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Convective Heat Transfer. Convective Heat Transfer Goals: By the end of today’s lecture, you should be able to: use empirical correlations to compute estimates for heat transfer coefficients distinguish between individual versus overall heat transfer coefficients
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Convective Heat Transfer Goals: By the end of today’s lecture, you should be able to: • use empirical correlations to compute estimates for heat transfer coefficients • distinguish between individual versus overall heat transfer coefficients • compute the overall heat transfer coefficient for a system • define limiting resistance and identify for a given scenario
Outline: Review Example empirical relationships for heat transfer coefficients Overall heat transfer coefficients Example problem 1 -- overall heat transfer coefficient Example problem 2 -- overall heat transfer coefficient
Convection where h is the individual heat transfer coefficient and Is obtained from purely EMPIRICAL methods.
Ludwig Prandtl (1875-1953) Wilhelm Nusselt (1882-1957)
T Bulk Film T Surface Object (pipe wall / plate, etc)
Condensing vapors on horizontal tubes: Low Temp Fluid Low Temp Fluid
NOTE: The wall temperature, Tw, is usually an unknown but can be estimated using h. However, we need Tw to compute h. Thus, computation of film coefficients (heat transfer coefficients) for condensing vapors on vertical or horizontal tubes is usually iterative.
III. Overall heat transfer coefficients For pure conduction, we considered resistances in series in which our primary interest was in the temperatures at the inner and outer walls (not the interior walls). For convection, a similar principle arises. Consider a pipe filled with hot fluid at temperature T1. Define intermediate temperatures as follows: T1 T2 Insulation Air T5 Hot liquid T3 Pipe T4
The heat transfer rate for each "step": q1->2 = conduction or convection? q2->3 = conduction or convection? q3->4 = conduction or convection? q4->5 = conduction or convection? Convection Conduction Conduction Convection
(T5–T1) = (T5-T4) + (T4-T3) + (T3-T2) + (T2-T1) Q = UoAo (T5-T1) Q = hoAo (T5-T4) Outside film (air) Q = kinsul Alm (T4-T3) / Drinsul Insulation Q = kpipe Alm (T3-T2) / Drpipe Pipe Q = hiAi (T2-T1) Inside film (hot liquid)
(T5–T1) = (T5-T4) + (T4-T3) + (T3-T2) + (T2-T1) (T5-T1) = Q / UoAo Similar substitutions yield:
With U known, it is a simple matter to calculate the overall heat transfer rate given the total temperature difference. Note that we have used the symbol Uo because the overall heat transfer coefficient was defined with respect to the outside area (Ao) of the pipe. This is the most common practice. We could equally have defined 1/Ui or UiAi. In this case, we would use Ai as the basis for calculations. In either case, the values of U would be slightly different, but UA and hence q are the same.
Equation for 1/Ui Limiting Resistance: • A very useful concept in heat transfer is that of limiting resistance. • What is limiting resistance? • What would the limiting resistance be for a hot liquid flowing inside an uninsulated pipe? • How would you increase the heat transfer rate?
IV. Example problem 1 -- Overall heat transfer coefficient Blood circulation is critical for thermal regulation in the body. Changes in blood flow and/or dilation of the blood vessels are regulated by sympathetic nerve activity controlled by the brain. Raynaud's disease is a condition that occurs frequently in young, healthy women and is characterized by cold extremities (e.g. hands and feet). Although typically just a nuisance, in severe cases, this condition can lead to the formation of ulcers and a possible need for amputation. Raynaud's disease results from overactive neural stimulation which causes constriction ("vasoconstriction") of small arteries and arterioles (blood vessels), reducing blood flow and therefore reducing the heat transfer rate.
For a normal 20-µm small artery (inner diameter) with a wall thickness of 3 µm, what is the heat transfer rate through the vessel wall if the temperature of the blood is 37°C and the local body temperature in cool weather is 29°C. Assume steady state heat transfer for your calculation. The thermal conductivity of the blood vessel wall is about 0.15 Btu/hr-ft-°F. The individual heat transfer coefficient on the blood side is roughly 120 Btu/hr-ft2-°F and that for the tissue side is approximately 50 Btu/hr-ft2-°F. Compute the heat transfer rate for: (a) A normal 20-µm diameter blood vessel in a healthy patient (b) A blood vessel that is constricted to 10-µm diameter (~same thickness) in a Raynaud's patient
Ten Minute Problem -- Overall heat transfer coefficient A standard 2-inch Sch 80 pipe carries saturated steam at 500°F. The pipe is insulated with a 0.5-inch layer of 85% magnesia pipe covering. The ambient air temperature is 80°F. The thermal conductivities of steel and magnesia are 26.0 and 0.034 Btu/ft-hr-°F, respectively. The individual heat transfer coefficients on the inside and outside are 2000 and 2 Btu/hr-ft2-°F, respectively. (a) Calculate the rate of heat loss (Btu/hr) per unit length. (b) Compute the temperature on the outer surface of the insulation. ID 2-inch Sch 80 pipe = 1.939 inches OD 2-inch Sch 80 pipe = 2.375 inches
80 F 187.2 F 499.6 F 499.8 F 500 F Gas Film Insulation Pipe Steam Film Q = 189.5 BTU / hr