10800 =2.95m/s AP3600×z×00362×1000 1.36m/s 3600×2×0.0532×1000 Pr-p2=pgR p-P2=981×01=0981/kg input the data into the equation ∑h=0981+ 2952-1.362 0981+343=44lJ/kg 2)4=P∑h=100441=440N/m2 19. The liquid is flowing laminarly in the straight pipe. If the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original, how many times is the energy loss produced by resistance as much as the original? Solution: According to the formula of Hagon-Poiseuil 32h se the subscriptl and subscript to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the viscosity of liquid and the length of the pipe have From the problem, we know the ratio between these two diameters d,/d,=2 4×22=16
m s A w u S 2.95 / 0.036 1000 4 3600 10800 2 1 1 = = = u 1.36m / s 0.053 1000 4 3600 10800 2 2 = = J k g p p p p gR 9.81 0.1 0.981 / 1 2 1 2 = = − − = input the data into the equation hf 0.981 3.43 4.41J / k g 2 2.95 1.36 0.981 2 2 = + = − = + 2） 2 p f = hf =1000 4.41= 4410N / m 19. The liquid is flowing laminarly in the straight pipe. If the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original, how many times is the energy loss produced by resistance as much as the original? Solution：According to the formula of Hagon-Poiseuille， 2 32 d lu p f = Use the subscript1 and subscript2 to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the viscosity of liquid and the length of the pipe have not changed. 2 2 1 1 2 2 2 1 ( ) d d u u p p f f = From the problem, we know the ratio between these two diameters 4 2 16 ( ) 2 4 / 2 2 2 1 2 2 2 1 1 2 1 2 1 2 = = = = = = = f f S S p p d d u u V V d d
Thus, if the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original the energy loss produced by resistance has increase to 16 times as much as the orgiN 20. The liquid is flowing turbulently in the straight pipe. If the physical property of the liquid the length of the pipe and the diameter are constant, but the flow rate is increased to twice of the original, how many times is the energy loss produced by resistance as much as the original? For two cases, Reynolds number is between 3x10-1x10, the friction coefficient 2 can be calculate by the formula“ Palasus Solution: According to the frictional formula in the straight pipe Use the subscript and subscript2 to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the diameter of the pipe and the length of the pipe have not changed According to the formula 0.3164 The coefficients of friction in these cases A2,R21 MR Because of the flow rate is 2 times as much as the original then ,=2V The viscosity and the density of the liquid are invariability, therefore Rel 1/2 R e Upon that 22 =()023=0.84
Thus, if the pipe length and the physical property stay constant, but the pipe diameter decreases to half of the original，the energy loss produced by resistance has increase to 16 times as much as the original. 20. The liquid is flowing turbulently in the straight pipe. If the physical property of the liquid, the length of the pipe and the diameter are constant, but the flow rate is increased to twice of the original, how many times is the energy loss produced by resistance as much as the original? For two cases, Reynolds number is between 310³~1105 , the friction coefficient λ can be calculate by the formula “Palasus”. Solution: According to the frictional formula in the straight pipe, 2 2 u d l hf = Use the subscript1 and subscript2 to denote the original diameter of the pipe. When the diameter has changed, in these two cases, the diameter of the pipe and the length of the pipe have not changed. 2 1 2 2 1 2 ( ) 1 u u h h f f = According to the formula， 0.25 0.3164 Re = The coefficients of friction in these cases 0.25 2 1 1 2 ( ) e e R R = Because of the flow rate is 2 times as much as the original, then VS 2 = 2VS1 2 1 2 1 = u u The viscosity and the density of the liquid are invariability，therefore 2 1 e e R R =1/2 Upon that ) 0.84 2 1 ( 0.25 1 2 = =
