Answer:
b)false
Explanation:
Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling is a metal forming process.
We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.
So given statement is wrong.
Answer:
When the renewal period comes around (with respect to the license expiry date).
Explanation:
Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;

The Biot number is given as;

< 0.1, thus apply lumped system approximation to determine the constant time for the process;

The time for the heating process is given as;

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
Fastest
(Known as the fast lane)
Answer:
19063.6051 g
Explanation:
Pressure = Atmospheric pressure + Gauge Pressure
Atmospheric pressure = 97 kPa
Gauge pressure = 500 kPa
Total pressure = 500 + 97 kPa = 597 kPa
Also, P (kPa) = 1/101.325 P(atm)
Pressure = 5.89193 atm
Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)
Temperature = 28 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K
⇒n = 595.76 moles
Molar mass of oxygen gas = 31.9988 g/mol
Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g