By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
<h3>How to determine the differential of a one-variable function</h3>
Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:
By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
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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
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Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:
For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²
The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C
Please add more details because I don’t know if you are using a book or passage, therefore I cannot help you unless you add more detail
