Answer:
Space velocity = 30 hr⁻¹
Explanation:
Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.
It is given mathematically as
Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)
Volumetric flowrate of the reeactants
= (molar flow rate)/(concentration)
Molar flowrate of the reactants = 300 millimol/hr
Concentration of the reactants = 100 millimol/liter
Volumetric flowrate of the reactants = (300/100) = 3 liters/hr
Reactor volume = 0.1 liter
Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹
Hope this Helps!!!
Explanation:
Air fuel ratio:
Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that

As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.
When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.
When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.
C. seems like the best answer. i may be wrong so don’t quote me on that
Answer:
True
Explanation:
A semicircular or circular torch movement should be used when depositing weld beads.
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

For ideal gas,
= 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.