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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!!!
Answer:
To fit text to a shape in Affinity Designer, make sure you have your text selected. Then, grab the Frame Text Tool and click on the shape. A blinking cursor will appear within the shape, indicating that you can begin typing. The text you type will be confined to the boundaries of the shape.
Explanation:
Answer:
The power developed by engine is 167.55 KW
Explanation:
Given that
Mean effective pressure = 6.4 bar
Speed = 2000 rpm
We know that power is the work done per second.
So
We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.
P=167.55 KW
So the power developed by engine is 167.55 KW
Answer:
Q=67.95 W
T=119.83°C
Explanation:
Given that
For air
Cp = 1.005 kJ/kg·°C
T= 20°C
V=0.6 m³/s
P= 95 KPa
We know that for air
P V = m' R T
95 x 0.6 = m x 0.287 x 293
m=0.677 kg/s
For gas
Cp = 1.10 kJ/kg·°C
m'=0.95 kg/s
Ti=160°C ,To= 95°C
Heat loose by gas = Heat gain by air
[m Cp ΔT] for air =[m Cp ΔT] for gas
by putting the values
0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )
T=119.83°C
T is the exit temperature of the air.
Heat transfer
Q=[m Cp ΔT] for gas
Q=0.95 x 1.1 x ( 160 -95 )
Q=67.95 W
