92=2260W=226W 14. As shown in the figure. the inside diameter d of the reservoir is 2m and its bottom is connected with a steel pipe whose inside diameter do is 32mm. There is no liquid supply to the reservoir, and the heigh of the liquid level h2 is 2m. The total energy loss of the liquid flowing through the pipe can be calculated by the formula zhf-20u, where u is the velocity of liquid (in m/s). Try to calculate the required time when the liquid level of the reservoir drops Im? Solution: The problem attributes to the instability flow, the time when the surface of the reservoir descend Im can be calculated from material balance Suppose F is instant feed into system, D'is instant discharge out of system, dA is the accumulated quantity in the de time, then in the de time, the material balance equation is Fde-D'de=dA And then suppose in the de time, the surface of reservoir descend dh, the instant flow rate of liquid in the tube is uo In the equation F’=0 dA'=-D-dh Then the equation reduces to D、2ah In order to work out the relation between the height of instant surface h take the cente of drainpipe as benchmark and the instant velocity u in equation (a), we can use the quation of Bernoulli
0 0 0 1 1 1 = = p u Z 4 2 2 2 2 9.807 10 / 2 / 14 p N m u m s Z m = = = 2 2 2 hf ,1 +hf ,2 = 2u +10u =12u from the data above, we get We 12.5 2 285.4J / k g 1000 9.807 10 9.81 14 2 4 + = = + the pump’s effective power is Ne = Wews = 285.47.92 = 2260W = 2.26kW 14. As shown in the figure, the inside diameter D of the reservoir is 2m and its bottom is connected with a steel pipe whose inside diameter do is 32mm.There is no liquid supply to the reservoir, and the heigh of the liquid level h2 is 2m.The total energy loss of the liquid flowing through the pipe can be calculated by the formula Σhf=20u²,where u is the velocity of liquid (in m/s).Try to calculate the required time when the liquid level of the reservoir drops 1m? Solution：The problem attributes to the instability flow，the time when the surface of the reservoir descend 1m can be calculated from material balance. Suppose F’ is instant feed into system，D’ is instant discharge out of system，dA’ is the accumulated quantity in the dθ time，then in the dθ time, the material balance equation is F’dθ―D’dθ=dA’ And then suppose in the dθ time，the surface of reservoir descend dh，the instant flow rate of liquid in the tube is u。 In the equation F’=0 D’= d u 2 0 4 dA D dh 2 4 = Then the equation reduces to d ud D dh 2 0 4 4 − = or u dh d D d 2 0 = −（ ） In order to work out the relation between the height of instant surface h（take the center line of drainpipe as benchmark）and the instant velocity u in equation（a），we can use the instant equation of Bernoulli
P2+>h z1+)+=822p In the equation Z1=h Z,=0 ll, 0 0 P1=P2 ∑h=20n2 substitute the above values into formula, we have 9.8h=202 u=07√h Put formula b into formula a 07h0.0320.7h integration range as follows h1=2m 、= h, db=-5580/=1dh =-5502lVh2-h] =5580×2(2-√)=4632s=1284h 15. A cooling cycle system of brine is shown in the figure. The density of the brine is 1100kg/m, and the circulation is 36m/h The diameter of the pipeline is the same. The loss from A and two heat exchangers to B is 98. 1J/kg, the loss from B TO A is 49J/kg Try to calculate 1) If the efficiency of the pump is 70%, what is the shaft power of it?(in kw) 2) If the reading of the gauge in position A is 2. 5kgf/cm, what is the reading of the
