Answer:
Considering the guidelines of this exercise.
The pieces produced per month are 504 000
The productivity ratio is 75%
Explanation:
To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21 making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.
Answer:
The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.
Explanation:
Diameter of pipe = 2 in = 0.0508 m
Steam temperature
= 300 F = 422.04 K
Duct temperature
= 70 F = 294.26 K
Emmisivity of surface 1 = 0.79
Emmisivity of surface 2 = 0.276
Net emmisivity of both surfaces ∈ = 0.25
Stefan volazman constant
= 5.67 ×

Heat transfer per foot length is given by
Q = ∈
A (
) ------ (1)
Put all the values in equation (1) , we get
Q = 0.25 × 5.67 ×
× 3.14 × 0.0508 × 1 × (
)
Q = 54.78 Watt per foot.
This is the value of heat transferred watt per foot length.
Answer:
Both Technician A and B are correct.
Explanation: A brake lathe is a special tool used to improve or work on the surface of brake pads it helps to smoothen the surface.
Brake lathe has been found to be very effective in removing rusts in rotors and unevenness in the brake pad surfaces in order to ensure the efficiency and effectiveness of the brake system of a vehicle. Hence, a brake lathe helps to make brake rotor surface as smooth as possible.
Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle =
...............1
duty cycle = 1.5 × 
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 ×
× 2
applied voltage = 3 mV
so current will be
current =
................3
current = 
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life = 
battery life = 80000 hours
battery life in year = 
battery life in year = 9.13 years
battery life in year = 9 years and 48 days
Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ =
+ x₄×
= 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ =
+ x₄×
= 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction,
, is given as follows;
= T₀ ×
= T₀ ×
× (s₄ + s₂ - s₁ - s₃)
= T₀ ×
×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴
= 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output,
=
+
≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.