4ChEB Group 6

Allan Gabriel Carasco Roent Dune Cayetano

Patrick Cristobal Nikko Zandro Reyes

Aaron Turingan

4.2-2 Heat Removal of a Cooling CoilA cooling coil of 1.0 ft of 304 stainless-steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40F and I 80F on the outside. The thermal conductivity of 304 stainless steel is a function of temperature:

k=7.75+7.78E-3T where k is in btu/h*ft*F and T is in F. Calculate the heat removal in btu/s and watts.

Given: k=7.75+7.78E-3T Din= 0.25in =0.0208ft

kin=8.0612btu/h*ft*F

Dout= 0.40in =0.0333ftkout=8.3724btu/h*ft*F

Sol’n:

**=1308W

5.3-9 Temperature of Oranges on Trees During Freezing Weather. In orange-growing areas, the freezing of the oranges on the trees during cold noghts is of serious economic concern. If the oranges are initially at a temperature of 21.1C, calculate the center temperature of orange is exposed to air at -3.9C for 6 hours. The oranges are 102 mm in diameter and the convective coefficient is estimated as 11.4W/m2-K. The thermal conductivity k is 0.431 W/m-K and α is 4.65E-4m2/h. Neglect any latent heat effects.

Given:k = 0.431 W/m-K α = 4.65E-

4m2/hTo=21.1C T1=-3.9C T(center)=?

Solution: assume m=0

From the graph we get value of Y=0.05

T at center = -2.65C

Problem #6A flat slab of rubber, 2.5 cm thick initially at 20°C is to be placed between two heated steel plates maintained at 140°C. The heating is to be discontinued when the temperature at the mid plane of the slab is 132°C. The rubber has a thermal conductivity of 0.16 W/mK and thermal diffusivity of 8.671x10-6 m2/s. Thermal resistance from metal to rubber may be neglected. Calculate: A) The length of the heating period, sec B) The temperature of the rubber 0.65cm

from the metal C) The time required for the rubber to reach

132.°C at the plane specified at b).

Given:k=0.16W/m-K α=8.671E-6 m2/sTo=20C T1=140C T(center)=132C

a)Time @ T(center)=132C, assume m=0

using the Gurney graph for plate we get value of X=1.3

b) Temp @ 0.65cm from metalGiven:

x=1.25-0.65 =0.6cm t=23.43 secn=0.6/1.25 =0.48 X=1.3

From the graph we get a value of Y=0.037

c) Time when the Temp @ 0.65cm from the metal is 132C.

Given:m=0 n=0.48 Y=0.0667

We get the value X=1.1 from the graph