Räkneövningar 5 av 5,

PG015 - X5 Kemiteknik - Värme- och strömningsteknik Processteknikens grunder (PG) 015 Räkneövningar 5 av 5, 0.10.015 1.0-15.00 kurs-assistent C. Haikarainen 1 frågor; 1-7 utan svar; 8-1 med svar 1. Old PG exam question (4 p.) En tvättsvamp som tillverkats av ett polymermaterial med densiteten ρp = 800 kg/m har densiteten ρsp = 640 kg/m, då porositeten (öppna volymen) är 0 %. Svampen är en kub med måtten 10 cm 10 cm 10 cm. Se figuren. Svampen läggs i vatten (densitet ρw = 1000 kg/m ) vilket resulterar i att porutrymmet fylls med vatten. Det visar sig att den vattendränkta svampen endast delvis sjunker ned i vattnet. Hur många cm av svampen kommer att vara ovanför vattenytan? Antag att volymen av svampen som är ovanför vattenytan även är genomdränkt av vatten. Densitet luft ρa = 1, kg/m. A sponge is made of a polymer material with density ρp = 800 kg/m, while the density of the sponge, with 0% porosity (open space) has a density ρsp = 640 kg/m. he sponge is a cube with dimensions 10 cm 10 cm 10 cm. See the figures. he sponge is put in water (density ρw = 1000 kg/m ) with the result that the pore space fills with water. It is found that the water-soaked sponge only partly sinks into the water. How many cm of the sponge will be above the water surface? Assume that the volume of the sponge that is above the water surface is soaked with water too. Density air ρa = 1, kg/m.. KJ05-Q4-4Water issues from a hole in a large tank, as shown. Assuming frictionless flow (i.e. negligible viscous friction losses, find.

PG015 - X5. Old PG exam question (4 p.) Rökgas från en processugn körs ut i atmosfären genom en cylindrisk skorsten med konstant inre diameter. Rökgasens temperatur är g = 7 C medan omgivningens temperatur är o = 0 C. Inne i skorstenen är trycket vid inloppet (på marknivå) p1, vilket är Δp under omgivningens tryck på marknivå po: Δp = po p1. Beräkna, ifall Δp = 50 Pa, skorstenens höjd H så att rökgasen kommer att strömma ut i luften. Värmeöverföring med omgivningen samt tryckförlust som resultat av gasströmmen i skorstensröret kan försummas. ips: hitta uttrycken för trycket p i skorstensröret samt i den omgivande luften, vid höjden H. Flue gas from a process furnace is emitted to the atmosphere from a cylindrical stack with constant inside diameter. he temperature of the flue gas is 7 C while the temperature of the surroundings is 0 C. Inside the stack, at the entrance (at ground level) the pressure p1 is Δp below the pressure po of the surroundings at ground level: Δp = po p1. Calculate, for Δp = 50 Pa, the height H of the stack so that the gas will flow freely into the atmosphere. Heat transfer with the surroundings and pressure drop inside the stack tubing as a result of the flow can be neglected. Hint: find expressions for the pressure p inside the stack as well as in the surrounding air, at height H. 4. KJ05-Q9-16 An air conditioning duct is 5 cm square and must convey 5 m /min of air at 100 kpa, 5 C. he duct is made of sheet metal that has a roughness of approximately 0.05 mm. Determine the pressure drop for 5 m of horizontal duct (in kpa and mm of water). Data air 100 kpa, 5 C: Density ρ = 1.169 kg/m, Dynamic viscosity η = 1.8 10-5 Pa.s

PG015 - X5 5. Old PG exam question (8p.) Figuren nedan visar hur en simbassäng är sammankopplad med ett vattenkonditioneringssystem (som kan beaktas som en cylinder med diametern d). Rörsystemet är av gjutjärn (cirkulärt, med väggskrovligheten 0, mm) och genom det strömmar 0,1 m /s vatten. Beräkna tryckförlusten (i Pa) i rörsystemet från punkt 1 till punkt 5. Data vatten: kinetiska viskositeten ν = 10-6 m²/s, densiteten ρ = 1000 kg/m³. Data för motståndstalen hittas i kursmaterialet. Obs: Figuren visar systemet ovanifrån. he Figure above shows how a swimming pool is linked to a water conditioning system (which can be described as a cylinder with diameter d ). he tubing system is made of cast iron (circular, wall roughness 0, mm) through which the flow is 0.1 m /s water. Calculate the pressure drop (in Pa) across the tubing system from point 1 to point 5. Data for water: kinematic viscosity ν = 10-6 m²/s, density ρ = 1000 kg/m³. Data for the friction coefficients can be found in the course material. Note: the Figure gives the view from above. 6. Old PG exam question (6+4 = 10p.) Vatten (densitet ρ = 1000 kg/m, dynamisk viskositet η = 0,001 Pa s) pumpas med en pump P från en reservoar A till en reservoar _ B som ligger på en högre nivå, Δz = zb za = m. Röret mellan A och B är cirkulärt med den totala längden = 1 m, diametern D = 0, m och väggskrovligheten x =,5 mm. Volymströmmen Φv = 0,06 m /s. I rörledningen finns 4 stycken 90 knän (eng: elbow), en strömningshastighetsmätare (motståndstalet ζ = 4) och en ventil (motståndtalet ζ = ζ (θ), som bäst med θ = 70 ). Se figuren. a. Beräkna tryckförlusten (i Pa) över ledningssystemet mellan punkten 1 och. b. Beräkna vilken pumpeffekt (i kw) som behövs. Röret mellan reservoar A och pumpen kan försummas.

