
Interpretation:
Thermal conductivity of laminatefor parallel and perpendicular to the layer should be calculated.
Concept introduction:
The relationship of thermal conductivity, parallel to fiber is defined as:
Thermal conductivity is defined as the ability of material to conduct heat per unit surface area.
Relation of thermal conductivity for parallel condition is given as:
Where,
Relation of thermal conductivity, perpendicular to fiber is given as:
Where,
Volume fraction of composite is defined as the ratio of volume of one component to sum of volume of all components.
The relationship of volume fraction with volume is stated as:

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Chapter 17 Solutions
Essentials of Materials Science and Engineering, SI Edition
- Prelab Information Laboratory Preliminary Discussion Second-order RLC Circuit Analysis The second-order RLC circuit shown in figure 1 below represents all voltages and impedances as functions of the complex variable, s. Note, of course, that the impedances associated with R, RL, and Rs are constant independent of frequency, so the 's' notation is omitted. Again, one of the advantages of s-domain analysis is that we can apply all of the circuit analysis techniques learned for AC and DC circuits. ZI(s) Zc(s) Rs w RL ww + + VRS(S) VRL(S) VL(s) Vc(s) VR(S) R Vs(s) Figure 1: A second-order RLC circuit represented in the s-domain. To generate the s-domain expression for the output voltage, Vout(s) = VR(S), for the circuit shown in figure 1, we can apply voltage division in the s-domain as shown in equation 1 below. For equation 1 we define the following circuit parameters. RT=RS + RL + R where: R₁ = Total series resistance Rs Signal generator output resistance (fixed) Inductor internal…arrow_forward5.137 The BJT in the circuit of Fig. 5.137 has ẞ = 100. (a) Find the de collector current and the de voltage at the collector. (b) Replacing the transistor by its T model, draw the small-signal equivalent circuit of the amplifier. Analyze the resulting circuit to determine the voltage gain vo/vi. V ww 0.3 mA 300 ΚΩ = 250 Ω Va 30 ΚΩ www|| Fig. 5.137arrow_forwardsolve this, show all steps, no ai pz, please draw it outarrow_forward
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