Different examples of heat transfer Convection heat transfer by hot metal plate.
It is well known that a hot plate of metal will cool faster (figure blow ) when placed in front of a fan than when exposed to still air, and heat is convected away, the process call convection heat transfer. For example, we know that the velocity at which the air blows over the hot plate obviously influences the heat-transfer rate. the temperature gradient is dependent on the rate at which the fluid carries the heat away; a high velocity produces a large temperature gradient, and so on. Newton’s law of cooling:
q = hA (Tw - T?)
Here the heat-transfer rate is related to the overall temperature difference between the wall and fluid and the surface area A. the quantity h (W/m2.K) is called the convection heat-transfer coefficient.
Example 1 Air at 20? blows over a hot plate 50 by 75 cm maintained at 250?. The convection heat-transfer coefficient is 25 W/m 2•?. Calculate the heat transfer.
Solution. From Newton’s law of cooling
q = hA(Tw-T?)
= (25)(0.50)(0.75)(250-20)
= 2.156 kW? ?[7356 Btu/h]
Example 2 Assuming that the plate in Ex. 1 is made of carbon steel (1%) 2 cm thick and that 300 W is lost from the plate surface by radiation, calculate the inside plate temperature.
Given , the thermal conductivity 43 Kw / M . ?
Solution. The heat conducted through the plate must be equal to the sum of convection and radiation heat losses:
Q cond = q conv + q rad
-kA = 2.156 + 0.3 = 2.456 kW
= = -3.05?? ?[-5.49?]
Ti = 250 + 3.05 = 253.05 ?
1- Heat Loss from a Person
Consider a person standing in a breezy room at 20°C. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m2 and 29°C, respectively, and the convection heat transfer coefficient is 6 W/m2 • °C (Fig. blow).
SOLUTION
The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at specified temperatures is to be determined.
Assumptions
1 Steady operating conditions exist. 2 The person is completely surrounded by the interior surfaces of the room. 3 The surrounding surfaces are at the same temperature as the air in the room. 4 Heat conduction to the floor through the feet is negligible.
Properties
The emissivity of a person is = 0.95 .
Analysis
The heat transfer between the person and the air in the room will be by convection (instead of conduction) since it is conceivable that the air in the vicinity of the skin or clothing will warm up and rise as a result of heat transfer from the body, initiating natural convection currents. It appears that the experimentally determined value for the rate of convection heat transfer in this case
is 6 W per unit surface area (m2) per unit temperature difference (in K or °C) between the person and the air away from the person. Thus, the rate of convection heat transfer from the person to the air in the room is
and the net rate of radiation heat transfer from the person to the surrounding walls, ceiling, and floor is
the rate of total heat transfer from the body is determined by adding these two quantities:
2- Heat Transfer between Two Isothermal Plates
Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 300 K and T2 = 200 K that are L = 1 cm apart, as shown in (Fig. Blow). Assuming the surfaces to be black (emissivity ), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, (b) evacuated, (c) filled with urethane insulation, and (d) filled with superinsulation that has an apparent thermal conductivity of 0.00002 W/m • °C.
SOLUTION
The total rate of heat transfer between two large parallel plates at specified temperatures is to be determined for four different cases.
Assumptions
1 Steady operating conditions exist. 2 There are no natural convection currents in the air between the plates. 3 The surfaces are black and thus .
Properties
The thermal conductivity at the average temperature of 250 K is k = 0.0219 W/m • °C for air 0.026 W/m • °C for urethane insulation , and 0.00002 W/m • °C for the superinsulation.
Analysis
(a) The rates of conduction and radiation heat transfer between the plates through the air layer are
And
Therefore ;
(b) When the air space between the plates is evacuated, there will be no conduction
or convection, and the only heat transfer between the plates will be by radiation. Therefore,
(c)An opaque solid material placed between two plates blocks direct radiation heat transfer between the plates. Also, the thermal conductivity of an insulating material accounts for the radiation heat transfer that may be occurring through the voids in the insulating material. The rate of heat transfer through the urethane insulation is
(d ) The layers of the superinsulation prevent any direct radiation heat transfer between the plates. However, radiation heat transfer between the sheets of superinsulation does occur, and the apparent thermal conductivity of the superinsulation accounts for this effect. Therefore,
المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .
|