Answer:
10.984mm
Explanation:
by elastic modulus
stress=modulus of elasticity*strain
stress=loading/area area" cross-section"
11mm=0.011m
area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter
stress=55000/(9.503*10^-5)=578.745 MPa
convert MPa and GPa to pascal.
strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463............axial strain
v=Poisson ratio
lateral strain=(-v)*axial strain= -0.31*0.00463
lateral strain= -1.4353*10^-3=change in diameter/ original diameter
change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m
negative indicates decrease in diameter.
decrease in dia.=0.01578mm
new diameter=11-0.01578= 10.984mm
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: 133.88 MPa approximately 134 MPa
Explanation:
Given
Plane strains fracture toughness, k = 26 MPa
Stress at which fracture occurs, σ = 112 MPa
Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m
Critical internal crack length, l' = 6 mm = 6*10^-3 m
We know that
σ = K/(Y.√πa), where
112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]
112 MPa = 26 MPa / Y.√(3.142 * 0.043)
112 = 26 / Y.√1.35*10^-2
112 = 26 / Y * 0.116
Y = 26 / 112 * 0.116
Y = 26 / 13
Y = 2
σ = K/(Y.√πa), using l'instead of l and, using Y as 2
σ = 26 / 2 * [√3.142 * (6*10^-3/2)]
σ = 26 / 2 * √(3.142 *3*10^-3)
σ = 26 / 2 * √0.009426
σ = 26 / 2 * 0.0971
σ = 26 / 0.1942
σ = 133.88 MPa
Answer:
(a) 1.003 MHz
(b) increase the signal-to-noise ratio
Explanation:
The relationship between channel capacity, bandwidth, and signal-to-noise ratio was described by Claude Shannon in the 1940s, based on previous work by Harry Nyquist and Ralph Hartley. The relationship is given by the formula ...
C = Blog₂(1 +S/N)
where C is the channel capacity (bits per second), B is the bandwidth (Hz), and S/N is the signal to noise power ratio.
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<h3>(a)</h3>
You want to find B for C=10^7 and 10·log(S/N) = 30. Solving for B, we find ...
B = C/log₂(1 +S/N)
B = 10^7/log₂(1 +10^(30/10)) = 10^7/log₂(1001) ≈ 1.003 MHz
The allocated bandwidth must be at least 1.003 MHz.
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<h3>(b)</h3>
Shannon's relation tells us the alternative to increasing bandwidth is increasing signal-to-noise ratio. If "sufficient" bandwidth is not available, the signal-to-noise ratio must be increased.