Convective Heat Transfer Assignment 3: Solution Q. 1 A. C. Sol. A fluid is flowing inside a circular tube of 5 cm radius. The maximum velocity of fluid inside tubeis 0.3 m/s and velocity profile at any section along the axis of the tube is parabolic. Determine the average fluid temperature in the tube if temperature profile is given as: T(r) = 300+150(r/R)3. Consider temperature scale in Kelvin. 212.5 K B. 106.25 K 167.14 K D. 334.28 K Given Radius, R = 5 cm = 0.05 m Maximum velocity, Umax = 0.3 m/s Assume parabolic velocity profile. Temperature profile, π π = 300 + 150 Determine Average temperature Solution Given assume parabolic velocity profile, π 3 π So velocity profile in pipe flow, π π = ππππ₯ × 1 − π π 2 π We know, average velocity, πππ£π = πππ₯ 2 π 2 πππ£π = π π × π π × π × ππ πππ£π × π 2 0 π =0.05 2 π 3 π πππ£π = 300 + 150 × 0.3 × 1 − 2 0.15 × π 0 π π πππ£π = 334.28 πΎ Q. 2 A. C. Sol. 2 × π × ππ Water enters inside a tube of closed feed water heater having 5 cm diameter at 30 °C at a rate of 0.25kg/s and is heated by the saturated steam withdrawn from high pressure turbine at 150 °C. Determine the length of the tube required to heat the water to 130 °C if the average heat transfer coefficient is 1200 W/m2-K. Properties of water at mean temperature are: ρ = 970.87 kg/m3, μ = 343x10-6 N-s/m2, k = 0.671 W/m-K, Pr = 2.14. 9.95 m B. 7.93 m 4.63 m D. 2.76 m Given Diameter, D = 5 cm = 0.05 m Inlet temperature, Ti = 30 ΛC = 303 K Surface temperature, TS = 150 ΛC = 423 K Mass flow rate, m = 0.25 kg/s Average heat transfer coefficient, havg = 1200 W/m2K Properties at mean temperature are given. Determine Length of tube. Solution For constant temperature cases −βππ£π π΄π ππ = ππ₯π ππ πππ −βππ£π π΄π ππ − ππ = ππ₯π ππ − ππ πππ 423 − 403 −1200 × π × 0.05 × πΏ = exp 0.671 ×2.14 423 − 303 0.25 × 343 × 10 −6 πΏ = 9.95 π Q. 3 A. C. Sol. Water is heated inside the circular tube of 5 cm internal diameter and 7 m length from 20 °C to 75 °C. An electric resistance heater is employed around the periphery of the tube that provides uniform heating. The outer surface of the heater is well insulated such that all the heat is transferred to water. Determine the power rating of resistance heater if the water flows inside the tube at a rate of 0.1 kg/s. Properties of water at mean temperature are: ρ = 989.12 kg/m3, μ = 577x10-6 N-s/m2, k = 0.640 W/m-K, Pr = 3.77. 2.3 kW B. 23 kW 2300 kW D. 230 kW Given Diameter, D = 5 cm = 0.05 m Length, L = 7 m Inlet temperature, Ti = 20 ΛC Exit temperature, Te = 75 ΛC Mass flow rate, m = 0.1 kg/s Properties are given at mean temperature. Determine Pumping rating. Solution Heat transfer rate, π = π πΆπ βπ 0.64 × 3.77 π = 0.1 × 75 − 20 577 × 10−6 π = 23 ππ Q. 4 A. C. Sol. Also, calculate the surface temperature of the tube at exit given in Q. 3 if