PG015 - X5 7. Old PG exam question (++5+1 = 10p.) Kylvatten ska pumpas från en reservoar upp till en processanläggning i en rörledning med rörkrökar och en ventil enligt nedanstående figur. Vid båda vattenytorna råder atmosfärstryck. Kylvattenflödet (0 C) ska vara 15 m /t (d.v.s. m /h). Höjdskillnaden är 105 m och den totala rörlängden är 166 m, varav 16 m före pumpen. a) Vilken rördiameter ska man välja om vattenhastigheten i röret inte får överstiga m/s och vad är Reynoldstalet? b) Vilken uppfordringshöjd måste pumpen ha för att den skall klara att pumpa vattnet och ge det önskade flödet? c) Beräkna effektförbrukningen (i kw) för pumpen om den har en verkningsgrad på 80%. d) Finns det risk för s.k. kavitation någonstans i rörsystemet? Antag att ventilens förlustkoefficient är ζ =,0, Antag två 45 krökar (ζ = 0,4) och en 90 krök (ζ = 0,9). Densitet vatten = 1000 kg/m ; dynamisk viskositet vatten = 0,001 Pa s. Vattnets ångtryck vid 0 C är 6,8 Pa. Uppskatta skrovligheten i röret till x = 4,7 10-4 m. Cooling water must be pumped from a reservoir up to a process through a tube system with a few bends and a valve as shown in the figure. At both liquid surfaces the pressure equals ambient atmospheric pressure. he cooling water (0 C) flow is 15 m /h. he height difference between reservoir and process is 105 m and the total tube length is 166 m, of which 16 m is upstream ( before ) of the pump. a) What tube diameter must be chosen so that the flow velocity does not exceed m/s, and what is the Reynolds number of the flow then? b) What pressure head should the pump be able to produce so that the flow objective is achieved? c) Calculate the pump power that is needed for a pump with an efficiency of 80%. d) Is there a risk of so-called cavitation somewhere in this tube system? Assume that the friction coefficient for the valve is ζ =,0, assume two 45 elbow bends (ζ = 0,4) and an 90 elbow bend (ζ = 0,9). Density water = 1000 kg/m ; dynamic viscosity water = 0,001 Pa s. Water vapour pressure at 0 C is 6,8 Pa. Assume the tube wall roughness to be x = 4,7 10-4 m. Source: KJ05 D. Kaminski, M. Jensen Introduction to hermal and Fluids Engineering, Wiley (005)

PG015 - X5 Comments to the questions given here, see also below: 1. he Moody chart is to be used for the tube friction factor ζ=4ƒ, see slides PG01 part 6. For ζ=4ƒ also the following expression by Haaland can be used: ζ f. ε / D log.. Re with wall roughness ε and hydraulic tube diameter D PG course material gives 0.5 ζ 4f x / D. log.. Re with wall roughness x and hydraulic tube diameter D. Pressure drop Δp (Pa) can also be expressed as pressure head loss Δh (m), with Δp = ρ g Δh with fluid density ρ (kg/m ) and gravity constant g his gives: horizontal p h p h gh p g gh p g ½ v ½ v g tube elements : tube : ½ v ½ v g ' Dh Dh ' If necessary use the following (from KJ05) for the element friction coefficients ζ = K :

