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.
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<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
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<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
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<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
The answer is D-all choices
Answer:
Explanation:
Reynolds number:
Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .
Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.
For plate can be given as
Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.
Flow on plate is a external flow .The values of Reynolds number for different flow given as
Answer:
flow(m) = 7.941 lbm/s
Q_in = 90.5184 Btu/lbm
Q_out = 56.01856 Btu/lbm
Explanation:
Given:
- T_1 = 60 F = 520 R
- T_6 = 940 = 1400 R
- Heat ratio for air k = 1.4
- Compression ratio r = 3
- W_net,out = 1000 hp
Find:
mass flow rate of the air
rates of heat addition and rejection
Solution:
- Using ideal gas relation compute T_2, T_4, T_10:
T_2 = T_1 * r^(k-1/k)
T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R
- Using ideal gas relation compute T_7, T_5, T_9:
T_7 = T_6 * r^(-k-1/k)
T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R
- The mass flow rate is obtained by:
flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)
flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)
flow(m) = 7.941 lbm/s
- The heat input is as follows:
Q_in = c_p*(T_6 - T_5)
Q_in = 0.24*(1400 - 1022.84)
Q_in = 90.5184 Btu/lbm
- The heat output is as follows:
Q_out = c_p*(T_10 - T_1)
Q_out = 0.24*(711.744 - 520)
Q_out = 56.01856 Btu/lbm
