Answer:
Supply, demand, global markets, imports and exports, and government Regulation.
Explanation:
Answer:
1. High friction
2. High extrusion temperature
Explanation:
Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.
Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.
Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.
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
Assumptions:
- Steady state.
- Air as working fluid.
- Ideal gas.
- Reversible process.
- Ideal Otto Cycle.
Explanation:
Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):
- Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
- Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).
![r =\frac{V_1}{V_2}](https://tex.z-dn.net/?f=r%20%3D%5Cfrac%7BV_1%7D%7BV_2%7D)
- Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
- Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.
![r = \frac{V_4}{V_3} = \frac{V_1}{V_2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7BV_4%7D%7BV_3%7D%20%3D%20%5Cfrac%7BV_1%7D%7BV_2%7D)
- Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
- Exhaust 1-0: the working fluid is vented to the atmosphere.
If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:
![\eta = 1-(\frac{1}{r^{\gamma - 1} } )](https://tex.z-dn.net/?f=%5Ceta%20%3D%201-%28%5Cfrac%7B1%7D%7Br%5E%7B%5Cgamma%20-%201%7D%20%7D%20%29)
where:
![\gamma = \frac{C_{p} }{C_{v}} : specific heat ratio](https://tex.z-dn.net/?f=%5Cgamma%20%3D%20%5Cfrac%7BC_%7Bp%7D%20%7D%7BC_%7Bv%7D%7D%20%3A%20specific%20heat%20ratio)
Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.
![\gamma = 1.4](https://tex.z-dn.net/?f=%5Cgamma%20%3D%201.4)
Answer:
See image attached.
...simplify devices, reducing weight and the chance of failure.