Answer:
the maximum length of specimen before deformation is found to be 235.6 mm
Explanation:
First, we need to find the stress on the cylinder.
Stress = σ = P/A
where,
P = Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4
A = 1.0752 x 10^-5 m²
σ = 2000 N/1.0752 x 10^-5 m²
σ = 186 MPa
Now, we find the strain (∈):
Elastic Modulus = Stress / Strain
E = σ / ∈
∈ = σ / E
∈ = 186 x 10^6 Pa/107 x 10^9 Pa
∈ = 1.74 x 10^-3 mm/mm
Now, we find the original length.
∈ = Elongation/Original Length
Original Length = Elongation/∈
Original Length = 0.41 mm/1.74 x 10^-3
<u>Original Length = 235.6 mm</u>
Answer:
![Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J](https://tex.z-dn.net/?f=Q%3DhA%28T_%7Bw%7D-T_%7Binf%7D%29%3D16.97%2A0.5%2827-300%29%3D-2316.4J)
Explanation:
To solve this problem we use the expression for the temperature film
![T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5](https://tex.z-dn.net/?f=T_%7Bf%7D%3D%5Cfrac%7BT_%7B%5Cinf%7D%2BT_%7Bw%7D%7D%7B2%7D%3D%5Cfrac%7B300%2B27%7D%7B2%7D%3D163.5)
Then, we have to compute the Reynolds number
![Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7BuL%7D%7Bv%7D%3D%5Cfrac%7B10%5Cfrac%7Bm%7D%7Bs%7D%2A0.5m%7D%7B16.96%2A10%5E%7B-6%7D%5Crfac%7Bm%5E%7B2%7D%7D%7Bs%7D%7D%3D2.94%2A10%5E%7B5%7D)
Re<5*10^{5}, hence, this case if about a laminar flow.
Then, we compute the Nusselt number
![Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77](https://tex.z-dn.net/?f=Nu_%7Bx%7D%3D0.332%28Re%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%28Pr%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D0.332%282.94%2A10%5E%7B5%7D%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%280.699%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D159.77)
but we also now that
![Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\](https://tex.z-dn.net/?f=Nu_%7Bx%7D%3D%5Cfrac%7Bh_%7Bx%7DL%7D%7Bk%7D%5C%5Ch_%7Bx%7D%3D%5Cfrac%7BNu_%7Bx%7Dk%7D%7BL%7D%3D%5Cfrac%7B159.77%2A26.56%2A10%5E%7B-3%7D%7D%7B0.5%7D%3D8.48%5C%5C)
but the average heat transfer coefficient is h=2hx
h=2(8.48)=16.97W/m^{2}K
Finally we have that the heat transfer is
![Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J](https://tex.z-dn.net/?f=Q%3DhA%28T_%7Bw%7D-T_%7Binf%7D%29%3D16.97%2A0.5%2827-300%29%3D-2316.4J)
In this solution we took values for water properties of
v=16.96*10^{-6}m^{2}s
Pr=0.699
k=26.56*10^{-3}W/mK
A=1*0.5m^{2}
I hope this is useful for you
regards
Answer:
ICP -OES stand for inductively coupled plasma optical emission spectroscopy
Explanation:
It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.
The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.
Answer:
0.08kg/s
Explanation:
For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.
The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.
finally you use the two previous equations to make a system and find the mass flows
I attached procedure
Answer:
(a) Increases
(b) Increases
(c) Increases
(d) Increases
(e) Decreases
Explanation:
The tensile modulus of a semi-crystalline polymer depends on the given factors as:
(a) Molecular Weight:
It increases with the increase in the molecular weight of the polymer.
(b) Degree of crystallinity:
Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.
(c) Deformation by drawing:
The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.
(d) Annealing of an undeformed material:
This also results in an increase in the tensile strength of the material.
(e) Annealing of a drawn material:
A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.