Solved Problems

SOLVED PROBLEMS Combined Convection and Radiation

Problem 1:  A surface is at 200°C and is exposed to surroundings at 60°C

and convects and radiates heat to the surroundings. The convection coefficient is 80W /m2K. The radiation factor is one. If the heat is conducted to the surface through a solid of conductivity 12 W/mK, determine the temperature gradient at  the surface in the solid. Solution:

Heat convected + heat radiated = heat conducted considering 1 1m m2, h(T 1 – T 2) + σ(T 14 – T 24) = – kdT /dx   Therefore, 80(200 – 60) + 5.67 {[(200 + 273)/100]4 – [(60 + 273)/100]4} = – 12 dT/dx   Therefore dT/dx = dT/dx = – (11200 + 2140.9)/12 = – 1111.7°C/m. Problem 2: Heat is conducted through a material with a temperature

gradient of – 9000 °C/ °C /m. The conductivity of the material is 25W/mK. If  this this heat heat is conv convec ecte ted d to su surr rrou ound ndin ings gs at 30 30°C °C with with a conv convec ecti tion on coefficient of 345W/m2K, determine the surface temperature. If the heat is radiated to the surroundings at 30°C determine the surface temperature. Solution: In this case only convection and conduction are involved.

– kAdT /dx = dx = hA hA((T 1 – T 2). Considering unit area, – 25 × 1 × (– 9000) = 345 × 1 ( T 1 – 30)  Therefore, T1 = 682.17°C In this case conduction and radiation are involved. Heat conducted = Heat radiated – 25 × 1 × (– 9000) = 5.67 [(T  [( T 1/100)4 – (303/100)4]  Therefore, T1 = 1412.14K = 1139°C. Problem 3: There is a heat flux through a wall of 2250W/m 2. The same is

dissipated to the surroundings by convection and radiation. The surroundings is at 30°C. The convection coefficient has a value of 75W/m 2K. For radiation F = 1. Determine the wall surface temperature. Solution: For the specified condition, Consider unit area.  The heat conducted = heat convected convected + heat radiated Using the rate equations, with absolute temperature 2250 = [(T2 – 303)/(1/75 × 1] + 5.67 × 1[(T  1[( T 2/100)4 – (303/100)4] = 75T  75T 2 – 22725 + 5.67(T  5.67(T 2/100)4 – 477.92 or, (T  (T 2/100)4 + 13.2275T  13.2275T 2 – 4489.05 = 0.

 This equation can be solved only by trial. It may be noted that the contribution of (T 2/100)4 is small and so the first choice of  T 2 can be a little less than 4489/13.227 = 340K. The values of the reminder for T 2 = 300, 310, 320, 330 are given below:  Assumed value of T 2 Remainder 

300

310

320

330

330.4

330.3

– 439.80

– 296.2

– 15.1

– 5.38

0.484

– 0.98

So, the temperature T 2 is near 330K. By one more trial T 2 is obtained as 330.4K or 57.4°C. Check: Q = 75(330.4 – 303) + 5.69(3.304 4 – 3.034)

= 2047.5 + 206 = 2253.5 W.

PROBLEM 1.1

The outer surface of a 0.2m-thick concrete wall is kept at a temperature of –5°C, while the inner surface is kept at 20°C. The thermal conductivity of the concrete is 1.2 W/(m K). Determine the heat loss through a wall 10 m long and 3 m high. GIVEN  10 m long, 3 m high, and 0.2 m thick concrete wall  Thermal conductivity of the concrete ( k ) = 1.2 W/(m K)  Temperature of the inner surface ( Ti) = 20°C  Temperature of the outer surface ( To) = –5°C FIND  The heat loss through the wall ( qk ) ASSUMPTIONS  One dimensional heat flow  The system has reached steady state SKETCH

SOLUTION  The rate of heat loss through the wall is qk = (AK/L) × (∆T ) qk = [(10m) (3m)(1.2W/(mK))/0.2m] × (20°C – (–5°C)) qk = 4500 W COMMENTS Since the inside surface temperature is higher than the outside temperature heat is transferred from the inside of the wall to the outside of the wall.

PROBLEM 1.9 Heat is transferred at a rate of 0.1 kW through glass wool insulation (density = 100 kg/m3) of 5 cm thickness and 2 m 2 area. If the hot surface is at 70°C, determine the temperature of the cooler surface. GIVEN 3  Glass wool insulation with a density ( ρ) = 100 kg/m  Thickness ( L) = 5 cm = 0.05 m 2  Area ( A) = 2 m  Temperature of the hot surface ( T h) = 70°C

Rate of heat transfer ( qk ) = 0.1 kW= 100 W FIND  The temperature of the cooler surface ( T C) ASSUMPTIONS  One dimensional, steady state conduction  Constant thermal conductivity SKETCH 