+ + = + + +hf u p gZ u p gZ 2 2 2 2 1 2 1 1 2 2 In the equation Z1 = h Z2 = 0 = = = 2 1 2 1 2 20 0 0 h u p p u u f substitute the above values into formula，we have 9.81h=20 2 u u = 0.7 h Put formula b into formula a， h dh h dh d D d 0.7 ) 0.032 2 ( 0.7 ( ) 2 2 0 = − = − =―5580 h dh integration range as flollows = = 2 1 0 h m h m 1 2 2 1 = = = = = = = − 1 0 2 2 1 2 1 5580 h h h dh d θ=―5580×2 1 2 1 2 2 1 [ ] = − = h h h h =5580×2 ( 2 − 1) = 4632s = 1.284h 15． A cooling cycle system of brine is shown in the figure. The density of the brine is 1100kg/m³,and the circulation is 36m³/h.The diameter of the pipeline is the same. The energy loss from A and two heat exchangers to B is 98.1J/kg,the loss from B TO A is 49J/kg.Try to calculate: 1) If the efficiency of the pump is 70%, what is the shaft power of it?(in kw) 2) If the reading of the gauge in position A is 2.5kgf/cm²,what is the reading of the
gauge in position B? Solution: 1)The efficiency of the pump take whichever section as 1-l and 2-2. it means liquid flow starts from sectionl-I and reaches section 2-2 after a circle. For sections 1-1 and 2-2, the Bernoulli equation will be PI ∑h since PI=p The equation can be simplified to W=∑h=∑hAB+∑hB=98.1+49=1471J/kg The mass flow rate is =Vp=36×110/3600=11kg/s Then, the power of the pump N=Ww,/n=1471×11/0.7=2312W≈231kW 2)The reading of the gauge in position B Set up Bernoulli equation between section A and section B, and take the center of e section ++∑ In the equation Z,=0 ZR=7m p4=2.5×9807×104=245×103N/m( gage pressure) ∑b,A=98Jkg nput the data into the bernoulli equation PB=245×103-(981×7+98.1)×1100=62×104N/m2( gage pressure The reading of the gauge in position B 6.2×10 =0.63kg/c 987×10
gauge in position B? Solution：1）The efficiency of the pump take whichever section as 1-1 and 2-2. it means liquid flow starts from section1-1 and reaches section 2-2 after a circle. For sections 1-1 and 2-2, the Bernoulli equation will be + + + e = + + +hf u p W gZ u p gZ 2 2 2 2 1 2 1 1 2 2 since 1 2 1 2 1 Z Z p p u u = = = The equation can be simplified to We =hf =hf ,AB +hf ,BA = 98.1+ 49 =147.1J / kg The mass flow rate is w V kg s s s = = 361100/ 3600 =11 / Then, the power of the pump is N = Wews / =147.111/ 0.7 = 2312W 2.31k W 2）The reading of the gauge in position B Set up Bernoulli equation between section A and section B，and take the center of the section A as the horizon. + + = + + + f AB B B B A A A h u p gZ u p gZ , 2 2 2 2 In the equation Z A = 0 ZB = 7m 4 5 2 , 2.5 9.807 10 2.45 10 / (gage pressure 98.1 / A B A f AB u u p N m h J kg = = = = ） input the data into the Bernoulli equation 5 4 2 pB = 2.4510 − (9.817 + 98.1)1100 = 6.210 N / m （gage pressure） The reading of the gauge in position B 2 4 4 0.63 / 9.87 10 6.2 10 pB = kg cm =
16.In a laboratory, the acetic acid(70%)at 20C is sent by the glass pipe. The inside diameter of the pipe is 1.5m, and the flow rate is 10kg/min. Calculate Reynolds number in SI system and pectively. what is the flow solution: 1)In SI system (70%,p=1069kg/m3(20℃) 25×10-3N.s/ d=1.5cm=0.0151 0.882m/ 60×2×0.0152×1069 dp0.015×0882×1069 2.5×10-2 5657(belong to the overfall) 2) In physical unit system p=1.0698/cm3 =0.025gA(cm·s d=1 u=88.2cm/s 1.5×88.2×1.069 R =5657 0.025 17. The liquid whose density is 850kg/m,, viscosity is 8cp, flows in the steel pipe(inside diameter is 14mm). The velocity of liquid is Im/s. Try to find 1)The Reynolds number, and the flow pattern is pointed out 2) The position at which the local velocity is equal to the average one 3) Suppose the pipeline is horizontal pipe, and the pressure of the upstream is 1.5kgf/cm, How long the liquid has to flow when the pressure drop to 1. 3kgf/cm Solution: 1 )The Reynolds number R=如=00141×850=1480do 1000 2) The position at which the local velocity is equal to the average one
