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1.  Determine the rate of heat transfer by conduction per unit area, by means me ans of conduction for a furnace wall made of fire clay. Furnace wall thickness is 6" or half a foot. Thermal conductivity of the furnace wall clay is 0.3 W/m·K. The furnace wall temperature can be taken to be same as furnace operating temperature which is 6500C and temperature of the outer wall of the furnace is 1500C. A )480.3 W/m^2

C) 492.13 W/m^2 *

B) 471.4 W/m^2

D) 501.13 W/m^2

Solution: Q/A = k × (T1 - T2)/L Q/A = 0.3×(650-150)/0.3048 W/m2 =

492.13 W/m^2

2.  A 2-in schedule 40 steel pipe (k= 36 W/m*K) is insulated i nsulated with a 2-cm-thick-layer of insulation (k=0.02 W/m*K). The inside surface of the pipe is maintained at a temperature of 60*C, and the outside surface of the insulation is maintained at a temperature at 20*C. Find the heat transfer from the pipe per meter of length. NOTE: From Appendix, 2-in steel pipe has an inside diameter of 0.0525m and an outside diameter of 0.0603m A.  6.90 W/m B.  10.90 W/m Given: Radius 1= 0.5(0.0525)m = 0.0263m Radius 2= 0.5(0.0603)m =0.0302m Radius 3= (radius 2 + 0.020)m= 0.0502m T1= 60*C T2= 20*C Solution: T2-T1 Q/L = ___________________________ ______________ ________________ ___ Ln (r2/r1) ln(r3/r2)  ______________  ____________ __ + __________ _______________ _____ 2πK 2πK 

Q/L =

(60-20)*C  ______________________________  _______________________ _______

D. 7.90 W/m C. 9.90 W/m *Ans

 

 

Ln (0.0302/0.0263)m ln(0.0502/0.0302)m  ______________  ____________ __ + __________ _______________ _____ 2π(0.02W/m*K)  2π(36W/m*K)

Q/L= 9.90 W/m Ans.

3. 

The roof of an electrically heated home is 6m long, 8m wide, and 0.25m thick, and is made of flat layer of concrete whose thermal conductivity is k=0.8 W/m-°C. The temperatures of the inner and outer surface of the roof one night are measured to be 15°C and 4°C, respectively, for a period of 10 hours. Determine (a) the rate of heat loss through the roof that night and (b) the cost of that heat loss to the home owner if the cost of electricity is $0.08/kWh.

a.1.73kW & $1.45

*b.1.69kW & $1.35

c.2kW & $2.5

Solution for (a) Q = kA

−   

Q = (0.8W/m-°C)(48m2)

−4°C .  

= 1690W = 1.69kW Solution for (b)

∆ ∆

Q = Q   = (1.69kW)(10h) = 16.9kWh Cost = (Amount of energy)(Unit cost of energy) = (16.9kWh)($0.08/kWh) = $1.3

d.1.58kW & $2

 

  4.  Consider a composite plane wall have a value shown below. Find the heat rate transfer through the wall. k1=20W/m-K k2=10W/m-K A1=1m^2 A2=1m^2 L1=L2=1m Sol.

T1=120C R1=L1/(k1xA1) =1m/(20W/m-K)(1m^2) = .05 K/W R2=L2/(k2xA2) =1m/(10W/m-K)(1m^2) =0.1 K/W Rt=R1+R2 =.05+.1 =0.15 K/W Q=(T1-T2)/Rt =(120-43)C/(0.15K/W)

T2=43C

 

Q=513.33W ans.

5.  A furnace wall is to be constructed of brick having standard dimensions 22.5 cm × 11 cm× 7.5 cm. Two kinds of material are available. One has a maximum usable temperature of 1040°C and a thermal conductivity of 1.7 W/(m K), and the other has a maximum temperature limit of 870°C and a thermal conductivity of 0.85 W/(m K). The bricks cost the same and can be laid in any manner, but we wish to design the most economical wall for a furnace with a temperature on the hot side of 1040°C and on the cold side of 200°C. If the maximum amount of heat transfer permissible is 950 W/m2 for each square foot of area, determine the most economical arrangements for the available bricks. GIVEN Furnace wall made of 22.5 cm × 11 cm × 7.5 cm bricks of two types Type 1 bricks Maximum useful temperature (T1, max) = 1040°C Thermal conductivity (k1) = 1.7 W/(m K) Type 2 bricks

Maximum useful temperature (T2, max) = 870°C

Thermal conductivity (k2) = 0.85 W/(m K) Bricks cost the same Wall hot side temperature (Thot) = 1040°C and wall cold side temperature (Tcold) = 200°C Maximum permissible heat transfer (qmax/A) = 950 W/m2

SOLUTION Q=k1(T1-T2)/L1 L1=k1(T1-T2)/Q/A L1=1.7W/m-K(1040-870)K/(950W/m^2) L1=30.42cm

Ans.