PROPERTIES AND CONSTANTS From Appendix 2, Table 11 The thermal conductivity of glass wool at 20°C ( k ) = 0.036 W/(m K) SOLUTION For one dimensional, steady state conduction, the rate of heat transfer, is

qk  = (Ak/L) × (T h – T c) Solving this for Tc Tc = T h – qk  L/A k  T c = 70°C – [(100W) (0.05m)]/[(2 m 2 )( 0.036W/mK)] T c = 0.6°C PROBLEM 1.10 A heat flux meter at the outer (cold) wall of a concrete building indicates that the heat loss through a wall of 10 cm thickness is 20 W/m 2. If a thermocouple at the inner surface of the wall indicates a temperature of 22°C while another at the outer surface shows 6°C, calculate the thermal conductivity of the concrete and compare your result with the value in Appendix 2, Table 11. GIVEN  Concrete wall  Thickness ( L) = 100 cm = 0.1 m 2  Heat loss (q/ A) = 20 W/m  Surface temperature  Inner (T i) = 22°C  Outer ( T o) = 6°C FIND  The thermal conductivity ( k ) and compare it to the tabulated value ASSUMPTIONS  One dimensional heat flow through the wall  Steady state conditions exist SKETCH

SOLUTION The rate of heat transfer for steady state, one dimensional conduction, is qk  = (k A/L) × (T hot – T cold) Solving for the thermal conductivity k = (qk  / A) × L/(T i – To  )

k = (20W/m2 ) × [0.1m2/(22°C – 6°C)] = 0.125 W/(m K)

This result is very close to the tabulated value in Appendix 2 , Table 11 where the thermal conductivity of concrete is given as 0.128 W/(m K).

PROBLEM 1.11 Calculate the heat loss through a 1-m by 3-m glass window 7 mm thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C. Comment on the possible effect of radiation on your answer. GIVEN  Window: 1 m by 3 m  Thickness ( L) = 7 mm = 0.007 m  Surface temperature  Inner (T i) = 20°C and outer (T o) = 17°C FIND  The rate of heat loss through the window ( q) ASSUMPTIONS  One dimensional, steady state conduction through the glass  Constant thermal conductivity SKETCH

PROPERTIES AND CONSTANTS From Appendix 2, Table 11 Thermal conductivity of glass (k ) = 0.81 W/(m K) SOLUTION The heat loss by conduction through the window is

qk  = (k A/L) × (T hot – T cold) qk  = [[(0.81W/(mK))(1m) (3m)]/(0.007m)] × (20°C – 17°C) qk  = 1040 W COMMENTS  Window glass is transparent to certain wavelengths of radiation, therefore some heat may be lost by radiation through the glass.  During the day sunlight may pass through the glass creating a net heat gain through the window.

PROBLEM 1.12 If in Problem 1.11 the outer air temperature is –2°C, calculate the convective heat transfer coefficient between the outer surface of the window and the air assuming radiation is negligible. Problem 1.11: Calculate the heat loss through a 1 m by 3 m glass window 7 mm thick if  the inner surface temperature is 20°C and the outer surface temperature is 17°C. Comment on the possible effect of radiation on your answer. GIVEN  Window: 1 m by 3 m  Thickness ( L) = 7 mm = 0.007 m  Surface temperatures  Inner (T i) = 20°C and outer (T o) = 17°C

The rate of heat loss = 1040W (from the solution to Problem 1.11)  The outside air temperature = –2°C FIND  The convective heat transfer coefficient at the outer surface of the window ( hc ) ASSUMPTIONS  The system is in steady state and radiative loss through the window is negligible SKETCH 

SOLUTION For steady state the rate of heat transfer by convection (Equation (1.10)) from the outer  surface must be the same as the rate of heat transfer by conduction through the glass qc = hc A ∆T = qk  Solving for hc hc = qk  / A (T o – T∞  )

hc = 1040W / (1m)(3m)( 17°C – – 2°C) hc = 18.2 W/(m 2 K) COMMENTS  This value for the convective heat transfer coefficient falls within the range given for the free convection of air in Table 1.4.

PROBLEM

 The wall of an industrial furnace is constructed from 0.15-m-thick fireclay brick having a thermal conductivity of  1.7 W/m. K. Measurements made during steady-state operation reveal temperatures of 1400 and 1150 K at the inner and outer surfaces, respectively. What is the rate of heat transfer through a wall that is 0.5 m by 1.2 m on a side? Solution  Known: Steady-state conditions with prescribed wall thickness, area, thermal conductivity, and surface temperatures. Find: Heat transfer rate through the wall.  Schematic and Given Data:

 Assumptions: 1. Steady-state conditions. 2. One-dimensional conduction through the wall. 3. Constant thermal conductivity.  Analysis: Since heat transfer through the wall is by conduction, the heat flux may be

determined from Fourier’s law. q x = k ∆T /L = 1.7 W/m. K × 250 K / 0.15 m = 2833 W/m 2 The heat  flux represents the rate of heat transfer through a section of unit area, and it is uniform across the surface of the wall. The heat rate through the wall of area A = H × W is then Q x = ( HW ) q x = (0.5 m × 1.2 m) 2833 W/m 2 = 1700 W Comments: Note the direction of heat flow and the distinction between heat flux and heat rate