9-1 At an oil tank farm, a vandal opens a valve at the end of a 5-cm diameter, 50-m long horizontal pipe from the bottom of a large diameter oil tank. he oil tank is open to the atmosphere, and the oil depth is 6.5 m. he oil has a SG = 0.85 and a kinematic viscosity of 6.8 10-4 m /s. Neglecting minor losses, determine the initial flow rate from the tank (in m /s). Approach: he flow rate depends on a balance between the head supplied by the oil depth and the head loss in the drain pipe. he steady, incompressible flow energy equation can be used to determine the flow. An iterative solution may be required because the friction factor is a function of the flow (velocity). Assumptions: 1. he system is steady.. he flow is fully developed.. Properties are constant. 4. Neglect minor losses. Solution: he steady, incompressible flow energy equation is: P1 V1 P V + + z1+ hp = + + z + h + h ρg g ρg g he areas at 1 and are large, so V1 = V 0. he pressures at 1 and are both atmospheric, so P1 = P. here is no pump or turbine, so hp = h = 0, and minor losses are neglected. herefore, V z1 z = f D g he friction factor is a function of Reynolds number and roughness. Assuming a smooth pipe: ρ VD VD ( V m s)( 0.05m) Re = = = = -4 7.5 V µ ν 6.8 10 m s For any reasonable velocity, the flow will be laminar (check this). herefore, for fully developed laminar flow in a straight circular tube: f = 64 Re = 64ν V D Substituting this into the pressure drop equation and solving for velocity: 64ν V g ( z1 z) D z1 z = V = VDDg 64ν ( 9.81m s )( 6.5m)( 0.05m) m V = =0.147 Answer 64-4 ( 6.8 10 m s )( 50m ) s Checking the Reynolds number: ( 0.147m s)( 0.05m) Re = = -4 10.8 6.8 10 m s aminar flow, so our assumption is correct. Comments: Note that the velocity calculated above is the initial velocity, assuming that the transient start-up of the flow is brief and the flow is quasi-steady. As the oil level drops, then the velocity will decrease. 9-

9-9 he reservoir behind a dam is connected to a hydroelectric power plant with a penstock (a large pipe to convey the water). At a particular plant, the elevation difference between the reservoir surface and the hydroturbine is 50 m, and the penstock is constructed of 150-m of 1-m diameter cast iron pipe. he turbine has a mechanical efficiency of 78% and the electric generator has an efficiency of 94%. For a 1 m /s flow of 10 C water, determine: a. the power output from the plant (in kw) b. the power output if a fully open gate valve and two long radius 45 flanged elbows also are in the pipe (in kw). Approach: he steady, incompressible flow energy equation is used to determine the turbine head. Once that is known, the power can be calculated. Assumptions: 1. he system is steady.. he flow is fully developed.. Properties are constant. Solution: P1 V1 P V a) he steady, incompressible flow energy equation is: + + z1+ hp = + + z + h + h ρg g ρg g here is no pump, so h P = 0. At points 1 and the areas are large and the pressure is atmospheric, so V1 = V = 0andP1 = P. Including the exit loss where, K exit = 1.0.: V h = z1 z f + Ke xit D g he friction factor is a function of Reynolds number and roughness, and from able 9- ε =0.6 mm. For 4 water from Appendix A-6 at 10 ºC, µ = 1.9 10 Ns m, ρ = 999.6kg m he velocity is: ( s) V 4V 41m m V = = = =1.7 s A π D π ( 1m) ( )( )( ) ρ VD 999.6kg m 1.7 m s 1m Re = = 4 µ 1.9 10 Ns m = 987,000 he flow is turbulent, so 1.11 1.11 1 ε D 6.9 0.6 1000 6.9 = 1.8log + = 1.8log + f.7 Re.7 987, 000 f = 0.015 herefore, ( ) h ( 1.7 m s) 150 = 50m- 0.015 +1.0 = 49.7m 1 9.81ms ( ) Power extracted from the water is kg m m kn s W = mgh = ρvgh = 999.6 1 9.81 ( 49.7m ) =488kW m s s 1000kg m aking into account the turbine and electric generator efficiencies: W elec = ηη genw = ( 0.78)( 0.94)( 488kW ) =58kW Answer b) Adding a gate valve, K = 0.15, and long radius 45º bends, K = 0. gate ( 1.7 m s) 150 h = 50m- ( 0.015 ) +1.0+0.15+( 0.) = 49.7m 1 9.81ms ( ) W elec = 58kW Answer Comments: Other losses dominate the process so adding a gate valve has essentially no effect. 9- bend.

