Answer:
Both Techs A and B
Explanation:
Electronic braking systems are controlled by the electronic brake control module. It is a microprocessor that processes information from wheel-speed sensors and the hydraulic brake system to determine when to release braking pressure at a wheel that's about to lock up and start skidding and activates the anti lock braking system or traction system when it detects it is necessary.
Some electronic brake control modules can be programmed to the size of the vehicle's new tires to restore proper electronic brake control performance. While others may require replacing the module to match the module's programming to the installed tire size. So, both technicians A and B are correct.
Answer: true
Explanation:
it flows faster over the top of the wing because the top is more curved than the bottom of the wing. However
A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined
Assumptions :
1. Steady operating conditions exist.
2. Kinetic and potential energy changes are negligible.
Properties: The specific heat of geothermal water (
[) is taken to be 4.18 kJ/kg.ºC.
Analysis (a) We need properties of isobutane, we can obtain the properties from EES.
a. Turbine
P![P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }](https://tex.z-dn.net/?f=P_%7B3%7D%20%3D%203.25mPa%20%3D%20%283.25%2A1000%29%20kPa%5C%5C%3D%203250kPa%5C%5Cfrom%20the%20EES%20TABLE%5C%5Ch_%7B3%7D%20%3D%20761.54%20kJ%2Fkg%5C%5Cs_%7B3%7D%20%3D%202.5457%20kJ%2Fkg%5C%5CP_%7B4%7D%20%3D%20410kPa%5C%5C%5C%5Cs_%7B4%7D%20%3D%20s_%7B3%7D%20%5C%5Ch_%7B4s%7D%20%3D%20470.40kJ%2Fkg%5C%5C%5C%5CT_%7B4%7D%20%3D%20179.5%5E%7B0%7D%20C%5C%5C%5C%5Ch_%7B4%7D%20%3D%20689.74%20kJ%2FKG%5C%5C%5C%5C%20The%20%20isentropic%20%20efficiency%2C%20n_%7BT%7D%20%3D%20%5Cfrac%7Bh_%7B3%7D-h_%7B4%7D%20%20%7D%7Bh_%7B3%7D-%20h_%7B4s%7D%20%7D)
=![=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788](https://tex.z-dn.net/?f=%3D%5Cfrac%7B761.54-689.74%7D%7B761.54-670.40%7D%20%5C%5C%3D%5Cfrac%7B71.8%7D%7B91.14%7D%20%5C%5C%3D0.788)
b. Pump
![h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\](https://tex.z-dn.net/?f=h_%7B1%7D%20%3D%20h_%7Bf%7D%20%40%20410kPa%20%3D%20273.01kJ%2Fkg%5C%5Cv_%7B1%7D%20%3D%20v_%7Bf%7D%20%40%20410kPa%20%3D%200.001842%20m%5E%7B3%7D%2Fkgw_%7Bp%2Cin%7D%20%3D%20%20%5Cfrac%7Bv_%7B1%7D%28P_%7B2%7D-P_%7B1%7D%29%20%20%20%7D%7Bn_%7Bp%7D%20%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B0.01842%283250-410%29%7D%7B0.9%7D%20%5C%5C%5C%5C%20%3D5.81kJ%2Fkg%5C%5Ch_%7B2%7D%20%3Dh_%7B1%7D%20%2B%20w_%7Bp%2Cin%7D%5C%5C%20%20%20%20%20%20%20%20%20%20%3D%20273.01%2B5.81%5C%5C%20%20%20%20%20%20%20%20%20%20%20%3D%20278.82%20kJ%2Fkg%5C%5C%5C%5Cw_%7BT%2Cout%7D%20%3D%20m%5E%7B.%7D%20%20%28h_%7B3%7D%20-h_%7B4%7D%20%29%5C%5C%3D%28305.6%29%28761.54-689.74%29%5C%5C%3D305.6%2871.8%29%5C%5C%3D21%2C942kW%5C%5C%5C%5C)
![W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\](https://tex.z-dn.net/?f=W%5E%7B.%7D%20_%20%7BP%2Cin%7D%20%3D%20m%5E%7B.%7D%20%28h_%7B2%7D%20-h_%7B1%7D%29%20%5C%5C%3Dm%5E%7B.%7D%20%20w_%7Bp%2Cin%20%5C%5C%3D305.6%285.81%29%5C%5C%5C%5C%3D1%2C777kW%5C%5CW%5E%7B.%7D%20%20_%7Bnet%7D%20%3D%20W%5E%7B.%7D%20_%7BT%2C%20out%7D%20-%20W%5E%7B.%7D%20%20_%7BP%2Cin%7D%20%5C%5C%3D%2021%2C942-1%2C777%5C%5C%3D20%2C166%20kW%5C%5C%5C%5CHEAT%20EXCHANGER%5C%5C%5C%5CQ_%7Bin%7D%20%3D%20m%5E%7B.%7D%20_%7Bgeo%7D%20c_%7Bgeo%7D%20%28T_%7Bin-T_%7Bout%7D%20%7D%20%29%5C%5C%3D555.9%284.18%29%28160-90%29%5C%5C%3D162.656kW%5C%5C)
c. ![The thermal efficiency of the cycle n_{th} =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%](https://tex.z-dn.net/?f=The%20thermal%20efficiency%20of%20the%20cycle%20%20n_%7Bth%7D%20%20%3D%5Cfrac%7BW%5E%7B.%7D%20_%7Bnet%7D%20%7D%7BQ%5E%7B._%7Bin%7D%20%7D%20%7D%20%5C%5C%5C%5C%3D%20%5Cfrac%7B20%2C166%7D%7B162%2C656%7D%20%5C%5C%3D0.124%5C%5C%3D12.4%25)
Answer:
a) The key issues are the sue for libel and the evidence.
b) I would make a deal with her and implement a security program in the company.
Explanation:
The main issue in this case is that Pam Jones sued the company for libel, and the company remains in a position in which it has to prove that the internal investigation followed the right steps and indeed, the proves reflected that she was guilty and the fact that she got fired was correct.
The important here is exactly that the theft can be proved.
As an HR Director, I would give the correct proves in order to win the case, and if that is impossible, because of the tools and evidence, I would make a deal with her where both parts can be adequate to the problem.
She can´t be working again in the company but she can get financed according to her working years; also I would use this case as a growing opportunity by implementing new security methods that give more confidence between the company and its employees.