Answer:
id
Explanation:
I WILL GIVE BRAINLEIST!! What are some challenges did Sunita williams (astronaut) faced throughout her life and career?
Charles, Clarissa, and Francine all work in Manufacturing. The table shows the breakdown of one day for each employee.
Charles
Clarissa
Francine
9:00 a.m.–11:00 a.m.
meeting with production team and first-line supervisor of production
11:00 a.m.–12:00 p.m.
meeting with finance team
1:00 p.m.–2:00 p.m.
meeting with materials engineer for new project
2:00 p.m.–5:00 p.m.
e-mails and reports
6:00 a.m.–7:00 a.m.
review lists of damaged equipment
8:00 a.m.–11:00 a.m.
fix damaged machines in the hospital laboratory
12:00 p.m.–2:00 p.m.
fix damaged machines on the hospital floor
8:00 a.m.–9:00 a.m.
decide on which products need assessment
9:00 a.m.–12:00 p.m.
test products on factory floor against standards
1:00 p.m.–4:00 p.m.
write reports and recommendations on tested products
Which correctly identifies the career of each person?
A.)Charles is a First Line Supervisor of Production, Clarissa is an Industrial Machinery Mechanic, and Francine is a Product Safety Engineer.
B.)Charles is a Product Safety Engineer, Clarissa is a First Line Supervisor of Production, and Francine is an Industrial Machinery Mechanic.
C.)Charles is an Industrial Production Manager, Clarissa is a Medical Equipment repairer, and Francine is a Quality Control Analyst.
D.)Charles is a Quality Control Analyst, Clarissa is an Industrial Production Manager, and Francine is a Medical Equipment Repairer.
Answer:
C.)Charles is an Industrial Production Manager, Clarissa is a Medical Equipment repairer, and Francine is a Quality Control Analyst.
Explanation:
A power cycle operates between hot and cold reservoirs at 600 K and 300 K, respectively. At steady state the cycle develops a power output of 0.45 MW while receiving energy by heat transfer from the hot reservoir at the rate of 1 MW. a. Determine the thermal efficiency and the rate at which energy is rejected by heat transfer to the cold reservoir, in MW. b. Compare the results of part (a) with those of a reversible power cycle operating between these reservoirs and receiving the same rate of heat transfer from the hot reservoir
Answer:
a. The thermal efficiency of the actual cycle is 45%
ii) The rate at which energy is rejected by heat transfer to the cold reservoir is 0.55 MW
b. The (maximum) cycle efficiency, is 50%
The rate of energy rejection to the cold reservoir by heat transfer is 0.55 MW
Therefore, there is an increased efficiency and reduced heat rejection in the reversible power cycle working between the given two reservoirs
Explanation:
The given parameters are;
The temperature of the hot reservoir, [tex]T_H[/tex] = 600 K
The temperature of the hot reservoir, [tex]T_C[/tex] = 300 K
The power output, [tex]\dot W[/tex] = 0.45 MW = The cycle work per second
The heat input from the hot reservoir, [tex]\dot Q_{in}[/tex] = 1 MW
a. The thermal efficiency of the actual cycle is given by the work done divided by the heat supplied as follows;
[tex]\eta_{actual} = \dfrac{\dot W}{\dot Q_{in}}[/tex]
Therefore, we have;
[tex]\eta_{actual} = \dfrac{0.45 \ MW}{1 \ MW} = 0.45 = 45 \%[/tex]
The thermal efficiency of the actual cycle, [tex]\eta_{actual}[/tex] = 45%
ii) The rate at which energy is rejected by heat transfer to the cold reservoir, [tex]\dot Q_{c}[/tex], is given as follows;
[tex]\dot Q_{in}[/tex] = [tex]\dot Q_{c}[/tex] + [tex]\dot W[/tex]
∴ [tex]\dot Q_{c}[/tex] = [tex]\dot Q_{in}[/tex] - [tex]\dot W[/tex]
Therefore, we have;
[tex]\dot Q_{c}[/tex] = 1 MW - 0.45 MW = 0.55 MW
The rate at which energy is rejected by heat transfer to the cold reservoir, [tex]\dot Q_{c}[/tex] = 0.55 MW
b. For a reversible power cycle, we have the maximum cycle efficiency, [tex]\eta_{maximum}[/tex], given as follows;
[tex]\eta_{maximum} = \dfrac{T_H - T_C}{T_H}[/tex]
[tex]\therefore \eta_{maximum} = \dfrac{600 \, K - 300 \, K}{600 \, K} = 0.5 = 50\%[/tex]
The maximum cycle efficiency, [tex]\eta_{maximum}[/tex] = 50%
The rate of work done by the cycle, [tex]\dot W[/tex], is therefore give as follows;
[tex]\eta_{maximum} = \dfrac{\dot W}{\dot Q_{in}}[/tex]
Therefore;
[tex]\dot Q_{in}[/tex] × [tex]\eta_{maximum}[/tex] = [tex]\dot W[/tex]
1 MW × 50% = 0.5 MW = [tex]\dot W[/tex]
[tex]\dot W[/tex] = 0.5 MW
[tex]\dot Q_{c}[/tex] = [tex]\dot Q_{in}[/tex] - [tex]\dot W[/tex]
∴ [tex]\dot Q_{c}[/tex] = 1 MW - 0.5MW = 0.5MW
[tex]\dot Q_{c}[/tex] = 0.5MW
The rate of energy rejection to the cold reservoir by heat transfer, [tex]\dot Q_{c}[/tex] = 0.55 MW
Therefore the frequency and the rate at which energy is rejected by heat transfer to the cold reservoir are both higher for the reversible power cycle