L2=k2(T1-T3)/Q/A L2=(.85W/m-K)(870-20)/950W/m^2 L2=.6m=60cm

Ans.

 

 

6.  The temperature at the inner and outer surfaces of a boiler wall made of 20 mm thick steel and covered with an insulating material of 5 mm thickness are 3000 C and 500 C respectively. If the thermal conductivities of steel and insulating material are 58W/m0C and 0.116 W/m0C respectively, determine the rate of flow through the boiler wall.

a. 5567.8 W

b. *5767.8 W c. 5878.7 W d. 5234.9 W

Given L1= 20 x 10^-3 m k1= 58 W/mC L2= 5x10^-3 K2= 0.116 W/mC T1= 300 C T2= 50 C Find: Q Solution: Q/A= Δ T/RT 

RT= R1+R2 R1= L1/K1A R1= (20 x10^-3)/ 58x1 R1= 3.45x10^-4 C/W R2= L2/K2A R2= (5 x10^-3)/ 0.116x1 R2=0.043 C/W Q/A= (300-50)/(3.45x10^-4 + 0.043) Q/A= 5767.8 W

 

 

7.  One side of a copper block 4 cm thick is maintained at 175◦C. The other side is covered with a layer of fiberglass 1.5 cm thick. The outside of the fiberglass is maintained at 80◦C, and the total heat flow through the composite slab is 300 W. What is the area of the slab? (K copper= 386 W/mC, K fiberglass= 0.038 W/mC) a.*1.247 m²

b.2.32 m²

c. 4.44 m² d. 0.99 m²

Solution: Q/A= Δ T/RT 

300W/A= (175-80)/(0.04/386)+(0.015/0.038) A= 1.247m²

 

8. A thick-walled tube of stainless steel ( 18% 19 W/m C) with 2 ccm m inner diameter (ID) and 4cm outer diameter (OD) is covered with a 3 cm layer of asbestos insulation (k 0.2 w/m-c) if the inside wall temperature of the pipe is maintained at 600°c, calculate the heat loss per meter of length, and the tube-insulation interface interface temperature. Stainless steel T1 600 C Asbestos T2 100 C a. *680 W/m b. 690 W/m c. 700 W/m d. 710 W/m Solution: Q/L= 2π (T1-T2)/ln(r2/r1)/ks+ln(r3/r2)/ka Q/L= 2π(600-100)/(ln2)/19+(ln5/2)/0.2 Q/L= 680 W/m

 

 

9.  Mild steel nails were driven through a solid wood wall consisting of two layers, each 2.5 cm thick, for reinforcement. If If the total cross-sectional ar area ea of the nails is 0.5% of the wall area, determine the unit thermal conductance of the composite wall when the temperature difference difference across the wall is 25°C. Neglect contact resistance between the wood layers. a.)7.2

c.)5.0

b.)7.1

d.)7.3

The individual resistances are Rw = Lw/AwKw = 0.05m/ (0.995Aw )[0.15W/(mK)] = 1/Aw[0.335(Km2/w)] Rs = Ls/KsAs= 0.05m/(0.005Aw[43W/mK]) = 1/Aw(0.233)Km2/W The total resistance of the wood and steel in parallel is R total = RwRs/Rw+Rs = 1/Aw{[0.335(0.233)]/[(0.335+0.233)]}=1/Aw[(0.1374 ) (K m2 )/W ] The unit thermal conductance (k/L) is: k /L = 1/Rtotal Awall = 1/[ 0.1374(Km2 )/W] = 7.3 W/(K m2)

 

 

10.  Determine the rate of heat transfer by conduction per unit area, by means of conduction for a furnace wall made of fire clay. Furnace wall thickness is 6" or half a foot. Thermal conductivity of the furnace wall clay is 0.3 W/m·K. The furnace wall temperature can be taken to be same as furnace operating temperature temperature which is 6500C and temperature of the outer wall of the furnace is 150 0C. C.  

For the given sample problem, T1= 6500C T2= 1500C L = 12" = 12 × 0.0254 m = 0.3048 m k = 0.3 W/m·K 

Q/A = k × (T1- T2)/L )/L   Q/A = 0.3×(650-150)/0.3048 W/m2= 492.13 W/m2 

A. 492.13 W/m2  B. 550 550 W/m  W/m2  C. 490 W/m2  D. 413.92 W/m2 

 

11.  The temperature at the inner and outer surfaces of a boiler wall made of 20 mm thick steel and covered with an insulating material of 5 mm thickness are 3000 C and 500 C respectively. If the thermal conductivities of steel and insulating material are 58W/m0C and 0.116 W/m0C W/m0C respectively, determine the rate of flow through the boiler wall.

