Heat transfer43. A heat exchanger is to be designed to transfer heat to a certain cold fluid. The given conditions are: ṁh=5 kg/min, ṁc=10 kg/min, Thi=85⁰C, Tci=27⁰C, Tco=25⁰C, Cpc=2.93 KJ/kg-K. What flow pattern will be best suitable to the system? Given: ṁh = 5 kg/min; Thi=85⁰C ṁc = 10 kg/min; Tci=27⁰C; Tco=25⁰C; Cpc=2.93 KJ/kg-K Required: Best suitable flow pattern Solution: Assume: hot fluid (water) 𝑄𝑔𝑎𝑖𝑛𝑒𝑑 = −𝑄𝑙𝑜𝑠𝑡 5𝑘𝑔/ min (4.184 𝐾𝐽⁄𝑘𝑔 ∙ 𝐾 ) (𝑇ℎ𝑜 − 85)𝐾 = −10𝑘𝑔/ min (2.93 𝐾𝐽⁄𝑘𝑔 ∙ 𝐾 ) (55 − 27) 𝐾 𝑇ℎ𝑜 = 45.7839 ℃ Parallel flow ∆𝑇1 = 85 − 27 = 58℃ ∆𝑇2 = 45.7839 − 55 = 9.2161℃ 58 − 9.2161 ∆𝑇𝑙𝑚 = = 𝟐𝟔. 𝟓𝟐𝟎𝟑 ℃ 58 𝑙𝑛 (9.2161) Countercurrent flow 𝑇1 = 85 − 55 = 30℃ ∆𝑇2 = 45.7839 − 27 = 18.7839℃ 30 − 18.7839 ∆𝑇𝑙𝑚 = = 23.9559 ℃ 30 𝑙𝑛 (18.7839) ∴ 𝑡ℎ𝑒 𝑏𝑒𝑠𝑡 𝑓𝑙𝑜𝑤 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑓𝑙𝑜𝑤 𝑠𝑖𝑛𝑐𝑒 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑡𝑜 ℎ𝑖𝑔ℎ𝑒𝑟 𝐿𝑀𝑇𝐷 015 mm to 0. 𝟏𝟖𝟕𝟓 𝒌𝒈/𝒎𝟑 Calc 1 & 2 7. A mixture of quartz and galena of a size range from 0.8125 𝑘𝑔/𝑚3 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑎𝑠 ℎ𝑒𝑎𝑣𝑖𝑒𝑟 0.93% O2.065 2650 − 𝜌𝑓𝑙𝑢𝑖𝑑 =√ 0. What is the minimum apparent density of the fluid that will give this separation? The density of galena is 7500 kg/m3 and the density of quartz is 2650 kg/ m3.065 mm Required: minimum apparent density of the fluid to give separation Solution: let A-galena.Momentum 15.32 % SO2 Burner 6.015 2650 − 𝜌𝑓𝑙𝑢𝑖𝑑 𝝆𝒇𝒍𝒖𝒊𝒅 = 𝟐𝟑𝟕𝟕.015 7500 − 𝜌𝑓𝑙𝑢𝑖𝑑 𝜌𝑓𝑙𝑢𝑖𝑑 = 77772. The charge fuel contains 48% sulfur.015 mm – 0. Impure sulfur is burned to SO2 for conversion to SO3 in a sulfuric acid plant.93 % O2 (83. Given: Ρgalena = 7500 kg/m3 Ρquartz = 2650 kg/m3 Size range = 0. 6.75 % N2) Feed 48 % S 52 % inerts .32% SO2. B-quartz 𝐷𝑝𝐴 𝜌𝐵 − 𝜌𝑓𝑙𝑢𝑖𝑑 =√ 𝐷𝑝𝐵 𝜌𝐴 − 𝜌𝑓𝑙𝑢𝑖𝑑 larger particles as heavier 0.065 7500 − 𝜌𝑓𝑙𝑢𝑖𝑑 =√ 0.065 mm is to be separated into two pure fractions using a hindered settling process. Orsat analysis of the burner shows 9. What percent of the sulfur fired leaves as SO3? Given: Burner Gas (Orsat) 9. 26 − 9.55 bar 𝑅𝑇 𝑎 𝑃= − 2 𝑉−𝑏 𝑉 .32 %𝑺 → 𝑺𝑶𝟑 = 𝟑𝟎.1 K Pc= 220.75𝑘𝑚𝑜𝑙 𝑁2 [ ] = 22.Required: %S converted to SO3 Solution: Basis: 100 kmol of Burner Gas (Orsat) 3 𝑆+ 𝑂 → 𝑆𝑂3 2 2 21 𝑂2 𝑂2 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 = 83.015 Tc = 647.32 − 6.01 𝑘𝑚𝑜𝑙 𝑆𝑂3 3⁄ 𝑂 2 2 4. 