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Quiz – 2014.02.05 An organic liquid enters a 0.834-in. ID horizontal steel tube, 3.5 ft long, at a rate of 5000 lb/hr. You are given that the specific heat, thermal conductivity, and viscosity of the liquid is 0.565 Btu/lb-°F, 0.0647 Btu/hr-ft-°F, and 0.59 lb/ft-hr, respectively. All these properties are assumed constant. If the liquid is being cooled, determine the inside-tube heat transfer coefficient using the Dittus-Boelter equation: N N u 0.023 N R e 0.8 N Pr n = 0.4 when fluid is heated n = 0.3 when fluid is cooled n TIME IS UP!!! Recall Convection Heat Transfer = ℎ − Where: Q = heat flow rate Driving force A = heat transfer area h = heat transfer coefficient − = Tw = temperature at solid wall 1 Tf = temperature at bulk fluid ℎ Thermal Resistance Combined Heat Transfer Outline 3. Conduction Heat Transfer 4. Convection Heat Transfer 5. Combined Heat Transfer 5.1. Overall Heat Transfer Coefficient 5.2. Log-Mean Temperature Difference 6. Overall Shell Heat Balances Overall Heat Transfer Coefficient Conduction Convection −∆ = ∆ −∆ = 1 ℎ Combined Heat Transfer (flat slab): −∆ = 1 ∆ 1 + + ℎ1 ℎ2 Overall Heat Transfer Coefficient Define: Overall Heat Transfer Coefficient, U 1 1 ∆ 1 = + + ℎ1 ℎ2 −∆ = 1 Combined Heat Transfer (flat slab): −∆ = 1 ∆ 1 + + ℎ1 ℎ2 Overall Heat Transfer Coefficient Define: Overall Heat Transfer Coefficient, U 1 1 ∆ 1 = + + ℎ1 ℎ2 Inside overall heat transfer coefficient, Ui 1 1 ∆ 1 = + + ℎ ℎ Outside overall heat transfer coefficient, Uo 1 1 ∆ 1 = + + ℎ ℎ Overall Heat Transfer Coefficient Define: Overall Heat Transfer Coefficient, U 1 1 ∆ 1 = + + ℎ1 ℎ2 1 1 ∆ 1 = + + ℎ ℎ 1 1 ∆ 1 = + + ℎ ℎ Relationship between the two: 1 1 = Overall Heat Transfer Coefficient Define: Overall Heat Transfer Coefficient, U −∆ = 1 = − = − Overall Heat Transfer Coefficient Exercise! Saturated steam at 267°F is flowing inside a steel pipe with an ID of 0.824 in. and an OD of 1.05 in. The pipe is insulated with 1.5 in. of insulation on the outside. The convective heat transfer coefficient inside and outside the pipe is hi = 1000 Btu/hr/ft2/°F and ho = 2 Btu/hr/ft2/°F, respectively. The mean thermal conductivity of the metal is 45 W/m/K or 26 Btu/hr/ft/°F, while that of the insulation material is 0.064 W/m/K or 0.037 Btu/hr/ft/°F. Calculate the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F. Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe Section) = − = − Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe Section) *The temperature of the fluid and immediate surroundings vary along the length. Let: TA1 = fluid temp. at pt.1 TB1 TB2 TA1 TA2 = fluid temp. at pt.2 TA2 1 2 TB1 = surr. temp. at pt.1 TB2 = surr. temp. at pt.2 Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Making a heat balance across the entire pipe for an area dA: = − = Let: TB1 TA2 = fluid temp. at pt.2 TA1 = fluid temp. at pt.1 TB2 TA1 TA2 1 2 TB1 = surr. temp. at pt.1 TB2 = surr. temp. at pt.2 Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Making a heat balance across the entire pipe for an area dA: = − = TB1 According to the combined heat transfer equation: TB2 TA1 TA2 1 2 = − Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Making a heat balance across the entire pipe for an area dA: = = − According to the combined heat transfer equation: 1 1 − = − + = − Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Equating the dq from the 2 equations below: − 1 1 = − + − 1 1 − = − + = − Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Equating the dq from the 2 equations below: − 1 1 = − + − Making a heat balance in the inlet and outlet: = 1 − 2 = 2 − 1 Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Equating the dq from the 2 equations below: − 1 1 = − + − Making a heat balance in the inlet and outlet: Adding the 2 equations: 1 1 2 − 1 + (1 − 2 ) + = Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Equating the dq from the 2 equations below: − 1 1 = − + − Substituting: ( − ) 2 − 1 + (1 − 2 ) = − − Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) 2 − 2 2 − 1 + (1 − 2 ) ln = − 1 − 1 Integrating: ( − ) 2 − 1 + (1 − 2 ) = − − Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) 2 − 2 2 − 1 + (1 − 2 ) ln = − 1 − 1 Rearranging: 2 − 2 − 1 − 1 = ln 2 − 2 1 − 1 Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Define: Logarithmic Mean Temperature Difference ∆ ∆2 − ∆1 = ∆2 ln ∆1 TB1 TB2 TA1 TA2 1 2 − 2 − 1 − 1 = ln 2 − 2 1 − 1 2 Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Define: Logarithmic Mean Temperature Difference ∆ ∆2 − ∆1 = ∆2 ln ∆1 TB1 TB2 TA1 TA2 1 2 − 2 − 1 − 1 = ln 2 − 2 1 − 1 2 Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Final Form: = ∆ TB1 TB2 TA1 TA2 1 2 − 2 − 1 − 1 = ln 2 − 2 1 − 1 2 Log-mean Temperature Difference Combined Heat Transfer (for Circular Pipe) Final Form: = ∆ But still: TB1 TB2 TA1 TA2 1 = ∆ = ∆ 2 Log-mean Temperature Difference Exercise! 250 kg/hr of fluid A (cp = 5.407 J/gK) is to be cooled from 150°C using a cooling fluid B which enters a countercurrent double-pipe heat exchanger at 50°C and leaves at 85°C. The total heat transfer area available is 5 m2 and the overall heat transfer coefficient is 230 W/m2K. Determine the outlet temperature of fluid A assuming no phase change. Log-mean Temperature Difference Solution! The heat used to increase the temperature of fluid A is the same heat transferred across the pipe. = ∆ = ∆ ∆ 150 − 85 − ( − 50) = 150 − 85 − 50 Log-mean Temperature Difference Solution! = ∆ = ∆ ∆ 150 − 85 − ( − 50) = 150 − 85 − 50 Substituting the values: 1000 250 5.407 150 − 1 ℎ 3600 2 = 230 5 ∆ 2 ∙ ∙ 1 ℎ Shift solve for T!