Answer:
First compute the characteristic length and the Biot number to see if the lumped analysis is applicable
Lc = V/A = (pie*D3/6) / (pie * D2)= 1.2/6 = 0.0012/6= 0.0002m
Bi = hLc/K = (110W/m2.oC)(0.0002)(moC/35W)= 110*0.0002/35 = 0.0006 less than 0.1
Since the Biot number is less than 0.1, we can use the lumped parameter analysis.
In such an analysis, the time to reach a certain temperature is given by
t = -1/bIn(T-Tinfinite/T - Tinfinite)
From the data in the problem we can compute the parameter, b, and then compute the time for the ratio (T – Tinfinite/(Ti – Tinfinite) to reach the desired value.
b = hA/pCpV = h/pCpLc = 110/8500*0.0002 *320*s
b = 110/544s = 0.2022/s
The problem statement is interpreted to read that the measured temperature difference T – Tinfinite has eliminated 98.5% of the transient error in the initial temperature reading Ti – Tinfinite so the value of value of (T – Tinfinite)/(Ti – Tinfinite) to be used in this equation is 0.015
t = -1/bIn(T-Tinfinite/T - Tinfinite)
t = -s/0.1654 (In0.015)
t = (-s*-4.1997)/0.2022
t = 20.77s
It will take the thermocouple 20.77s to reach 98.5% of the initial temperature
Answer:
d=4.63 mm
Explanation:
Given that
Load ,P = 10,100 N
FOS = 1.5
Sy= 900 MPa
Therefore the working stress will be
Lets take the diameter of wire = d mm
The area
For design purpose
A=16.83 mm²
d=4.63 mm
Answer:
a = 0.07m or 70mm
b = 0.205m or 205mm
Explanation:
Given the following data;
Modulus of rigidity, G = 14MPa=14000000Pa.
c = 80mm = 0.08m.
P = 46kN=46000N.
Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.
Deflection (d) of the plate is to be at least 7mm = 0.007m.
From shearing strain;
[
Making a the subject formula;
Substituting into the above formula;
a = 0.07m or 70mm.
Also, shearing stress;
Making b the subject formula;
Substituting into the above equation;
b = 0.205m or 205mm
Explanation:
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Answer:
The electric current from the batteries installed in a radio supplies direct current (DC) electricity to the radio components directly as an alternative source to the Alternating Current (AC) converted to DC by the power unit located at the radio end of the cable plugged into the wall outlet.
Explanation:
Part of the power unit in a radio includes an AC to DC converter, which is an electrical circuit that is able to convert the alternating current power input from the wall outlet into a direct current output to the radio with which the radio can work
The alternative source of electric current from the batteries installed in a radio bypasses the AC to DC converter and supplies power directly to the radio so it can also work.