Examples for wall construction let . 11
“Two-by-four” wood studs have actual dimensions of 4.13×9.21 cm and a thermal conductivity
of 0.1W/m. ?C. A typical wall for a house is constructed as shown Figure blow. Calculate
the overall heat-transfer coefficient and R value of the wall.
(a) Construction of a dwelling wall; (b) thermal resistance model.
Example for Cooling Cost Savings with Extra Insulation:
Asmall metal building is to be constructed of corrugated steel sheet walls with a total wall surface
area of about 300m2. The air conditioner consumes about 1kW of electricity for every 4kW of cooling supplied. Two wall constructions are to be compared on the basis of cooling costs. Assume that electricity costs $0.15/kWh. Determine the electrical energy savings of using 260 mm of fiberglass batt insulation instead of 159 mm of fiberglass insulation in the wall. Assume an overall temperature difference across the wall of 20?C on a hot summer day in Texas.
Solution :
The overall heat transfer coefficients for the two selected wall constructions are
U (260-mm fiberglass) =0.17W/m2 ? ?C
U (159-mm fiberglass) =0.31W/m2 ? ?C
The heat gain is calculated from q=UA?T, so for the two constructions
q (260-mm fiberglass) = (0.17)(300)(20)=1020W
q (159-mm fiberglass) = (0.31)(300)(20)=1860W
Savings due to extra insulation=840W
The energy consumed to supply this extra cooling is therefore
Extra electric power required= (840)(1/4)=210W and the cost is
Cost =(0.210kW)(0.15$/kWh)=0.0315 $/hr
Assuming 10-h/day operation for 23 days/month this cost becomes
(0.0315)(10)(23)=$7.25/month
Both of these cases are rather well insulated. If one makes a comparison to a 2×4 wood stud wall
with no insulation fill in the cavity (U =1.85W/m2 . ?C), the heating load would be
q = (1.85)(300)(20)=11,100W and the savings compared with the 260-mm fiberglass insulation would be 11,100 ?1020=10,080W
Producing a corresponding electric power saving of $0.378/h or $86.94/month. Clearly, the insulated wall will pay for itself. It is a matter of conjecture whether the 260-mm of insulation will
pay for itself in comparison to the 159-mm insulation.
Example for Overall Heat-Transfer Coefficient for a Tube :
Water flows at 50?C inside a 2.5-cm-inside-diameter tube such that hi =3500 W/m2. ?C. The
tube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m.?C. The outside of
the tube loses heat by free convection with ho =7.6W/m2 ? ?C. Calculate the overall heat-transfer
coefficient and heat loss per unit length to surrounding air at 20?C.
Solution
There are three resistances in series for this problem, as illustrated in Equation (59). With L=1.0 m, di =0.025 m, and do =0.025+(2)(0.0008)=0.0266 m, the resistances may be calculated as
Clearly, the outside convection resistance is the largest, and overwhelmingly so. This means that it
is the controlling resistance for the total heat transfer because the other resistances (in series) are
negligible in comparison. We shall base the overall heat-transfer coefficient on the outside tube
area and write
(a)
or a value very close to the value of ho =7.6 for the outside convection coefficient. The heat
transfer is obtained from Equation (a), with
q=UAo ?T =(7.577)?(0.0266)(1.0)(50?20)=19 W (for 1.0 m length)
CRITICAL THICKNESS OF INSULATION
Let us consider a layer of insulation which might be installed around a circular pipe, as
shown in Figure 18. The inner temperature of the insulation is fixed at Ti, and the outer
Figure 18 Critical insulation thickness
surface is exposed to a convection environment at T?. From the thermal network the heat
transfer is
(62)
Now let us manipulate this expression to determine the outer radius of insulation ro, which
will maximize the heat transfer. The maximization condition is
which give the result (63)
Equation (63) expresses the critical-radius-of-insulation concept. If the outer radius is less
than the value given by this equation, then the heat transfer will be increased by adding more
insulation. For outer radii greater than the critical value an increase in insulation thickness
will cause a decrease in heat transfer. The central concept is that for sufficiently small values
of h the convection heat loss may actually increase with the addition of insulation because
of increased surface area.
Example for the Critical Insulation Thickness
Calculate the critical radius of insulation for asbestos [k =0.17 W/m. ?C] surrounding a pipe
and exposed to room air at 20?C with h=3.0 W/m2 ? ?C. Calculate the heat loss from a 200?C,
5.0-cm-diameter pipe when covered with the critical radius of insulation and without insulation.
Solution
From Equation (63) we calculate ro as
The inside radius of the insulation is 5.0 / 2=2.5 cm, so the heat transfer is calculated from Equation (62) as
Without insulation the convection from the outer surface of the pipe is
So, the addition of 3.17 cm (5.67?2.5) of insulation actually increases the heat transfer by
25 percent. As an alternative, fiberglass having a thermal conductivity of 0.04 W/m? ?C might be employed as the insulation material. Then, the critical radius would be
Now, the value of the critical radius is less than the outside radius of the pipe (2.5 cm), so addition
of any fiberglass insulation would cause a decrease in the heat transfer. In a practical pipe insulation problem, the total heat loss will also be influenced by radiation as well as convection from the outer surface of the insulation.
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 .
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