From Problem 1.71, consider the wire leads connecting the transistor to the circuit board. The leads are of thermal conductivity k, thickness t, width w, and length L. One end of a lead is maintained at a temperature T c corresponding to the transistor case, while the other end assumes the temperature T b of the circuit board. During steady-state operation, current flow through the leads provides for uniform volumetric heating in the amount q . , while there is convection cooling to air that is at T ∞ and maintains a convection coefficient h . Derive an equation from which the temperature distribution in a wire lead may be determined. List all pertinent assumptions. Determine the temperature distribution in a wire lead, expressing your results in terms of the prescribed variables.
From Problem 1.71, consider the wire leads connecting the transistor to the circuit board. The leads are of thermal conductivity k, thickness t, width w, and length L. One end of a lead is maintained at a temperature T c corresponding to the transistor case, while the other end assumes the temperature T b of the circuit board. During steady-state operation, current flow through the leads provides for uniform volumetric heating in the amount q . , while there is convection cooling to air that is at T ∞ and maintains a convection coefficient h . Derive an equation from which the temperature distribution in a wire lead may be determined. List all pertinent assumptions. Determine the temperature distribution in a wire lead, expressing your results in terms of the prescribed variables.
Solution Summary: The author explains the equation from which the temperature distribution in a wire lead may be determined.
From Problem 1.71, consider the wire leads connecting the transistor to the circuit board. The leads are of thermal conductivity k, thickness t, width w, and length L. One end of a lead is maintained at a temperature
T
c
corresponding to the transistor case, while the other end assumes the temperature
T
b
of the circuit board. During steady-state operation, current flow through the leads provides for uniform volumetric heating in the amount
q
.
,
while there is convection cooling to air that is at
T
∞
and maintains a convection coefficient h.
Derive an equation from which the temperature distribution in a wire lead may be determined. List all pertinent assumptions.
Determine the temperature distribution in a wire lead, expressing your results in terms of the prescribed variables.
Meh
Battery operated train
Coll CD Af Pair
160,000kg 0.0005 0.15 5m² 1.2kg/m³
19
7et nong
0.98 0.9 0.88
Tesla Prated
Tesla Trated Ywheel ng Jaxle.
270kW
440NM
0.45m 20
2
8.5kgm²
Consider a drive cycle of a 500km trip with 3 stops in
the middle. Other than the acceleration and deceleration
associated with the three stops, the tran maintains.
constant cruise speed velocity of 324 km/hr. The
tran will fast charge at each stop for 15 min at a
rate Peharge = 350 kW
(ผม
τ
(MN
15MIN
Stop
w charging
(350kW
GMIJ
restored during 15
minutes of fast charging at
Calculate the battery energy Pcharge = 350kW
Calculate the net energy gain per stop
t
64
Determice the total battery energy required Ebat
to complete the 500km trip with 3 stops.
etc
DO NOT COPY SOLUTION
The differential equation of a cruise control system is provided by the following equation:
Find the closed loop transfer function with respect to the reference velocity (vr) .
a. Find the poles of the closed loop transfer function for different values of K. How does the poles move as you change K?
b. Find the step response for different values of K and plot in MATLAB. What can you observe?
c. For the given transfer function, find tp, ts, tr, Mp . Plot the resulting step response. G(s) = 40/(s^2 + 4s + 40)
Aswatan gas occupies a space of 0.3 millike cube at a pressure of 2 bar and temperature of 77 degree Celsius it is indicate at constant volume at pressure of 7 parts determine temperature at the end of process mass of a gas changing internal energy change in enthalpy during the process assume CP is equal to 10 1.005 CV is equal to 0.712 is equal to 287
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