Answer:
W = 12.8 KW
Explanation:
given data:
mass flow rate = 0.2 kg/s
Engine recieve heat from ground water at 95 degree ( 368 K) and reject that heat to atmosphere at 20 degree (293K)
we know that maximum possible efficiency is given as
![\eta = 1- \frac{T_L}{T_H}](https://tex.z-dn.net/?f=%5Ceta%20%3D%201-%20%5Cfrac%7BT_L%7D%7BT_H%7D)
![\eta = 1 - \frac{ 293}{368}](https://tex.z-dn.net/?f=%5Ceta%20%3D%201%20-%20%5Cfrac%7B%20293%7D%7B368%7D)
![\eta = 0.2038](https://tex.z-dn.net/?f=%5Ceta%20%3D%200.2038)
rate of heat transfer is given as
![Q_H = \dot m C_p \Delta T](https://tex.z-dn.net/?f=Q_H%20%3D%20%5Cdot%20m%20C_p%20%5CDelta%20T)
![Q_H = 0.2 * 4.18 8(95 - 20)](https://tex.z-dn.net/?f=Q_H%20%3D%200.2%20%2A%204.18%208%2895%20-%2020%29)
![Q_H = 62.7 kW](https://tex.z-dn.net/?f=Q_H%20%3D%2062.7%20kW)
Maximuim power is given as
![W = \eta Q_H](https://tex.z-dn.net/?f=W%20%3D%20%5Ceta%20Q_H)
W = 0.2038 * 62.7
W = 12.8 KW
A lot of manufacturer often uses 5G machines. How these capabilities could help improve safety of the operators is that it does includes an emergency switch for the operator so that one can manually shut off when needed.
<h3>Edge computing with 5G</h3>
- The edge computing along with 5G network and IoT devices can help put together different safety features and limitations and on can use them to known the unsafe action and also data can be communicated.
Edge computing when use with 5G produces good opportunities in all industry. It is known to help bring computation and data storage close to where data is been produced and it enable good data control, reduced costs, etc.
Learn more about 5G network from
brainly.com/question/24664177
Answer:
Metal wireways are sheet metal "U"s with removable housing for protecting electrical equipment, wires, and cables.
Explanation:
These are especially used to run wire in manufacturing environments.
Complete Question
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
The elongation is ![=21.29mm](https://tex.z-dn.net/?f=%3D21.29mm)
Explanation:
In order to gain a good understanding of this solution let define some terms
True Stress
A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as
.
True Strain
A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as
.
The mathematical relation between stress to strain on the plastic region of deformation is
![\sigma _T =K\epsilon^n_T](https://tex.z-dn.net/?f=%5Csigma%20_T%20%3DK%5Cepsilon%5En_T)
Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows
![K = \frac{\sigma_T}{(\epsilon_T)^n}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5Csigma_T%7D%7B%28%5Cepsilon_T%29%5En%7D)
No substituting
from the question we have
![K = \frac{345}{(0.02)^{0.22}}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B345%7D%7B%280.02%29%5E%7B0.22%7D%7D)
![= 815.82MPa](https://tex.z-dn.net/?f=%3D%20815.82MPa)
Making
the subject from the equation above
![\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }](https://tex.z-dn.net/?f=%5Cepsilon_T%20%3D%20%28%5Cfrac%7B%5Csigma_T%7D%7BK%7D%20%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%20%7D)
![Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n](https://tex.z-dn.net/?f=Substituting%20%5C%20411MPa%20%5C%20for%20%5C%20%5Csigma_T%20%5C%20815.82MPa%20%5C%20for%20%5C%20K%20%20%5C%20and%20%20%5C%20%200.22%20%5C%20for%20%5C%20n)
![\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }](https://tex.z-dn.net/?f=%5Cepsilon_T%20%3D%20%28%5Cfrac%7B411MPa%7D%7B815.82MPa%7D%20%29%5E%7B%5Cfrac%7B1%7D%7B0.22%7D%20%7D)
![=0.0443](https://tex.z-dn.net/?f=%3D0.0443)
From the definition we mentioned instantaneous length and this can be obtained mathematically as follows
![l_i = l_o e^{\epsilon_T}](https://tex.z-dn.net/?f=l_i%20%3D%20l_o%20e%5E%7B%5Cepsilon_T%7D)
Where
is the instantaneous length
is the original length
![Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T](https://tex.z-dn.net/?f=Substituting%20%20%5C%20470mm%20%5C%20for%20%5C%20l_o%20%5C%20and%20%5C%200.0443%20%5C%20for%20%20%5C%20%5Cepsilon_T)
![l_i = 470 * e^{0.0443}](https://tex.z-dn.net/?f=l_i%20%3D%20470%20%2A%20e%5E%7B0.0443%7D)
![=491.28mm](https://tex.z-dn.net/?f=%3D491.28mm)
We can also obtain the elongated length mathematically as follows
![Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i](https://tex.z-dn.net/?f=Substituting%20%5C%20470mm%20%5C%20for%20l_o%20and%20%5C%20491.28%20%5C%20for%20%5C%20l_i)
![Elongated \ Length = 491.28 - 470](https://tex.z-dn.net/?f=Elongated%20%5C%20Length%20%3D%20491.28%20-%20470)
![=21.29mm](https://tex.z-dn.net/?f=%3D21.29mm)