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4 years ago

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KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown

mesh currents, ixix and iyiy. Write the KVL equation for the supermesh by summing the voltages around the supermesh in the direction of the mesh currents and setting the sum to zero.

1 answer:

kaheart [24]4 years ago

8 0

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

$i_x+50mA=i_y\\i_x-i_y=0.05A$

Also

$V_{\Delta}=1K*i_y$

Now applying KVL on loop 1 as indicated in the attached figure

$1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0$

Similarly for loop 2

$2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0$

So the system of equations become

$i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0$

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

$V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V$

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

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3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

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Answer:

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Explanation:

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3 years ago

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Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

$P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}$

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

$\eta_t$ = The efficiency of the pump = 78%

$\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}$

$Q_{max}$ = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump $Q_{max}$ ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

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