4-40 Water flows at 5 kg/s through a gradual contraction in a pipe. he upstream diameter is 8 cm and the downstream diameter is 5.6 cm. If the exit pressure is 60 kpa, find the entrance pressure. Assume frictionless flow. Approach: Apply the Bernoulli equation. Find velocities from the known mass flow rate. Assumptions: 1. Density is constant.. he flow is frictionless and isothermal.. he system is at room temperature. Solution: aking station 1 to be the entrance of the contraction and station to be the exit, and applying Bernoulli s equation: P1 V1 P V + gz1+ = + gz + ρ ρ Since there are no elevation changes, P1 V1 P V + = + ρg g ρg g Velocities can be determined using the mass flow rate, since m = ρv A kg 5 m s m V1 = = = 4.99 ρ A kg 1 s 997.5 π (0.04) m m kg 5 m m V = = s = 10. ρ A kg 0.056 s 997.5 π m m o find the pressure, rearrange the Bernoulli equation P1 = P + ρ( V V1 ) kg m P1 = 60 10 Pa + 997.5 (10. 4.99 ) m s P = = Answer 1 99. 10 Pa 99. kpa 4-8

9-8 Consider a fully developed laminar flow of 0 C water down an inclined plane that is 0 to the horizontal. he water thickness is 1-mm. he water is exposed to atmosphere everywhere, and the air exerts zero shear on the water. Using a differential analysis similar to that used for fully developed laminar flow through an inclined pipe, determine the volume flow rate per unit width (in m /s m). Approach: he water flows down the inclined plane due to the influence of gravity. A force balance between gravity and shear forces is solved to find the velocity profile, and then integrated to find the volume flow rate. Assumptions: 1. he system is steady.. he flow is fully developed.. Properties are constant. Solution: A force balance in the x-direction on the differential element shown above, assuming fully developed laminar flow with constant properties, is: Fx = 0= τa τbab + dwsinθ where the weight is W = ρvg. Note there is no net pressure force. Atmospheric pressure acts all along the surface of the oil, and the infinitesimal difference in hydrostatic pressure is negligible. etting D be the depth of the plane into the page: τddx τbddx + ρddxdyg sinθ = 0 τ τb Canceling Ddx and rearranging: = ρ g sinθ dy dτ Recognizing the left hand side is a derivative: = ρ g sinθ dy Using Newton s law of viscosity τ = µ dv dy d dv d V µ = µ ρg sin = θ dy dy dy Separating variables and integrating twice: d V ρ g ρ = sinθ + C g 1 V = sinθ y + C1y+ C dy µ µ Boundary conditions are: 1) at y = 0, V = 0 ) at y = h there is no shear, so dv dy = 0 Applying the boundary conditions: 0= 0+ 0+ C C = 0 ρg ρg 0= sinθ h+ C1 C1 = sinθ h µ µ herefore: ρ g sin ρ g ρ sin hy sin hy y V = θ + θ = θ µ µ µ Volume flow rate is h h g y V ρ = VdA= VDdy = ty Dd 0 y 0 A µ V ρ gsinθ h = D µ 4 For water at 0 ºC from Appendix A-6, µ = 9.85 10 Ns m, ρ = 998. kg m V D = ( )( ) ( )( ) ( ) -4 ( 9.85 10 Ns m ) o 998. kg m 9.81m s sin 0 0.001m Ns kgm m =0.0011 sm Answer 9-8

9-9 A biomedical device start-up company is developing a liquid drug injection device. he device uses compressed air to drive the plunger in a piston-cylinder assembly that will push the drug (viscosity and density similar to water at 10 C) through the hypodermic needle (inside diameter 0.5 mm and length 50 mm). If the flow must remain laminar in the hypodermic needle, determine: a. the maximum flow possible (in cm /s) b. the required air pressure for the maximum flow if the pressure at the end of the needle must be 105 kpa (in kpa). (Assume fully developed flow.) Approach: Using a transition Reynolds number of 100 and the Reynolds number definition, we can calculate the maximum allowable velocity and, hence, flow rate. We can use Eq. 9-1, which relates pressure drop and velocity, to calculate the required air pressure. Assumptions: 1. he system is steady.. he flow is fully developed laminar flow.. Properties are constant. Solution: a) Volume flow rate is defined as: V = VA x = V π D 4 Reynolds number is defined as: ρ VD Re = µ Using Re = 100 for the transition between laminar and turbulent flow, and water properties from Appendix A-6 4 at 10 ºC, µ = 1.9 10 Ns m, ρ = 999.6 kg m : 4 µ Re ( 1.9 10 Ns m )( 100)( kgm Ns ) m V = = =10.8 ρd ( 999.6kg m )( 0.0005m ) s ( ) m π 0.0005m -7 m V = 10.8 =5.0 10 Answer s 4 s b) From Eq. 9-1 for a horizontal flow: 8µ V µ V P = = R D 4 µ V 1.9 10 Ns m 0.05m 10.8 m s 1kN 1000N P = P1+ = 105kPa+ D ( 0.0005m) = 105kPa+57kPa=46kPa Answer ( )( )( )( ) 9-9