• No results found

SKETCH: Figure 1.5(a) shows a solder-ball attachment of a microprocessor to a printed circuit board.

N/A
N/A
Protected

Academic year: 2022

Share "SKETCH: Figure 1.5(a) shows a solder-ball attachment of a microprocessor to a printed circuit board."

Copied!
7
0
0

Loading.... (view fulltext now)

Full text

(1)

PROBLEM 1.5.FAM GIVEN:

The attachment of a microprocessor to a printed circuit board uses many designs. In one, solder balls are used for better heat transfer from the heat generating (Joule heating) microprocessor to the printed circuit board.

This is shown in Figure Pr.1.5(a).

SKETCH:

Figure 1.5(a) shows a solder-ball attachment of a microprocessor to a printed circuit board.

Heat Sink or Coverplate

Solder Balls

Microprocessor

Physical Model of Microprocessor and Circuit Board with Solder Balls

Adhesive

Thermal Adhesive

Printed Circuit Board Se,J

Figure Pr.1.5(a) Solder-ball connection of microprocessor to the printed circuit board.

OBJECTIVE:

Track the heat flux vector from the microprocessor to the heat sink (i.e., bare or finned surface exposed to moving, cold fluid) and the printed circuit board.

SOLUTION:

Figure Pr.1.5(b) shows the heat flux vector starting from the microprocessor. Within the solid phase, the heat transfer is by conduction. From the solid surface to the gas (i.e., air), the heat transfer mechanism is surface radiation. If the gas is in motion, heat is also transferred by surface convection.

qr

qu

qr

qku qk qk

Se,J

qu

qr

qu

qr

qku

qu

qr

qku

qk

qu

qu

qr

qr

qku

qk

qk

qk

qk

qu

qr

qku

qk

qk

qk

qk

qk

qk

qr

qr

qu

qu qu

qu qku

qku

qu

Figure Pr.1.5(b) Heat flux vector tracking in microprocessor and its substrate.

COMMENT:

If the heat generation ˙Se,J is large, which is the case for high performance microprocessors, then a heat sink (e.g., a finned surface) is needed. We will address this in Section 6.8.

(2)

PROBLEM 1.8.FAM GIVEN:

A bounded cold air stream is heated, while flowing in a tube, by electric resistance (i.e., Joule heating). This is shown in Figure Pr.1.8(a). The heater is a solid cylinder (ceramics with the thin, resistive wire encapsulated in it) placed centrally in the tube. The heat transfer from the heater is by surface convection and by surface-radiation emission (shown as ˙Se,ǫ). This emitted radiation is absorbed on the inside surface of the tube (shown as ˙Se,α) and then leaves this surface by surface convection. The outside of the tube is ideally insulated. Assume that no heat flows through the tube wall.

SKETCH:

Figure Pr.1.8(a) shows the tube, the air stream, and the Joule heater. The surface radiation emission and absorption are shown as ˙Se,ǫ and ˙Se,α, respectively.

???

yyy

Tube Wall

Bounded Cold

Air Stream In Bounded Hot

Air Stream Out Ideal Insulation (No Heat

Flows in Tube Wall)

Electric Resistance Heating (Thin Wires Encapsulated in Ceramic Cover)

Se,a Se,J

Se,'

Figure Pr.1.8(a) A bounded air stream flowing through a tube is heated by a Joule heater placed at the center of the tube.

OBJECTIVE:

Draw the steady-state heat flux tracking showing the change in fluid convection heat flux vector qu, as it flows through the tube.

SOLUTION:

Figure Pr.1.8(b) shows the inlet fluid convection heat flux vector qu entering the tube. The heat transfer by surface convection qkufrom the heater contributes to this convection heat flux vector. The surface radiation emission from the heater is absorbed by the inner surface of the tube. Since no heat flows in the tube wall, this heat leaves by surface convection and further contributes to the air stream convection heat flux vector. These are shown in Figure Pr.1.8(b).

???

yyy

No Heat Flow

(Ideal Insulation) Se,a (Surface Radiation Absorption)

Se,J

qu

qku

Se, (Surface Radiation Emission)

qk

qku

qu

qr

'

Figure Pr.1.8(b) Tracking of heat flux vector.

COMMENT:

In practice, the heater may not be at a uniform surface temperature and therefore, heat flows along the heater.

The same may be true about the tube wall. Although the outer surface is assumed to be ideally insulated, resulting in no radial heat flow at this surface, heat may still flow (by conduction) along the tube wall.

