4bgroup1-1231161941459375-2(1)

Group 14ChE BAlmazar, Karina

Ascalon, Kristine

Malana, Rebecca

Rozario, Vera Cresta

Tancio, Romar

Problems1.) A steam pipe 5 cm outside diameter has an

outside surface temperature of 175 °C. The pipe is covered with a coating material 5 cm, thick. The thermal conductivity of the coating varies with the temperature such that, k= 0.89 + 0.0015T where T is in °C and k in W/mK. The outside surface of the coating is at 38 °C. Calculate the heat loss per meter of pipe length.

SolutionGiven: D= 5cm L= 1m T1= 175°C T2= 38°C

k2= 0.89 + 0.0015T

K2 = 0.89 + 0.0015(38°C) = 0.947 W/mK

Do = 0.05 + 0.05(2) = 0.15 m

Di = 0.05 m

q=

Problems9.) A small electric furnace is 15 by 15 by 30 cm

inside dimensions and has fire-brick walls (k= 1.12 W/mK) 2m thick. The front of the furnace is movable wall which permits entry into the furnace. In this section is by a 5 by 5 by 0.6 cm quartz observation windows (k= 0.07 W/mK). The inner temperature for all sides is 1100 °C, and the outer surface temperature is 121°C. Assuming all joints perfectly made and neglecting the influence of the corners on the temperature distribution, what is the heat loss from this furnace?.

Solution

1

2

4

53

0.15 m

2m

0.30m

GIVEN:q

T=11000C

T=1210C

3 is .05 x .05 x .006

Solution

Solution

k= 1.12 W/mK

R3

Solution

Problems5.2-1. Temperature Response in Cooling a Wire. A

small copper wire with a diameter of 0.792 mm and initially at 366.6 K is suddenly immersed in a liquid held constant at 311 K. The convection coefficient h=85.2W/m2. K. The physical properties can be assumed constant and are k=372 W/m.K , Cp=0.389 kJ/kg.K ρ = 8890kg/m3

a. Determine the time in seconds for the average temperature of the wire to drop to 338.8K (one-half the initial temperature difference).

b. Do the samew but =11.36W/m2.K.

c. For part (b), calculate the total amount of heat removed for a wire 1.0 m long.

Given:

D=0.792mm T1=366.5 K T2 =311K h=85.2W/m.K

K=374 W/m.K Cp=0.389 kJ/kg.K ρ = 8890kg/m3

a.

T’ = 338.8 K

X1 = D/4 = (0.792/4) x (1/1000) = 1.98x10-4

NBi= hX1 /K = (85.2 x 1.98x10-4 )/ 374 = 4.51x10-5 <<0.1

Solution

Lumped capacity method can be used

t = 5.56 sec

Solution

b.)

if h = 11.36W/m.K

t = 41.65 sec

Solution

t = 41.65 sec

Solution L = 1.0m

q = (0.398x 1000) (8890) (4.927X-10-7) (366.5-311)

(1-e-(0.0166)(41.65) )

q = 47.20 J

4.3-4. Heat Loss from Steam Pipeline. A steel pipeline, 2-in. schedule 40 pipe, contains saturated steam at 121.1 0C. The line is covered with 25.4 mm of insulation. Assuming that the inside surface temperature of the metal wall is at 121.1 0C and the outer surface of insulation is at 26.7 0C , calculate the heat loss for 30.5 m of pipe. Also, calculate the kg of steam condensed per hour in the pipe due to the heat loss. The average k for steel from Appendix A.3 is 45 W/m. K and the k for the insulation is 0.182.

Problems

Given: 2” Sch 40 steel pipeL = 30.5m Ks = 45W/m.K

Insulation thickness = 25.4mm Ki = 0.182 W/m.K

T1 = 121.1C T2 = 26.7C

a.)

2” Sch 40:

Do = 2.375” (0.0254) = 0.0603m

Di = 2.067” ((0.0254) =0.0525m

X = 0.154” (0.0254) =3.9116 x 10-3m

Solution

Solution

Di = 0.0603m

Di = 0.0603m