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liubo4ka [24]
3 years ago
6

Add a capacitor, C2 = 6 pF, in parallel with R3, and a capacitor, C3 = 2 pF in parallel with R4. Use PSpice to plot the magnitud

e of the voltage gain 12 V vs. frequency from 10 Hz to 10 MHz.
Engineering
1 answer:
Ganezh [65]3 years ago
6 0
Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember
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The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

3 0
3 years ago
A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The
Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

       P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

      h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

       s1 = s-P1 = 0.6492 KJ/kg.K

       v1 = v-P1 = 0.001010 m^3 / kg

       

       P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

       s2 = s1 = 0.6492 KJ/kg.K   .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

   

                           w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}

- From the following relation we can determine ( h2 ) as follows:

                          h2 = h1 + wp

                          h2 = 191.81 + 8.0699

                          h2 = 199.88 KJ/kg

                           

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

              P3 = 8 MPa

              T3 = ?  ( assume fluid exist in the saturated vapor phase )

              h3 = hg-P3 = 2758.7 KJ/kg

              s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

                          q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

             P4 = 10 KPa

             s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

             

- Compute the quality of the mixture at condenser inlet by the following relation:

                           x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947

- Determine the isentropic ( h4s ) at this state as follows:

                          h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}        

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

                         h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

                        w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

                       W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

                        n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31

Answer: The thermal efficiency of the cycle is 0.31

       

   

7 0
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In a food processing facility, a spherical container of inner radius r1 = 40 cm, outer radius r2 = 41 cm, and thermal conductivi
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Answer:

attached below

Explanation:

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When measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial re
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Answer:

10.5

Explanation:

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