(3)

PROBLEM 1.13.FAM.S GIVEN:

Popcorn can be prepared in a microwave oven. The corn kernels are heated to make the popcorn by an energy conversion from oscillating electromagnetic waves (in the microwave frequency range) to thermal energy designated as ˙se,m(W/m3). With justifiable assumptions for this problem, (1.23) can be simplified to

Q |A= −ρcvVdT

dt + ˙se,mV, integral-volume energy equation,

where the corn kernel temperature T is assumed to be uniform, but time dependent. The control volume for a corn kernel and the associated energy equation terms are shown in Figure Pr.1.13(a).

The surface heat transfer rate is represented by

Q |A= Q +T (t) − T

Rt

,

where, Q (prescribed heat flow) = 0, and T is the far-field ambient temperature and Rt(K/W) is the constant heat transfer resistance between the surface of the corn kernel and the far-field ambient temperature.

ρ = 1,000 kg/m3, cv= 1,000 J/kg-K, V = 1.13×107m3, ˙se,m= 4×105W/m3, T (t = 0) = 20C, T= 20C, Rt= 5 × 103 K/W.

SKETCH:

Figure Pr.1.13(a) shows the corn kernel and the thermal circuit diagram.

Rt (K/W)

− ρcvV

se,mV dT

dt

T1

Qt

V

Corn Kernel, T(t) ee

Oscillating Electric Field Intensity

Q = 0

Figure Pr.1.13(a) Thermal circuit model for a corn kernel heated by microwave energy conversion.

OBJECTIVE:

(a) For the conditions given below, determine the rise in the temperature of the corn kernel for elapsed time t up to 5 min. Use a software for the time integration.

(b) At what elapsed time does the temperature reach 100C?

SOLUTION:

The energy equation

Q|A=T − T

Rt = −ρcvVdT

dt + ˙se,mV

is an ordinary differential equation with T as the dependent variable and t as the dependent variable. The solution requires the specification of the initial condition. This initial condition is T (t = 0). This energy equation has a steady-state solution (i.e., when the temperature no longer changes). The solution for the steady temperature is found by setting dT /dt = 0 in the above energy equation. Then, we have

T − T

Rt

= ˙se,mV or T = T+ ˙se,mV Rt. Here we are interested in the transient temperature distribution up to t = 5 min.

(4)

(a) The solution for T = T (t), up to t = 1,000 s, is plotted in Figure Pr.1.13(b). Examination shows that initially (T − T)/Rtis small (it is zero at t = 0) and the increase in T is nearly linear. Later the time rate of increase in T begins to decrease. At steady-state (not shown), the time rate of increase is zero.

As will be shown in Section 3.6.2, the exact analytical solution is

T (t) = T (t = 0) + ˙se,mV Rt(1 − et/τ), τ = ρcvV Rt (1) Here ˙se,mV Rt= 2.260 × 102, and τ = 565.0 s. The derivation is as follows:

T − T

Rt

= −ρcvVdT

dt + ˙se,mV, Define θ = T − T, the equation can be rewritten as

θ

Rt = −ρcvVdθ

dt + ˙se,mV,

dθ dt = ˙se,m

ρcv

− θ

ρcvV Rt

, Define τ = ρcvV Rt, the equation can be rewritten as

dt =V Rt˙se,m− θ

τ ,

d(V Rt˙se,m− θ) V Rt˙se,m− θ = −dt

τ,

dln(V Rt˙se,m− θ) = d(−t τ ),

dln(V Rt˙se,m− θ) = d(−t τ ),

V Rt˙se,m− θ = Ae−tτ . Using the initial condition

t = 0, θ = T (t = 0) − T,

in the previous equation, the coefficient A can be determined, and thus θ and T (t).

(b) The time at which T = 100C is t = 247 s and is marked in Figure Pr.1.13(b).

COMMENT:

The pressure rise inside the sealed corn kernel is due to the evaporation of the trapped water. This water absorbs most of the electromagnetic energy. Once a threshold pressure is reached inside the corn kernel, the sealing membrane bursts.

(5)

0 200 400 600 800 1,000

t , s

20 60 100 140 180 220

T ,

o

C

Figure Pr.1.13(b) Variation of corn kernel temperature with respect to time.

