Answer:
a)Cycle time = 2.37 min
b)Numbers of workers =21
c)Stations on the line =24
Explanation:
Given that
Total work content time(TWC) = 50 min
Production rate Rp= 24 units/hr
manning level will be close =1.5
Line balancing efficiency =0.94
a)
Cycle time
Cycle time = 2.37 min
b)
Numbers of workers ,W
W= 21
Numbers of workers =21
c)
Stations on the line(n)
Lets find service time Ts
Ts = Cycle time - Time for repositioning
Ts = Tc- Tr
Ts= 2.37 - 9/ 60 min
Ts= 2.22 min
We know that efficiency
n=23.94 ⇒n=24
n=24
Stations on the line =24
Answer:
Explanation:
internal combustion engine.
Answer:
asking questions, drawing conclusions, and gathering information.
Explanation:
hope this helps!
Answer:
79 kW.
Explanation:
The equation for enthalpy is:
H2 = H1 + Q - L
Enthalpy is defined as:
H = G*(Cv*T + p*v)
This is specific volume.
The gas state equation is:
p*v = R*T (with specific volume)
The specific gas constant for air is:
287 K/(kg*K)
Then:
T1 = 60 + 273 = 333 K
T2 = 200 + 273 = 473 K
p1*v1 = 287 * 333 = 95.6 kJ/kg
p2*v2 = 287 * 473 = 135.7 kJ/kg
The Cv for air is:
Cv = 720 J/(kg*K)
So the enthalpies are:
H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW
H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW
Ang the heat is:
Q = 34 kW
Then:
H2 = H1 + Q - L
381 = 268 + 34 - L
L = 268 + 34 - 381 = -79 kW
This is the work from the point of view of the air, that's why it is negative.
From the point of view of the machine it is positive.
Answer:
The pressure inside the cooker is 1.0804 atm.
Explanation:
Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.
<u>For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.</u>
So, Standard conditions:
T₁ = 373.15 K
P₁ = 1 atm
Given ,
The water boils at Temperature = 130 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So, the temperature, T₂ = (130 + 273.15) K = 403.15 K
To find pressure inside the cooker (P₂) :
<u>Applying Amontons's Law as:</u>
So,
<u>Thus, The pressure inside the cooker is 1.0804 atm.</u>
