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!!!
The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.
<h3>What are modern Brayton engines?</h3>
Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.
<h3>What is the ranking cycle?</h3>
A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.
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To know more about Rankine cycle, visit: brainly.com/question/13040242
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Answer:
brake fade. loss of brake effectiveness due to overheating.
Explanation:
The correct question;
An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?
Answer:
h'_2 = 40 W/K.m²
Explanation:
We are given;
L1 = 1m
L2 = 5m
T_s = 400 K
T_(∞) = 300 K
V = 100 m/s
q = 20,000 W/m²
Both objects have the same shape and density and thus their reynolds number will be the same.
So,
Re_L1 = Re_L2
Thus, V1•L1/v1 = V2•L2/v2
Hence,
(h'_1•L1)/k1 = (h'_2•L2)/k2
Where h'_1 and h'_2 are convection coefficients
Since k1 = k2, thus, we now have;
h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)
Thus,
h'_2 = [20,000/(400 - 300)]•(1/5)
h'_2 = 40 W/K.m²