the Nusselt number is estimated using the Dittus-Boelter correlation as Nu = 0.023Re0.8Pr0.4. B. 125.72 °C 398.72 °C D. 80.07 °C 98.22 °C Given Same as above in Q3. Determine Surface temperature at exit. Solution 4π = 4413.3086 πππ· ππ’ = 0.023 π π 0.8 ππ 0.4 π π = β × 0.05 = 0.023 4413.3086 0.8 3.77 0.4 0.640 β = 412.36878 π/π2 πΎ π = βπ΄π ππ − ππ ππ₯ππ‘ = 412.36878 × π × 0.05 × 7 × ππ − 75 ππ = 125.72 °πΆ ππ’ = Q. 5 A. C. Sol. Air flows inside a 10 m long square duct of 0.25 m side at 70 °C and atmospheric pressure. The duct passes through a room which is maintained at 40 °C. Consider the surface of duct attains isothermal conditions at 40 °C when air flows at a rate of 0.12 m3/s. Determine the temperature of air at the exit of duct. Estimate average Nusselt number using Dittus-Boelter correlation as Nu = 0.023Re0.8Pr0.3. Use hydraulic diameter 4π΄ of duct, π·β = π π as length scale to calculate Reynolds number. Here, AC and p are the cross-sectional area and perimeter of duct cross-section, respectively. Use following properties of air: ρ = 1.023 kg/m3, μ = 200.2x10-7 N-s/m2, k = 0.0297 W/m-K, Pr = 0.701. B. 61.32 °C 62.5 °C D. 56.06 °C 57.47 °C Given Side, a = 0.25 m Length, L = 10 m Inlet temperature, Ti = 70 ΛC = 343 K Constant Surface temperature, TS = 40 ΛC = 313 K Volumetric flow rate, π = 0.12 m3/s Determine Temperature at exit of duct. Solution π = π΄πππ£π 0.12 πππ£π = = 1.92 π/π (0.25 × 0.25) 4π΄ 4(0.25×0.25) Hydraulic Diameter, π·β = π = = 0.25 π π 2(0.25+0.25) ππππ£π π·β 1.023 × 1.92 × 0.25 π π = = = 24527.47253 π 200.2 × 10−7 ππ’ = 0.023 π π 0.8 ππ 0.3 β × 0.25 ππ’ = = 0.023 24527.47253 0.8 0.701 0.3 0.0297 β = 7.98 π/π2 πΎ For constant temperature cases: ππ −β π΄π = ππ₯π ππ πππ ππ − ππ −β π΄π = ππ₯π ππ − ππ πππ 40 − ππ = exp 40 − 70 −7.98 × 2(0.25 + 0.25) × 10 1.023 × 0.12 × ππ = 56.06 °πΆ 0.0297 ×0.701 200.2 × 10 −7 Q. 6 A. C. Sol. Air flows through an unisulated circular duct of 12 cm diameter at a mass flow rate of 0.03 kg/s. Air enters the duct at 95 °C and leaves the duct at a temperature of 70 °C. Length of the duct is 7 m and it is exposed to ambient at 3 °C. Determine the surface temperature of the duct at exit if heat transfer coefficient between duct surface and ambient is 5 W/m2-K. Use Dittus-Boelter correlation as Nu = 0.023Re0.8Pr0.3 to calculate the heat transfer coefficient between duct surface and inside air. Properties of air at mean temperature are: ρ = 0.993 kg/m3, μ = 210x10-7 N-s/m2, k = 0.0304 W/m-K, Pr = 0.699. B. 56.7 °C 82.5 °C D. 42.75 °C 49.78 °C Given Diameter, D = 12 cm = 0.12 m Length, L = 7 m Inlet temperature, Ti = 95 ΛC Exit temperature, Te = 70 ΛC Mass flow rate, π = 0.03 kg/s Determine Surface temperature at exit. Solution 4π 4 × 0.03 π π = = = 15157.6 π π π· 210 × 10−7 × π × 0.12 ππ’ = 0.023 π π 0.8 ππ 0.3 βπ × 