12=0.84×2 3.36 21. The rectangular chimney(inside cross section is 1000x 1200mm) is 30m high. The effluent gases at 400C is flowing from bottom to top, the average molecular weight of gases is 30kg/kmol. The bottom of the chimney keeps 5mm H2O(vacuum. The density of the air can be considered as constant in the range of the height of the chimney. The environmental temperature is 20C, and the atmospheric pressure is latm. The frictional coefficient of fluid flowing through the himney can be taken as 0.05, try to calculate the flow rate of the effluent gases(in kg/h) Solution: Suppose the bottom of the chimney is the upstream section l-l, and the inside cross section of the top is 2-2, and take section 1-l as horizon, then set up bernoulli equation between two sections PI gz2 +∑h z1=0 In this equation Z,=30m u,u,( the sections of the chimney are the same, and the pressure of gas in the chimney does not change a lot Because the pressure of gas in the chimney doesnt change a lot, the temperature is 400C and the atmospheric pressure is latm pM 10133×103×30 =0.534kg/m RT8316×103(273+400) pis atmosphere density P,andp,, indicate atmosphere of the bottom and top of the chimney respective P1=pa1-5×9798=P1-49N/m Because the pressure on the top inside of chimney is equal to the pressure of air in the same P2=pa2=pal -pgz2 The density of the air in the standard status is 1.293kg/m, so when the environmental temperature is 20C, and the atmospheric pressure is latm, the density of the air
0.84 2 3.36 2 1 2 = = f f h h 21. The rectangular chimney (inside cross section is 10001200mm) is 30m high. The effluent gases at 400℃ is flowing from bottom to top, the average molecular weight of gases is 30kg/kmol.The bottom of the chimney keeps 5mm H2O (vacuum. The density of the air can be considered as constant in the range of the height of the chimney. The environmental temperature is 20℃, and the atmospheric pressure is 1atm .The frictional coefficient of fluid flowing through the chimney can be taken as 0.05, try to calculate the flow rate of the effluent gases (in kg/h). Solution：Suppose the bottom of the chimney is the upstream section 1——1’，and the inside cross section of the top is 2——2’, and take section 1——1’ as horizon， then set up Bernoulli equation between two sections. + + = + + + hf u p gZ u p gZ 2 2 2 2 1 2 1 1 2 2 In this equation Z m Z 30 0 2 1 = = ( u1 u2 the sections of the chimney are the same，and the pressure of gas in the chimney does not change a lot） Because the pressure of gas in the chimney doesn’t change a lot, the temperature is 400℃, and the atmospheric pressure is 1atm. 3 3 5 0.534 / 8.316 10 (273 400) 1.0133 10 30 k g m RT pM = + = = 1 2 2 1 1 1 is atmosphere density, indicate atmosphere of the bottom and top of the chimney respective 5 9.798 49 / a a a p andp p p p N m = − = − Because the pressure on the top inside of chimney is equal to the pressure of air in the same height, so we have p2 pa2 pa1 gZ2 = = − The density of the air in the standard status is 1.293kg/m³，so when the environmental temperature is 20℃, and the atmospheric pressure is 1atm，the density of the air
273 p’=1.293× =1.2kg/m 273+20 upon that p2=pa1-12×981×30=pa-353N/m then input the data into the equation +(-0-(Pa-353) 981×30 0.543 =560-294=266J/kg d 2 12 therefore d.=4× 1.09m 2×(1+1.2) The velocity of the flue gas is 266×1.09×2 19.7m/s 0.05×30 The mass flow rate of the flue gas is Wh=36004=3600×197×1.2×1×0.543 =46210kg/h 22. The liquid is sent to the elevated tank from the reactor by pump at 2x10'kg/h(as shown the figure). The pressure keeps 20mmHg(vacuum) above the surface of the reactor, and the pressure above the surface of the elevated tank is atmosphere. The pipeline is steel pipe((76x4mm), and its length is 50m. There are two wide open brake valves, a orifice meter(the resistance coefficient is 4), five standard elbows in the pipeline. The distance from the surface of the reactor to the exit of the pipeline is 1.5m. If the efficiency of the pump is 0.7, try to calculate the shaft power of the pump. It is given that the density of the liquid is 1073kg/m, the viscosity is 0.63cP, and the absolute roughness of pipe walle is 0. 3mm Solution: Set up Bernoulli equation between the surface of reactorl-l'and the inner surface of pipe 2-2, and take section I-las referring section 2+z8 P In this equation z2=15m