A.  B.  C.  D. 

5767.8 W* 345.6 W 7658 W 4579.2 W

 

12. One side of a copper block bloc k 5 cm thick is maintained at 250 d degree egree C. The other side is covered with a layer of fiberglass 2.5 cm thick. T The he outside of the fiberglass is maintained at 35 degree C, and the total heat flow through the copper-fiberglass combination is 44 kW. What is the area of the slab?(Kcopper= 386 W/mC, Kfiberglass= 0.038 W/mC) W/ mC)

a. 150.6 m² b. 143.7 m² c.*134.7 m² d. 135.6 m²

Solution: Q/A= Δ T/RT 

44 000W/A= (250-35)/(.05/386)+(.025/.038) A= 134.7 m² answer

13. An outside wall for a building consists of a 10-cm 10 -cm layer of common brick and a 2.5-cm layer of fiberglass (k= .05 W/mC). Calculate the heat flow through the wall for a 75 degree C temperature difference.(kbrick= 0.69 W/mC) a. 39.81 W/m² b. 40.5 W/m² c.*38.76 W/m² d. 37.8 W/m² Solution: Q/A= Δ T/RT 

Q/A= 75/(0.10/0.69)+(0.025/0.05) Q/A= 38.76 W/m²

14. Determine the heat transfer through the composite wall shown in the figure below. Take the conductives of A, B, C, D & E as 50, 10, 6.67, 20& 30 W/mK respectively and assume one dimensional heat transfer. Take of area of A =D= E = 1m2 and B=C=0.5 m2. Temperature entering at wall A is 800 o C and leaving at wall E is 100 o C

 

*a. 3137.61 W b. 2368.92 W c. 4263.78 W d. 3992.67 W

 

 

15. A 5 cm layer of loosely packed asbestos is placed between two plates at 100 and 200 degree celcius. Calculate the heat transfer across the layer. K=0.161 W/m-◦C A. 231 W/m2 B. 432 W/m2 *C.322 W/m2 

solution

D.329 W/m2

  0.161200100     = 0.05 = 322 W/m2

 

16.  What is the rate of heat flow through a glass window that is 2m x 3m and 1.4 cm thick if the outside temperature is 22◦C and the indide temperature is 25 ◦C? k= 0.84W / m-◦C  A. 1000 W C. 1170 W

*B.1080 W D. 990 W

Solution

  ()3○     .   = .4   Q= 1080 W

.

17.  A 30 ft. long and 2 inch nominal diameter and outside diameter of  

2.375 in pipe is covered by   a 1 inch thick insulation (k = 0.0375 Btu/hr.ft.°F). Determine  the heat loss if the inner and outer temperatures of the   insulation are 380°F and 80 F, respectively. a. 3245 Btu/hr   b. 2894 Btu/hr   c. 3480 Btu/hr d. 4100 Btu/hr  

Solution: 

inside diameter of the insulation = 2.375 in..  outer diameter of the insulation = 2 + 2.375 = 4.375 in. L = length of pipe  Q/L = [{2πk(t1 – t2)} / {ln(r1 / r2)}] = [{2πk(t1 – t2)} / {ln(D1 / D2)}]  where D1 and D2 are the inner and outer diameters.   Substituting the values in the equation for Q/L   Q/L = [{2π 0.0375(380 – 80)} / {ln[{(4.375) / (12)} / {(2.375) / (12)}]}]  

 

 

= 116 Btu/hr ft  This is the heat loss for 1 foot length of the pipe. Therefore, for   30 feet of pipe, the heat loss is   q = 116 × 30 = 3480 Btu/hr

18.  Calculate the rate of heat loss through the vertical walls of a boiler furnace of size 4 m by 3 m by 3 m high. The walls are constructed from an inner fire brick wall 25 cm thick of thermal conductivity 0.4 W/mK, a layer of ceramic blanket insulation of thermal conductivity 0.2 W/mK and 8 cm thick, and a steel protective layer of thermal conductivity 55 W/mK and 2 mm thick. The inside temperature of the fire brick layer was measured at 600o C and the temperature of the outside of the insulation 600 C

A) 6982.94 W

B) 4892.8

C) 6320.96 W

D) 1437.44

Given: Composite Wall l= 4m b= 3m h= 3m Area of rectangular wall lb = 4x3 = 12m^2 t1 = .25 m

