R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
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The skills developed during a materials science degree allow graduates enter a range of sectors, including working as engineering professionals and in design and marketing roles.
Destination Percentage
Employed 60.4
Further study 24.5
Working and studying 5.1
Unemployed 4
Other 6.1
Graduate destinations for materials science and engineering
Type of work Percentage
Engineering and building 23.9
Marketing, PR and sales 11.9
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Technicians and other professionals 10
Other 42.5
Answer:
Judgement
Explanation:
Gilbert is required by the Judgement Principle to "disclose those conflicts of interest that cannot reasonably be avoided or escaped." Since Gilbert professionally believes that the software meets specifications, secures documents, and satisfies user requirements, it is not clear if he violated any principle. However, he could have informed his client of his interest in the software and also presented other software packages of different companies from which the client could make its independent choice.
Answer:
The minimum mass flow rate for the water is 14.39kg/s
Explanation:
In this question, we are asked to calculate the minimum mass flow rate for the water in kg/s.
Please check attachment for complete solution and step by step explanation
Answer:
D) 1.04 Btu/s from the liquid to the surroundings.
Explanation:
Given that:
flow rate (m) = 2 lb/s
liquid specific enthalpy at the inlet (
Btu/lb)
liquid specific enthalpy at the exit (
Btu/lb)
initial elevation (
)
final elevation (
)
acceleration due to gravity (g) = 32.174 ft/s²
= 3 Btu/s
The energy balance equation is given as:
![Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0](https://tex.z-dn.net/?f=Q_%7Bcv%7D-W%7Bcv%7D%2Bm%5B%28h_1-h_2%29%2B%28%5Cfrac%7BV_1%5E2-V_2%5E2%7D%7B2%7D%29%2Bg%28z_1-z_2%29%5D%3D0)
Since kinetic energy effects are negligible, the equation becomes:
![Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0](https://tex.z-dn.net/?f=Q_%7Bcv%7D-W%7Bcv%7D%2Bm%5B%28h_1-h_2%29%2Bg%28z_1-z_2%29%5D%3D0)
Substituting values:
![Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\](https://tex.z-dn.net/?f=Q_%7Bcv%7D-%28-3%29%2B2%5B%2840.09-40.94%29%2B%5Cfrac%7B32.174%280-100%29%7D%7B778%2A32.174%7D%20%5D%3D0%5C%5CQ_%7Bcv%7D%2B3%2B2%5B-0.85-0.1285%20%5D%3D0%5C%5CQ_%7Bcv%7D%2B3%2B2%28-0.9785%29%3D0%5C%5CQ_%7Bcv%7D%2B3-1.957%3D0%5C%5CQ_%7Bcv%7D%2B1.04%3D0%5C%5CQ_%7Bcv%7D%3D-1.04%5C%5C)
The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.
