Resoluções - Transcal

Resoluções - Transcal

(Parte 1 de 11)

PROBLEM 3.1 respectively.

and L.

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties, (4) Negligible radiation, (5) No generation.

ANALYSIS: For the foregoing conditions, the general solution to the heat diffusion equation is of the form, Equation 3.2,

The constants of integration, C1 and C2, are determined by using surface energy balance conditions at x = 0 and x = L, Equation 2.23, and as illustrated above,

dTdTkhTT0khTLT.
−=−−=−(2,3)

dt dx∞∞ For the BC at x = 0, Equation (2), use Equation (1) to find

Multiply Eq. (4) by h2 and Eq. (5) by h1, and add the equations to obtain C1. Then substitute C1 into Eq. (4) to obtain C2. The results are

CCT

1 L 1 1 Lkh h k h h k

TTx1Tx T. kh11 L h k

From Fourier’s law, the heat flux is a constant and of the form

TTdTqkk C. dx 1 L

h k

PROBLEM 3.2

KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces of a rear window.

FIND: (a) Inner and outer window surface temperatures, Ts,i and Ts,o, and (b) Ts,i and Ts,o as a function of the outside air temperature T∞,o and for selected values of outer convection coefficient, ho.

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation effects, (4) Constant properties.

PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.1 and 3.12,

22oi h k h 1.4W m K65W m K 30W m K

Hence, with ()i,i,oqhTT∞∞ ′′=−, the inner surface temperature is

Similarly for the outer surface temperature with ()os,o,oqhTT∞ ′′=− find

(b) Using the same analysis, Ts,i and Ts,o have been computed and plotted as a function of the outside air temperature, T∞,o, for outer convection coefficients of ho = 2, 65, and 100 W/m2⋅K. As expected, Ts,i and

Ts,o are linear with changes in the outside air temperature. The difference between Ts,i and Ts,o increases with increasing convection coefficient, since the heat flux through the window likewise increases. This difference is larger at lower outside air temperatures for the same reason. Note that with ho = 2 W/m2⋅K,

Ts,i - Ts,o, is too small to show on the plot. Continued …..

PROBLEM 3.2 (Cont.)

Surface temperatures, Tsi or Tso (C)

Tsi; ho = 100 W/m^2.K Tso; ho = 100 W/m^2.K Tsi; ho = 65 W/m^2.K Tso; ho = 65 W/m^2.K Tsi or Tso; ho = 2 W/m^.K

COMMENTS: (1) The largest resistance is that associated with convection at the inner surface. The values of Ts,i and Ts,o could be increased by increasing the value of hi.

(2) The IHT Thermal Resistance Network Model was used to create a model of the window and generate the above plot. The Workspace is shown below.

// Thermal Resistance Network Model: // The Network:

// Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 q43 = (T4 - T3) / R43

// Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 + q43 = 0 q4 - q43 = 0

/* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinfo// Outside air temperature, C //q1 =// Heat rate, W T2 = Tso// Outer surface temperature, C q2 = 0// Heat rate, W; node 2, no external heat source T3 = Tsi// Inner surface temperature, C q3 = 0// Heat rate, W; node 2, no external heat source T4 = Tinfi// Inside air temperature, C //q4 =// Heat rate, W

// Thermal Resistances:

R21 = 1 / ( ho * As )// Convection thermal resistance, K/W; outer surface R32 = L / ( k * As )// Conduction thermal resistance, K/W; glass R43 = 1 / ( hi * As )// Convection thermal resistance, K/W; inner surface

// Other Assigned Variables:

Tinfo = -10// Outside air temperature, C ho = 65// Convection coefficient, W/m2.K; outer surface L = 0.004// Thickness, m; glass k = 1.4// Thermal conductivity, W/m.K; glass Tinfi = 40// Inside air temperature, C hi = 30// Convection coefficient, W/m2.K; inner surface As = 1// Cross-sectional area, m2; unit area

PROBLEM 3.3

KNOWN: Desired inner surface temperature of rear window with prescribed inside and outside air conditions.

FIND: (a) Heater power per unit area required to maintain the desired temperature, and (b) Compute and plot the electrical power requirement as a function of ,oT∞ for the range -30 ≤ ,oT ∞ ≤ 0°C with ho of 2,

20, 65 and 100 W/m2⋅K. Comment on heater operation needs for low ho. If h ~ Vn, where V is the vehicle speed and n is a positive exponent, how does the vehicle speed affect the need for heater operation?

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Uniform heater flux, hq′′, (4) Constant properties, (5) Negligible radiation effects, (6) Negligible film resistance.

PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K.

ANALYSIS: (a) From an energy balance at the inner surface and the thermal circuit, it follows that for a unit surface area,

,i s,i s,i ,o h

T T T q

1h L k 1h

oi 2

exterior convection coefficients is shown in the plot. When ho = 2 W/m2⋅K, the heater is unecessary, since the glass is maintained at 15°C by the interior air. If h ~ Vn, we conclude that, with higher vehicle speeds, the exterior convection will increase, requiring increased heat power to maintain the 15°C condition.

COMMENTS: With hq′′ = 0, the inner surface temperature with ,oT∞ = -10°C would be given by

,i s,i i ,i ,o i o

(Parte 1 de 11)

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