0.12 ππ’ = = 0.023 15157.6 0.8 0.699 0.3 0.0304 βπ = 11.57 π/π2 πΎ Apply energy balance at wall, Energy taken by wall from hot fluid = Energy released by wall to surrounding βπ π΄π ππ − ππ = βπ π΄π (ππ − π∞ ) 11.57 343 − ππ = 5 ππ − 276 ππ = 322.78 πΎ = 49.78 °πΆ Q. 7 Water enters inside a circular duct of 12 cm diameter and 7 m length at a temperature of 20 °C. The initial 2 m long section of the duct is maintained at ambient temperature of 20 °C. Remaining section of the duct is wrapped with and electric resistance heater which maintains the duct surface at constant temperature of 60 °C. Flow is assumed to be hydrodynamically fully developed before it enters the heated section of the duct. Determine the exit temperature of water, neglecting the axial conduction between heated and unheated portion of the tube if mass flow rate of water is 0.1 kg/s. Use Edwards et al. (1979) correlation to calculate the heat transfer coefficient: ππ’ = 3.66 + 0.065 1+0.04 A. C. Sol. π· π πππ πΏ 2 3 π· π πππ πΏ . Use following properties of water: ρ = 997 kg/m3, μ = 855x10-6 N- s/m2, k = 0.613 W/m-K, Pr = 5.83. 28.79 °C 26.50 °C Given Diameter, D = 12 cm = 0.12 m Length, L = 7 m B. 27.26 °C D. 29.78 °C Inlet temperature, Ti = 20 ΛC Mass flow rate, m = 0.1 kg/s Determine Exit temperature. Solution π = ππ΄π 0.1 π= π 997 × 4 × (0.12)2 π = 8.87 × 10−3 π/π 4π π π = = 1240.974 πππ· π· 0.065 πΏ π π ππ ππ’ = 3.66 + π· 1 + 0.04 πΏ π π ππ 0.12 5 0.065 ππ’ = 3.66 + 0.12 5 1 + 0.04 ππ’π π· 2 3 × 1240.974 × 5.83 × 1240.974 × 5.83 2 3 = 8.69 ο¨ β= = 44.39 π/π2 πΎ For constant temperature cases ππ −β π΄π = ππ₯π ππ πππ ππ − ππ −β π΄π = ππ₯π ππ − ππ πππ 60 − ππ −44.39 × π × 0.12 × 5 = exp 0.613 ×5.83 60 − 20 0.1 × 855 × 10 −6 ππ = 27.26 ΛπΆ Q. 8 A. C. Sol. Engine oil flows through a circular tube of diameter 100 mm and 10 m length whose surface is maintained at uniform heat flux of 200 W/m2. Determine the temperature of the tube surface at exit if the inlet temperature and mass flow rate of engine oil are 25 °C and 0.1 kg/s. Use following properties of engine oil: ρ = 868.8 kg/m3, μ = 11.36x10-2 Ns/m2, k = 0.142 W/m-K, Pr = 1585. B. 28.17 °C 52.86 °C D. 60.48 °C 66.65 °C Given Diameter, D = 100 mm = 0.1 m Heat flux, q = 200 W/m2 Inlet temperature, Ti = 25 ΛC = 298 ΛK Determine Exit temperature. Solution Mass flow rate, m = 0.3 Kg/s π = ππ΄π π= 0.1 868.8 × π 4 × 0.1 2 = 0.0146 π π π = πππ (ππ − ππ ) 0.142 × 1585 200 × π × 0.1 × 10 = 0.1 × × (ππ − 25) 11.36 × 10−2 ππ = 28.17 ΛπΆ πππ· 868.8×0.0146 ×.1 Reynolds number, π π = = = 11.16 −2 π 11.36 × 10 ο Flow is laminar, therefore Nusselt number, Nu = 4.36 (Constant heat flux condition) ππ’π and β = π· = 6.19 π/π2 πΎ Applying energy balance at tube exit, π = β(ππ − ππ ) ο 200 = 6.19 × (ππ − 28.17) ο Temperature of tube surface at exit, ππ = 60.48 ΛπΆ
