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Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
Answer:
![250\ \text{lbm/min}](https://tex.z-dn.net/?f=250%5C%20%5Ctext%7Blbm%2Fmin%7D)
![625\ \text{ft/min}](https://tex.z-dn.net/?f=625%5C%20%5Ctext%7Bft%2Fmin%7D)
Explanation:
= Area of section 1 = ![10\ \text{ft}^2](https://tex.z-dn.net/?f=10%5C%20%5Ctext%7Bft%7D%5E2)
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 = ![4\ \text{ft}^3/\text{lbm}](https://tex.z-dn.net/?f=4%5C%20%5Ctext%7Bft%7D%5E3%2F%5Ctext%7Blbm%7D)
= Density of fluid = ![0.2\ \text{lb/ft}^3](https://tex.z-dn.net/?f=0.2%5C%20%5Ctext%7Blb%2Fft%7D%5E3)
= Area of section 2 = ![2\ \text{ft}^2](https://tex.z-dn.net/?f=2%5C%20%5Ctext%7Bft%7D%5E2)
Mass flow rate is given by
![m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}](https://tex.z-dn.net/?f=m%3D%5Crho%20A_1V_1%3D%5Cdfrac%7BA_1V_1%7D%7Bv_1%7D%5C%5C%5CRightarrow%20m%3D%5Cdfrac%7B10%5Ctimes%20100%7D%7B4%7D%5C%5C%5CRightarrow%20m%3D250%5C%20%5Ctext%7Blbm%2Fmin%7D)
The mass flow rate through the pipe is ![250\ \text{lbm/min}](https://tex.z-dn.net/?f=250%5C%20%5Ctext%7Blbm%2Fmin%7D)
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1
![m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}](https://tex.z-dn.net/?f=m%3D%5Crho%20A_2V_2%5C%5C%5CRightarrow%20V_2%3D%5Cdfrac%7Bm%7D%7B%5Crho%20A_2%7D%5C%5C%5CRightarrow%20V_2%3D%5Cdfrac%7B250%7D%7B0.2%5Ctimes%202%7D%5C%5C%5CRightarrow%20V_2%3D625%5C%20%5Ctext%7Bft%2Fmin%7D)
The speed at section 2 is
.
Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K ![= 98.9 MPa \sqrt{m}](https://tex.z-dn.net/?f=%20%3D%2098.9%20MPa%20%5Csqrt%7Bm%7D%20)
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:
![oc = \frac{K}{Y \sqrt{pi * a}}](https://tex.z-dn.net/?f=%20oc%20%3D%20%5Cfrac%7BK%7D%7BY%20%5Csqrt%7Bpi%20%2A%20a%7D%7D%20)
We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
![a= \frac{1}{pi} * [\frac{k}{y*a}]^2](https://tex.z-dn.net/?f=%20a%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7Bk%7D%7By%2Aa%7D%5D%5E2%20)
Substituting figures in the expression above, we have:
![= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7B98.9%20MPa%20%5Csqrt%7Bm%7D%7D%20%7B10%20%2A%20%5Cfrac%7B860MPa%7D%7B2%7D%7D%5D%5E2)
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection