If the system is to provide hot water at a rate of 10

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: m FIGURE 8–29 Schematic for Example 8–5. SOLUTION Water is to be heated in a tube equipped with an electric resistance heater on its surface. The power rating of the heater and the inner surface temperature are to be determined. Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform. 3 The inner surfaces of the tube are smooth. Properties The properties of water at the bulk mean temperature of Tb (Ti Te)/2 (15 65)/2 40°C are (Table A-9). k 992.1 kg/m3 0.631 W/m °C / 0.658 10 Cp Pr 6 4179 J/kg °C 4.32 m2/s Analysis The cross sectional and heat transfer surface areas are Ac As D2 1 (0.03 m)2 7.069 4 pL DL (0.03 m)(5 m) 1 4 · The volume flow rate of water is given as V the mass flow rate becomes · m · V (992.1 kg/m3)(0.01 m3/min) 10 4 m2 0.471 m2 0.01 m3/min. Then 10 L/min 9.921 kg/min 0.1654 kg/s To heat the water at this mass flow rate from 15°C to 65°C, heat must be supplied to the water at a rate of · Q · m Cp(Te Ti) (0.1654 kg/s)(4.179 kJ/kg °C)(65 34.6 kJ/s 34.6 kW 15)°C cen58933_ch08.qxd 9/4/2002 11:29 AM Page 447 447 CHAPTER 8 All of this energy must come from the resistance heater. Therefore, the power rating of the heater must be 34.6 kW. The surface temperature Ts of the tube at any location can be determined from · qs h(Ts Tm) → Ts Tm · qs h where h is the heat transfer coefficient and Tm is the mean temperature of the fluid at that location. The surface heat flux is constant in this case, and its value can be determined from · qs · Q As 34.6 kW 0.471 m2 73.46 kW/m2 To determine the heat transfer coefficient, we first need to find the mean velocity of water and the Reynolds number: · V Ac 0.010 m3/min 14.15 m/min 7.069 10 4 m2 (0.236 m/s)(0.03 m) mD 10,760 0.658 10 6 m2/s m Re 0.236 m/s which is greater than 10,000. Therefore, the flow is turbulent and the entry length is roughly Lh Lt 10D 10 0.03 0.3 m which is much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire pipe and determine the Nusselt number from hD k Nu 0.023 Re0.8 Pr0.4 0.023(10,760)0.8 (4.34)0.4 69.5 Then, k Nu D h 0.631 W/m · °C (69.5) 0.03 m 1462 W/m2 °C and the surface temperature of the pipe at the exit becomes Ts Tm · qs h 65°C 73,460 W/m2 1462 W/m2 · °C 115°C Discussion Note that the inner surface temperature of the pipe will be 50°C higher than the mean water temperature at the pipe exit. This temperature difference of 50°C between the water and the surface will remain constant throughout the fully developed flow region. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 448 448 HEAT TRANSFER EXAMPLE 8–6 Ts = 60°C 0.2 m Te Air 1 atm 80°C Heat Loss from the Ducts of a Heating System Hot air at atmospheric pressure and 80°C enters an 8–m-long uninsulated square duct of cross section 0.2 m 0.2 m that passes through the attic of a house at a rate of 0.15 m3/s (Fig. 8–30). The duct is observed to be nearly isothermal at 60°C. Determine the exit temperature of the air and the rate of heat loss from the duct to the attic space. 0.2 m 8m FIGURE 8–30 SOLUTION Heat loss from uninsulated square ducts of a heating system in the attic is considered. The exit temperature and the rate of heat loss are to be determined. Schematic for Example 8–6. Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the duct are smooth. 3 Air is an ideal gas. Properties We do not know the exit temperature of the air in the duct, and thus we cannot determine the bulk mean temperature of air, which is the temperature at which the properties are to be determined. The temperature of air at the inlet is 80°C and we expect this temperature to drop somewhat as a result of heat loss through the duct whose surface is at 60°C. At 80°C and 1 atm we read (Table A-15) k 0.9994 kg/m3 0.02953 W/m °C 2.097 10 5 m2/s Cp Pr 1008 J/kg °C 0.7154 Analysis The characteristic length (which is the hydraulic diameter), the mean velocity, and the Reynolds number in this case are Dh m Re 4Ac 4a2 a 0.2 m p 4a · V 0.15 m3/s 3.75 m/s Ac (0.2 m)2 (3.75 m/s)(0.2 m) m Dh 2.097 10 5 m2/s 35,765 which is greater than 10,000. Therefore, the flow is turbulent and the entry lengths in this case are roughly Lh Lt 10D 10 0.2 m 2m which is much shorter than the total length of the duct. Therefore, we can assume fully developed turbulent flow in the entire duct and determine the Nusselt number from Nu hDh k 0.023 Re0.8 Pr0.3 0.023(35,765)0.8 (0.7154)0.3 91.4 cen58933_ch08.qxd 9/4/2002 11:29 AM Page 449 449 CHAPTER 8 Then, 0.02953 W/m · °C k Nu (91.4) 13.5 W/m2 °C Dh 0.2 m pL 4aL 4 (0.2 m)(8 m) 6.4 m2 · V (1.009 kg/m3)(0.15 m3/s) 0.151 kg/s h As · m Next, we determine the exit temperature of air from Te Ts · Ti) exp ( hAs /m Cp) (Ts 60°C [(60 (13.5 W/m2 · °C)(6.4 m2) (0.151 kg/s)(1008 J/kg · °C) 80)°C] exp 71.3°C Then the logarithmic mean temperature difference and the rate of heat loss from the air become Ti Te 80 71.3 15.2°C Ts Te 60 71.3 ln ln 6...
View Full Document

Ask a homework question - tutors are online