Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
__
<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
__
<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
Answer:
The head loss in Psi is 0.390625 psi.
Explanation:
Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.
Given:
Factor cv is 48.
Flow rate of water is 30 GPM.
GPM means gallon per minute.
Calculation:
Step1
Expression for head loss for the water is given as follows:

Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.
Step2
Substitute 48 for cv and 30 for Q in above equation as follows:


h = 0.390625 psi.
Thus, the head loss in Psi is 0.390625 psi.
This an example solved please follow up with they photo I sent ok
Solution :
Methods for selling pressure of a distillation column :
a). Set,
the overhead stream using cooling water.
(minimum of approximate 45°C condenser temperature)
b). Set,
of bottom product that avoids decomposition or reaction.
c). Set,
not utility for reboiler.
Running the distillation column above the ambient pressure because :
The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.
Run the reactor at an evaluated temperature because :
a). The rate of reaction is taster. This results in a small reactor or high phase conversion.
b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.
Run the reaction at an evaluated pressure because :
The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.
The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.
Answer:
<h2>the answer of sols brother is correct</h2><h3>hope it helps you have a good day</h3><h2 />