Tải bản đầy đủ (.pdf) (1 trang)

Truyền khối (bài tập) Các quá trình Truyền Khối

Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (10.82 KB, 1 trang )

019:006 Exercises on heat transfer by thermal radiation Solutions

1. In outer space heat transfer by convection and conduction cannot occur. How is
thermal energy removed from a spacecraft?
The only mechanism for heat removal on a spacecraft is by radiation.

2. You remove a roasted chicken from the oven but will not be ready to serve it for 15
minutes. What can you do to keep it from getting cold too fast?
The chicken should be covered immediately with aluminum foil. The rate of heat loss
by radiation is proportional to T
4
. When it is hottest it radiates at its highest rate, so
this heat loss must be reduced as soon as possible.

3. Often when cooking a pie, the edges of the pie crust will get brown before the rest of
the crust. What can be done to prevent the edges of the pie crust from burning?
You should wrap the edges of the pie crust with Al foil which will reflect the heat.

4. If you want to keep an object cool, is it better to paint it with silver paint or black
paint?
You want to allow it to radiate as much heat as possible and dark (black) objects
radiate more than light objects

5. Two objects have the same temperature and the same area but object A has an
emissivity e
A
= 0.2 and object B has an emissivity e
B
= 0.8. How much radiation is
emitted per second from object B compared to object A?
The rate of radiation emission is proportional to the emissivity, e. So Q


A
~ e
A
, and Q
B

~ e
B
. Therefore Q
B
/Q
A
= 0.8/0.2 = 4.

6. Two objects have the same temperature and the same emissivities but object A has an
area of 2 m
2
and object B has an area of 10 m
2
. How much radiation is emitted per
second from object A compared to object B?
The rate of radiation emission is ~ Area of the object. Therefore, Q
A
/Q
B
= A
A
/A
B
=

10/2 = 5.

7. Two objects have the same areas and the same emissivities but object A has a
temperature of 150 K and object B has a temperature of 450 K. How much radiation
is emitted per second from object B compared to object A?
Q ~ T
4
, so that Q
B
/Q
A
= (T
B
)
4
/(T
A
)
4
= (450/150)
4
= (3)
4
= 81.

×