Consider a 20-cm-thick large concrete plane wall (k = 0.77 W/mK) subjected to
convection on both sides with T∞1 = 27°C and h1 = 5 W/m2 K on the inside, and
T∞2 = 8°C and h2 = 12 W/m2 K on the outside. Assuming constant thermal
conductivity with no heat generation and negligible radiation,
(a) Express the differential equation and the boundary conditions for steady one-dimensional
heat conduction through the wall,
(b) Obtain a relation for the variation of temperature in the wall by solving the
differential equation,
(c) Evaluate the temperatures at the inner and outer surfaces of the wall
About heat transfer (熱傳)?
2017-12-05 3:43 pm
回答 (1)
2017-12-09 11:55 am
(a) Let Q be the amount of heat conducted through unit cross-sectional area of the wall in unit time.
Then, Q = -k(dT/dx) -------------- (1)
where k is the thermal conductivity of the wall (= 0.77 W/mK)
T is the temperature at distance x in the wall.
x is the distance from the inside surface of the wall.
[The -ve sign indicates that heat flows through the wall from the inside to outside, i.e. there is heat loss.]
The boundary conditions are:
At x = 0 m, T = T1, the inside surface temperature of the wall
At x = 0.2 m, T = T2, the outside surface temperature of the wall.
(b) From (1), dT = -(Q/k)dx
i.e. T = -integral { (Q/k).dx }, with the limits of integration from 0 to x
since Q/k is a constant, we have, T = -(Q/k). integral {dx}
i.e. T = -(Q/k)x
That means the temperature decreases linearly with distance from the inside surface of the wall.
(c) On the inside surface, Q = 5.(27 - T1) --------- (2)
On the outside surface, Q = 12.(T2 - 8) ---------- (3)
Equating (2) and (3), and solve for T2. This gives
T2 = 19.25 - 0.4167(T1) ------------- (4)
Consider heat conduction through the wall,
Q = 0.77(T1 - T2)/0.2 ------------ (5)
Equating (3) and (5),
12(T2 - 8) = 0.77(T1 - T2)/0.2 ---------- (6)
Substitute T2 from (4) into (6) and solve for T1, we have,
T1 = 20 C
hence, substitute T1 = 20 C into (4), we have T2 = 10.92 C
Then, Q = -k(dT/dx) -------------- (1)
where k is the thermal conductivity of the wall (= 0.77 W/mK)
T is the temperature at distance x in the wall.
x is the distance from the inside surface of the wall.
[The -ve sign indicates that heat flows through the wall from the inside to outside, i.e. there is heat loss.]
The boundary conditions are:
At x = 0 m, T = T1, the inside surface temperature of the wall
At x = 0.2 m, T = T2, the outside surface temperature of the wall.
(b) From (1), dT = -(Q/k)dx
i.e. T = -integral { (Q/k).dx }, with the limits of integration from 0 to x
since Q/k is a constant, we have, T = -(Q/k). integral {dx}
i.e. T = -(Q/k)x
That means the temperature decreases linearly with distance from the inside surface of the wall.
(c) On the inside surface, Q = 5.(27 - T1) --------- (2)
On the outside surface, Q = 12.(T2 - 8) ---------- (3)
Equating (2) and (3), and solve for T2. This gives
T2 = 19.25 - 0.4167(T1) ------------- (4)
Consider heat conduction through the wall,
Q = 0.77(T1 - T2)/0.2 ------------ (5)
Equating (3) and (5),
12(T2 - 8) = 0.77(T1 - T2)/0.2 ---------- (6)
Substitute T2 from (4) into (6) and solve for T1, we have,
T1 = 20 C
hence, substitute T1 = 20 C into (4), we have T2 = 10.92 C
收錄日期: 2021-04-11 21:48:56
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