𝟎𝟖𝟐𝟓% Thermo 1 13.01 %𝑆 → 𝑆𝑂3 = 𝑥100% 4. Given: steam T = 500⁰C ρ = 24kg/m3 Required: Pressure in kPa Solution: From Appendix A MW = 18. Calculate the pressure (in kPa) of a steam at a temperature of 500⁰C and a density of 24 kg/m3 using the Van der Waals equation.01 + 9.01 𝑘𝑚𝑜𝑙 𝑂2 𝑂2 𝑑𝑖𝑠𝑎𝑝𝑝𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 𝑂2 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑒 𝑆𝑂3 1 𝑆𝑂3 𝑆𝑂3 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 = 6.26 𝑘𝑚𝑜𝑙 𝑂2 79 𝑁2 𝑂2 𝑑𝑖𝑠𝑎𝑝𝑝𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 22.01 𝑘𝑚𝑜𝑙 𝑂2 [ ] = 4.93 = 6. 5 𝑠2 𝑚 𝑢 = 7. 𝟓𝟏𝟔𝟏 𝑳 .01325 𝑅𝑇𝑐 0. A hose shoots water straight up a distance of 2.750625 − 0.5m.465020296 𝑃=( − ) 𝑥101.0304908512 8𝑃𝑐 8𝑥220.01325 𝑀𝑊 18 𝑉= = = 0.75 𝑚3 /𝑚𝑜𝑙 𝜌 24 0.752 𝑷 = 𝟕𝟗𝟒𝟗.0036 𝑠 𝑚 2 1𝑚 2 1000𝐿 𝑉𝑜𝑙𝑢𝑚𝑒 = 7. How much water comes at in 1 minute? Given: D = 2.5 m A = 0.15 5. The end opening on the hose has an area of 0. 12 ) 𝑎= = = 5.55𝑥 1.082052 𝑥647. 27𝑅2 𝑇𝑐 2 27(0. 𝟎𝟖𝟖𝟐 𝒌𝑷𝒂 Transpo 30.75 cm2.08205𝑥647.08205𝑥773.0036 (0.465020296 64𝑃𝑐 1 64𝑥220.2439268096 0.75 𝑐𝑚 ) ( ) (60𝑠) ( ) 𝑠 100 𝑐𝑚 1𝑚3 𝑽𝒐𝒍𝒖𝒎𝒆 = 𝟑𝟏.1 𝑏= = = 0.75 cm2 t = 1 minute Required: Volume of H2O Solution: 𝐾𝐸 = 𝑃𝐸 1 𝑚𝑢2 = 𝑚𝑔ℎ 2 2 𝑚2 𝑢 = √2𝑥9.325 0.81𝑥2.55𝑥 1 1. Superheated steam H1=3702. The inlet to the turbine will be steam at 600⁰C and 10 bar with a velocity of 100 m/s and a flowrate of 2.5 (3281.9153 KJ/kg 𝑘𝑔 𝑘𝐽 𝑊𝑜𝑟𝑘 = 2. Given: Turbine: steam ṁ = 2.5 kg/s.2834 KJ/kg Outlet: @ 0.5 kg/s Inlet Outlet T1 = 600⁰C T2 = 400⁰C P1 = 10 bar P2 = 1 bar u1 = 100 m/s u2 = 30 m/s Required: Work in kW Solution: using Table 2-305 (HB) Inlet: @ 1 MPa Tsat = 450. The conditions at the turbine exit are 400⁰C. Superheated steam H1=3281.Thermo 2 46.4703 K T2>Tsat. An adiabatic steady state steam turbine is being designed to serve as an energy source for a small electrical generator.8728 K T1>Tsat. 1 bar and a gas velocity of 30 m/s. the rate at which work (kW) can be obtained from this turbine is.2834) 𝑠 𝑘𝑔 𝑾𝒐𝒓𝒌 = −𝟏𝟎𝟓𝟎.9153 − 3702.1 MPa Tsat = 372. 𝟗𝟐𝟎𝟑 𝒌𝑾 .