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2 years ago

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Explain how you would solve for total resistance in a parallel circuit versus a series circuit. How would you apply and solve fo

r Kirchhoff’s current law in a parallel circuit, assuming you know the values of each resistor?

1 answer:

Mekhanik [1.2K]2 years ago

8 0

Answer:

No, series parallel, as it implies has components of the circuit configured in both series and parallel. This is typically done to achieve a desired resistance in the circuit. A parallel circuit is a circuit that only has the components hooked in parallel, which would result in a lower total resistance in the circuit than if the components were hooked up in a series parallel configuration.

Explanation:

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The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm

oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

$m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s$

The exit temperature is:

$T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25$ °C

b) The mean fluid velocity is:

$V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s$

The hydraulic diameter is:

$D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m$

The Reynold´s number is:

$Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36$

Assuming fully developed turbulent flow, the Nusselt number is:

$Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69$

$h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C$

The highest component surface temperature is:

$T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22$ °C

6 0

3 years ago

Using any of the bilinear transform, matched pole-zero, or impulse invariance techniques in converting a continuous-time system

leonid [27]

Answer:

A. True

The bilinear transform is employed in digital signal processing and discrete-time control theory which helps in transforming continuous-time system representations to discrete-time

4 0

3 years ago

Read 2 more answers

Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,

Therefore,

$Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

Finally,

$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

(C) Total heat flux transfer to the plate is found out by,

$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

8 0

2 years ago

Coal fire burning at 1100 k delivers heat energy to a reservoir at 500 k. Find maximum efficiency.

Marizza181 [45]

Answer:

<em>55%</em>

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

<em>This is a Carnot system</em>

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - $\frac{Tc}{Th}$

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - $\frac{500}{1100}$

Eff = 1 - 0.45 = 0.55 or<em> 55%</em>

7 0

2 years ago

If the same type of thermoplastic polymer is being tensile tested and the strain rate is increased, it will: g

Serggg [28]

Answer:

It would break I think need to try it out

Explanation:

3 0

2 years ago