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3 years ago

Find the current Lx in the figure

Engineering

1 answer:

AleksandrR [38]3 years ago

6 0

Explanation:

$\frac{1}{8} + \frac{1}{2} \\ 1.6 + 1.4 = 3 \\ \frac{1}{3} + \frac{1}{9} \\ 2.25 + 2 = 4.25 \: ohm$

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

$\frac{9}{9 + 3} \times 4 = 3\\ \frac{2}{2 + 8} \times 3 = 0.6a \\ = 0.6 \: milli \: amper$

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Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.

saul85 [17]

Answer:

Explanation:

Given that:

At state 1:

Pressure P₁ = 20 bar

Volume V₁ = 0.03 $\mathbf{m^{3}}$

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C ; $v_1 = vg_1$ = 0.0996 $\mathbf{m^{3}}$ / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

At state 2:

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 $\mathbf{m^{3}}$ / kg

From temperature T₂ = 200⁰ C

$v_f_2 = 0.0016 \ m^3/kg$

$vg_2 = 0.127 \ m^3/kg$

Since $vf_2 < v_2$ , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality $x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}$

$x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}$

$x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}$

$\mathsf{x_2 =0.78}$

At temperature T₂, the specific internal energy $u_f_2 = 850.6 \ kJ/kg$ , also $ug_2 = 2594.3 \ kJ/kg$

Thus,

$u_2 = uf_2 + x_2 (ug_2 -uf_2)$

$u_2 =850.6 +0.78 (2594.3 -850.6)$

$u_2 =850.6 +1360.086$

$u_2 =2210.686 \ kJ/kg$

At state 3:

Temperature $T_3=T_2 = 200 ^0 C ,$

$V_3 = 2V_1 = 0.06 \ m^3$

Specific volume $v_3 = 0.2 \ m^3/kg$

Thus; $vg_3 =vg_2 = 0.127 \ m^3/kg$ ,

SInce $v_3 > vg_3$ , therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at $v_3 = 0.2 \ m^3/kg$ and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 $\ m^3/kg$

The specific internal energy $u_3$ at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

$u_2-u_1$

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

$u_3-u_2$

= (2622.3 - 2210.686) kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0

3 years ago

A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c

Paraphin [41]

Answer:

$10.8\ \text{lb/ft^2}$

$101.96\ \text{lb/ft}^2$

Explanation:

$v_1$ = Velocity of car = 65 mph = $65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}$

$\rho$ = Density of air = $0.00237\ \text{slug/ft}^3$

$v_2=0$

$P_1=0$

$h_1=h_2$

From Bernoulli's law we have

$P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}$

The maximum pressure on the girl's hand is $10.8\ \text{lb/ft^2}$

Now $v_1$ = 200 mph = $200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}$

$P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2$

The maximum pressure on the girl's hand is $101.96\ \text{lb/ft}^2$

5 0

3 years ago

Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in

Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H $_{2}o$ ( YH $_{2}o$ )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

= 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH $_{2}o$ = molar flow rate of water

Nair = molar flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541 moles /min

Molar flowrate of water

= n - Nair

= 311.541 - 300 = 11.541 mol/min

4 0

3 years ago

What is control flow in programming?

MakcuM [25]

Answer:

The control flow is the order in which the computer executes statements in a script. Code is run in order from the first line in the file to the last line, unless the computer runs across the (extremely frequent) structures that change the control flow, such as conditionals and loops.

Explanation:

hope it helps.

brainliest pls

7 0

2 years ago

Hard woods and tight curves should be cut using what blade speeds.

erica [24]

Answer:

slower

Explanation:

When it comes to using scroll saws, it is important to consider the type of wood in order to choose the correct speed. The blades of the saw tends to break easily, thus it is essential to choose a "slower speed" when shaping hard woods. Such speed is also used for "tight curves." Furthermore, it is important not to force the wood to the blade because it will also cause breakage.

4 0

3 years ago