The controller determines if a(n) error exists by calculating the difference between the SP and the PV.
<h3>How does a
controller work in control system?</h3>
The Control system is one where it entails if the output is one that has an effect on the input quantity.
So it uses the PV(Process Variable) set against the SP(Setpoint) to know if an error exists.
So, The controller determines if a(n) error exists by calculating the difference between the SP and the PV.
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Answer:
Condition A
Heat flux is 1400 W/M^2
Condition B
Heat flux is 12800 w/m^2
Explanation:
Given that:
is given as 30 degree celcius
condition A
Air temperature = - 5 degree c
convection coefficient h = 40 w/m^2. k

condition A
water temperature = 10 degree c
convection coefficient = 800 w/m^2.k

Answer:0.1898 Pa/m
Explanation:
Given data
Diameter of Pipe
Velocity of water in pipe
We know viscosity of water is
Pressure drop is given by hagen poiseuille equation

We have asked pressure Drop per unit length i.e.

Substituting Values

=0.1898 Pa/m
Answer:
Option E
Explanation:
All the given statements are true except the velocity gradients normal to the flow direction are small since these are not normally small. It's true that viscous effects are present only inside the boundary layer and the fluid velocity equals the free stream velocity at the edge of the boundary layer. Moreover, Reynolds number is greater than unity and the fluid velocity is zero at the surface of the object.
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))