Fire brick k1= 0.4 W/mK

t2 =0.002m

Steel

t3 = 0.08 m

insulation k3 = 0.2 W/mK

T1 = 6000 C

T4 = 600 C

k2 = 54 W/mK

We know that,

Q= (T1-T2 ) /RT R 1= t1/ (k1) (A) = .25m/(0.4 W/m.K) (12m^2) = 0.0521K/W R 2 = t2/(k2)(A) = (.002m)/(54 w/mk)(12m^2) = 0.0333K/W R 3= t3/(k3)(A)= (0.08m)/(0.2w/ mk)(12m^2) = 0.0000031 K/W R T= R 1+ R 2 + R 3

 

Q= (6000-600)k/(0.0521 k/W + 0.0333 k/W + 0.0000031 k/W) Q= 6320.96 W

-ANS

19. Calculate insulation thickness (minimum value) required for a  pipe carrying steam at 1800C. The pipe size is 8" and the maximum allowable temperature of outer wall of insulation is 500C. Thermal conductivity of the insulation material for the temperature range of the pipe can be taken as 0.04 W/m·K. The heat loss from steam per meter of pipe length has to be limited to 80 W/m.

For the given sample problem, T1 = 500C T2 = 1800C r1 = 8" = 8 × 0.0254 m = 0.2032 m k = 0.04 W/m·K N = length of the cylinder Q/N = Heat loss per unit length of pipe Q/N = 80 W/m Sol’n: 

80 = 2π × 0.04 × (180-50) ÷ ln(r 2/0.2032) ln(r 2/0.2032) = 2π × 0.04 × (180-50) / 80 = 0.4084 Hence, r 2/= r 1 × e0.4084  r 2/= 0.2032 × 1.5044 = 0.3057 m Hence, insulation thickness = r 2 - r 1  thickness = 305.7 - 203.2 = 102.5 mm

ANS.

 

20.  A steel pipe soaked in water 10cm below having an inside diameter of 6cm and outside diameter of 8cm. Given heat per length is at 600w/m; Ks = 50w/m - °C. Find the temperature in a pipe when water starts to boil kw at 20°C. GIVEN: r₁ = 3cm

Tout = 100°C

r₂ = 8cm 

Tin = ?

r₃ = 10cm + 8cm = 18cm

Q/L = 600w/m

Q/L = ΔT/ Rt

600 =Tin – 100_______________ In (0.18/0.08) + In (0.08/0.03) 2  (0.6) 2  (50)





Tin==230.93°C 230.93°C Tin

21. Heat flows inside a cold storage by conduction. The amount of heat flowing is 250 kJ in one hour. Compute the amount of heat flowing to the storage in 3 hours, assuming conditions are the same. Solution: Q=

  x ℎ

3hr =750 kJ ANS.

22. A room is 10 feet long, 12 feet wide and 8 feet high. The inside and outside temperature are 8 degrees F and 65 degree, respectively. Compute the heat transmitted through the walls if the coefficient of the heat transmission is 0.12 BTU/square-degree F-hour.

Solution: Area, A= 10(8)(2)= 336 ft Q=

Δ .33−8  = 4596.48 Btu/hr   = .

ANS.

 

 

23. A 48 ft² wall has an average thermal conductivity of 1.2 Btu-ft/ft²-°F-hr. If the temperature difference between two sides is 18°F and thickness is 0.7 ft, what is the total heat loss in 5 hours? Solution: Q=

Δ .488  = 1481.14 Btu/hr   = .7

In 5 hours: Q= 1481.14(5)= 7405.7 Btu Btu ANS.

24. A tank containing a liquid nitrogen at -120°C is suspended in a vacuum shell by three stainless steel rods 0.9 cm in diameter and 3 meters long with a thermal conductivity of 18.3 W/m-°C. If the ambient air outside the vacuum shell is 28°C, calculate the magnitude of the conductive condu ctive heat flow in watts along the supportive rods. Solution: Q= hA(T2-T1)

 4

Q= 18.3[  (  (0.009)²][28-(-120)]= 0.009)²][28-(-120)]= 0.172W ANS.

25. One side of a 40 cm stock wall (k= 1.5 W/m-k) is exposed to the air whose temperature is 18°C (h= 15 W/m²-K) and the other side is exposed to the

 

furnace gases whose temperature is 900°C (h= 15 W/m²-K). Estimate the heat transfer through the wall per m² area.

Solution:

T − T  Δ X + K X   +  K X  Q=  = K  9 − 8   = 2205 W   =   +. .  +