16.In a laboratory, the acetic acid (70%) at 20℃ is sent by the glass pipe. The inside diameter of the pipe is 1.5m, and the flow rate is 10kg/min. Calculate Reynolds number in SI system and physical unit system, respectively. What is the flow pattern? solution：1）In SI system From appendix 17 of the book, for the acetic acid (70%), 3 = 1069kg / m (20℃)， 3 2 = 2.510 N s/ m − d=1.5cm=0.015m u 0.882m /s 0.015 1069 4 60 10 2 = = 2 2.5 10 0.015 0.882 1069 − = = du Re =5657 (belong to the overfall) 2）In physical unit system 5657 0.025 1.5 88.2 1.069 88.2 / 1.5 0.025 /( ) 1.069 / 3 = = = = = = Re u cm s d cm g cm s g cm 17. The liquid whose density is 850kg/m³, viscosity is 8cp, flows in the steel pipe (inside diameter is 14mm). The velocity of liquid is 1m/s. Try to find : 1) The Reynolds number, and the flow pattern is pointed out. 2) The position at which the local velocity is equal to the average one. 3) Suppose the pipeline is horizontal pipe, and the pressure of the upstream is 1.5kgf/cm², How long the liquid has to flow when the pressure drop to 1.3kgf/cm²? Solution：1）The Reynolds number 1488 1000 8 0.014 1 850 = = = du Re (belong to the overfall) 2）The position at which the local velocity is equal to the average one
According to the formula 1-38 and 1-39 n=4(R2-r2 For the position at which the local velocity is equal to the average one,then Sor=0.707R=0.707×7=495mm The position at which the local velocity is equal to the average one is 4.95mm 3) The section of upstream is 1-1, the section of downstream is 2- cause pipeline is horizontal pipe, so P1-P2=Ap, According to the formula of Hagon-Poiseuille l =(P-P2 (1.5-1.3)9.807×104×00142 15 32 32× 1000 18. As shown in the figure, an inverse"U-shape differential pressure meter is fixed between the cross sections of the two horizontal pipes with different diameters in order to measure th pressure difference of the two cross sections. When the flow rate is 10800kg/h, the reading of the “U”- shape differential meter is 100mm. The diameters of the two pipe areΦ60×3.5 mm and (42x3mm, respectively. Calculate ) The energy loss of lkg water flowing through the two cross sections; 2)The pressure drop which is equal to the energy loss(in N/m2) Solution: 1) The energy loss of lkg water flowing through the two cross sections Set up Bernoulli equation between section 1-l and section 2-2, and take the axis of pipe as horizon 2,+4+P+W= p2 2 Z1=22=0
According to the formula 1——38 and 1——39， ( ) 4 2 2 R r l p u f r − = 2 8 R l p u f = For the position at which the local velocity is equal to the average one，then 2 2 2 2 1 R − r = R So r=0.707R=0.707×7=4.95mm The position at which the local velocity is equal to the average one is 4.95mm。 3）The section of upstream is 1——1’，the section of downstream is 2——2’，because the pipeline is horizontal pipe , so p1 − p2 = p f According to the formula of Hagon-Poiseuille， 2 32 d lu p f = then , m u p p d l 15 1 1000 8 32 (1.5 1.3)9.807 10 0.014 32 ( ) 2 4 2 1 2 = − = − = 18. As shown in the figure, an inverse “U”-shape differential pressure meter is fixed between the cross sections of the two horizontal pipes with different diameters in order to measure the pressure difference of the two cross sections. When the flow rate is 10800kg/h, the reading of the “U”-shape differential meter is 100mm.The diameters of the two pipe are Φ603.5mm and Φ423mm,respectively. Calculate : 1) The energy loss of 1kg water flowing through the two cross sections; 2) The pressure drop which is equal to the energy loss (in N/m²). Solution：1）The energy loss of 1kg water flowing through the two cross sections Set up Bernoulli equation between section 1——1’ and section 2——2’，and take the axis of pipe as horizon. + + + e = + + +hf u p W gZ u p gZ 2 2 2 2 1 2 1 1 2 2 In the equation Z1 = Z2 = 0