(6)

PROBLEM 1.15.FAM GIVEN:

In spark-ignition engines, the electrical discharge produced between the spark plug electrodes by the ignition system produces thermal energy at a rate ˙Se,J(W). This is called the Joule heating and will be discussed in Section 2.3. This energy conversion results in a rise in the temperature of the electrodes and the gas surrounding the electrodes. This high-temperature gas volume V , which is called the plasma kernel, is a mixture of air and fuel vapor. This plasma kernel develops into a self-sustaining and propagating flame front.

AboutR

AQ |A dt = −1 mJ is needed to ignite a stagnant, stoichiometric fuel-air mixture of a small surface area A and small volume V , at normal engine conditions. The conventional ignition system delivers 40 mJ to the spark.

(ρcvV )g= 2 × 107 J/C, Tg(t = 0) = 200C.

SKETCH:

Figure 1.15(a) shows the spark plug and the small gas volume V being heated.

Electrical Insulator

Igniting Gas Kernel by Spark Plug

Center Electrode

Ground Electrode

Insulator Nose Plasma Kernel, Se,J

.

Figure Pr.1.15(a) Ignition of a fuel-air mixture by a spark plug in a spark-ignition engine. The plasma kernel is also shown.

OBJECTIVE:

(a) Draw the heat flux vector tracking for the region around the electrodes marked in Figure Pr.1.15(a). Start from the energy conversion source ˙Se,J.

(b) Assume a uniform temperature within the gas volume V . Assume that all terms on the right-hand side of (1.22) are negligible, except for the first term. Represent this term with

∂E

∂t

¯

¯

¯

¯V

= (ρcvV )g

dTg

dt .

Then for the conditions given below, determine the final gas temperature Tg(tf), where the initial gas temperature is Tg(t = 0).

(c) What is the efficiency of this transient heating process?

SOLUTION:

(a) Figure Pr.1.15(b) shows the heat flux vector tracking for conduction and heat storage in the electrodes. The gas kernel, where the energy conversion occurs, also stores and conducts heat.

(b) Using (1.22), with all the right-hand side terms set to zero except for energy storage, we have Q|A= −∂E

∂t

¯

¯

¯

¯V

= −(ρcvV )gdTg

dt .

Integrating this with respect to time, from t = 0 to a final time where t = tf, we have Z tf

Q|Adt = −(ρcvV )g[Tg(tf) − Tg(t = 0)].

(7)

Plasma Kernal

- (ρcvV)gdTg

dt Qr

Qk

Qu

Qk

Qku

Qk

Qk

Q A

- (ρcvV)e

dTe

dt

- (ρcvV)e

dTe

dt

Plasma Kernal Heated by Joule Heating

Figure Pr.1.15(b) Thermal circuit diagram.

Solving for Tg(tf), we have

Tg(tf) = Tg(t = 0) − Z tf

0

Q|Adt (ρcvV )g . Using the numerical values, we have

Tg(tf) = 200(C) − −103(J) 2 × 107(J/C)

= 200(C) + 5,000(C) = 5,200C.

(c) The efficiency is

η = Z tf

0

Q|Adt Z tf

0

e,Jdt

= 103(J)

40 × 103(J) = 2.5%.

COMMENT:

Very high temperatures are reached for the plasma kernel for a short time. The efficiency may even be smaller than 2.5%. There are several regimes in the short sparking period (order of milliseconds). These are breakdown, arc, and glow-discharge regimes. The gas heat up occurs during the glow-discharge regime. Most of the energy is dissipated during the first two regimes and does not lead to the gas heat up.

References

Related documents

Consafe Logistics is one of Europe’s leading suppliers of production logistics, enterprise mobility and warehouse management solutions to customers using logistics to improve

If index futures is trading above 5099, we can buy index stocks in cash market and simultaneously sell index futures to lock the gains equivalent to the difference between

We gave examples of the relation between the police and citizens, state and society respectively, the bureaucratic dimensions of policing, and the production of order, or at least

For the poorest farmers in eastern India, then, the benefits of groundwater irrigation have come through three routes: in large part, through purchased pump irrigation and, in a

14 When black, Latina, and white women like Sandy and June organized wedding ceremonies, they “imagine[d] a world ordered by love, by a radical embrace of difference.”

_ommissioned jointly by the Barlow Endowment for Music Composition at Brigham Young University and the Buffalo Philharmonic Orchestra, Lukas Foss's flute concerto

Before setting the time, check if all the chronograph hands - 1/10 second (where applicable), second, minute -are at the “0”(12 o’clock) position. * If the stopwatch is in use,

Blog writer used assonance are some simple examples of these literary device to share them so quick, but get the poem is the newest fiction and can.. Four expert tips to bsser, but