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4 years ago

15

Identify each statement as referring to a series or parallel circuit.

1 answer:

jolli1 [7]4 years ago

7 0

Answer:

1. Parallel circuit

2. Parallel circuit

3. Series circuit

4. Series circuit

5. Parallel circuit

6. Parallel circuit

Explanation:

1. In a parallel circuit, there are multiple paths for current to flow. The path each current takes depends on the resistance of the resistors on that path.

2. In a parallel circuit, current splits up into various paths to get the total current through the circuit, the current flow through each resistor is added.

3. In a series circuit the voltage across each resistor is not the same. to get back the total voltage, the voltages across each resistor needs to be added.

4. Series circuits have voltage drops across each resistor. this makes the voltage across each resistor depend on the resistance of the resistor.

5. In parallel circuits voltage is the same across each resistor because they are all connected directly to the same source.

6. In parallel circuits, the power is the same in each resistor of equal resistance since the voltage across each resistor is the same

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Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A

uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

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7 0

3 years ago

Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat

attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

$COP_{HP}=\frac{T_H}{T_H-T_L}$

Therefore, the temperature relationship, $T_H=1.15\;T_L$

Then, we should apply the values in the COP.

$=\frac{1.15\;T_L}{1.15-1}$

$=7.67$

The number of heat rejected by the heat pump must then be calculated.

$Q_H=COP_{HP}\times W_{nst}$

$=7.67\times5=38.35$

We must then calculate the refrigerant mass flow rate.

$m=0.264\;kg/s$

$q_H=\frac{Q_H}{m}$

$=\frac{38.35}{0.264}=145.27$

The $h_g$ value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

$T_L=\frac{T_H}{1.15}$

$=\frac{64+273}{1.15}=293.04$

$=19.89\°C$

the lowest reservoir temperature = 258.703 kpa

So, the pressure ratio should be = 7.15

8 0

3 years ago