3 1.2 / 273 20 273 1.293 = kg m + = upon that 3 2 1 1 p = pa −1.2 9.81 30 = pa − 353N / m then input the data into the equation， 2 560 - 294 266J/kg 9.81 30 0.543 ( 49) ( 353) 2 1 1 u d l h p p h e f a a f = = = − − − − = therefore de 1.09m 2 (1 1.2) 1 1.2 4 = + = The velocity of the flue gas is u 19.7m / s 0.05 30 266 1.09 2 = = The mass flow rate of the flue gas is k g h wh uA 46210 / 3600 3600 19.7 1.2 1 0.543 = = = 22. The liquid is sent to the elevated tank from the reactor by pump at 2104kg/h (as shown in the figure). The pressure keeps 20mmHg (vacuum) above the surface of the reactor, and the pressure above the surface of the elevated tank is atmosphere. The pipeline is steel pipe(Φ764mm), and its length is 50m.There are two wide open brake valves, a orifice meter (the resistance coefficient is 4), five standard elbows in the pipeline. The distance from the surface of the reactor to the exit of the pipeline is 1.5m. If the efficiency of the pump is 0.7, try to calculate the shaft power of the pump. It is given that the density of the liquid is 1073kg/m³, the viscosity is 0.63cP, and the absolute roughness of pipe wallε is 0.3mm. Solution：Set up Bernoulli equation between the surface of reactor1——1’ and the inner surface of pipe 2——2’, and take section 1——1’as referring section + + + 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 15 0 2 1 = =
2×10 u2xd2p3600×2×0.0682×1073 =143m/s 200 P1 10133×103=-2.67×10N/m( page pre p,=0(page pressure input the data into the equation and we will get 267×104 143 W 9.81×15 1073 +∑h 173+ ∑ h,=( +∑5) dlap0.068×143×1073 =1.66×10 063×10-3 E_0.3 0.0044 According to the value of Re and - check the fig 1-24 and find the coefficient of the friction A=0.03, and from the fig 1--26, we can find that the equivalent lengths of the pipe and the valve are Brake valve(wide open) 0.43x2=0.86m Standard elbow 2.2×5=1lm so∑h=00×x-0068 50+0.86+11 32 +0.5+4) 232.5J/kg W=173+32.5=205.5J/kg p WO,2055×2×104 =163W≈163kW 3600×0.7 23. There are a few solvents in the exhaust gas discharged in the equipment. Before vented to the atmosphere, the gas is sent through a washing tower in order to recover the solvents and avoid the environmental pollution. The flow rate of gas is 3600m/h(under the operating condition), and
1 4 2 2 2 5 4 2 1 2 0 2 10 1.43 / 3600 0.068 1073 4 4 200 1.0133 10 2.67 10 / (page pressure 760 0(page pressure s u w u m s d p N m p = = = − = − = ） ） input the data into the equation and we will get = + + + + = f e f h W h 173 2 1.43 9.81 15 1073 2.67 104 2 2 ( )u d l l h e f + + = 5 3 1.66 10 0.63 10 0.068 1.43 1073 = = = − du Re 0.0044 68 0.3 = = d According to the value of Re and d ，check the fig.1——24 and find the coefficient of the friction = 0.03，and from the fig.1——26, we can find that the equivalent lengths of the pipe and the valve are Brake valve（wide open） 0.43×2=0.86m Standard elbow 2.2×5=11m So hf 32.5J / k g 2 1.43 0.5 4) 0.068 50 0.86 11 (0.03 2 + + = + + = We =173 + 32.5 = 205.5J / kg The power of the pump W k W W N e s 1631 1.63 3600 0.7 205.5 2 104 = = = 23. There are a few solvents in the exhaust gas discharged in the equipment. Before vented to the atmosphere, the gas is sent through a washing tower in order to recover the solvents and avoid the environmental pollution. The flow rate of gas is 3600m³/h (under the operating condition), and