It is proposed to absorb acetone from air using water as a solvent. Operation is at 10 atm and is isothermal at 20°C. The total flow rate of entering gas is 10 kmol /h. The entering gas is 1.2 mol% acetone. Pure water is used as the solvent. The water flow rate is 15 kmol/h. The desired outlet gas concentration should be 0.1 mol % acetone. For this system, Henry's law holds and Ye = 1.5 X where Ye is the mol fraction of acetone in the vapour in equilibrium with a mol fraction X in the liquid.
KGa = 0.4 kmol*m^-3*s^-1
1. Draw a schematic diagram to represent the process.
2. Determine the mole fraction of acetone in the outlet liquid.

Answer:

The answer is "26.55 V"

Explanation:

Given values:

[tex]d_i= 0.96m\\d_o= 1m\\\epsilon = 0.05\\T_0= 280k\\T_i= 95k\\[/tex]

For Answer  (a) please find the attachment.

Answer (b):

[tex]q_{i-0}= \frac{\sigma (T_{0}^4)-(T_{i}^4)}{\frac{1-\epsilon i }{\epsilon_{i} A_{i}}+ \frac{1 }{\ f_{i o} A_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0} A_{0}}}[/tex]

[tex]f_{i0}= 1 \ it \ is \ fully \ inside \ the \ large \ sphero \\[/tex]

[tex]q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} - 1+ 1 +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times \frac{A_i}{A_0}}\\\\q_{i-0}= \frac{\sigma A_i (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{d_i}{d_0})^2}\\\\q_{i-0}= \frac{\sigma (\pi d^2_i) (T_{0}^4)-(T_{i}^4)}{\frac{1 }{\epsilon_{i}} +\frac{1-\epsilon_{0}}{\epsilon_{0}} \times (\frac{r_i}{r_0})^2}\\\\[/tex]

[tex]q_{i-0}= \frac{5.67 \times 10^{-8} \times 3.14 \times 9.62 \times 9.62 \times (280^4-94^4)}{\frac{1 }{0.05} +\frac{1-0.05}{0.05} \times (\frac{0.96}{1})^2}\\\\\ After \ solve the \ equation \ the \ answer \ is:\\\\q_{i-0} = 26.55 \ V[/tex]

Answer & Explanation:

Circular Flow model denotes how goods & services, factor incomes & prices move within sectors of economy.

A closed economy has two sectors - households & firms, having following features of circular flow between them:

In case of open economy - with rest of world & foreign country, exports & imports also come in circular flow.

Answer:

  B) leads the voltage

Explanation:

One way to think about it is that the current causes charge to be accumulated on the capacitor, changing its voltage. The current must be non-zero before the voltage can change. Hence current leads voltage.

Answer: a. 0.4667

b. 0.4667 and C 0.0667

Explanation:

Given Data:

N = population size (10)

n = random selection (2)

r = number of observations = 7

Therefore

f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )

When y = 1

f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )

= 7 / 15

= 0.4667

When y = 2

f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )

= 7 / 15

= 0.4667

When y = 0

f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )

= 1 / 15

= 0.0667

Answer:

1.96 kg/s.

Explanation:

So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;

=> Superheated water vapor at a pressure = 20 MPa,

=> temperature = 500°C,

=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."

=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."

K1 = 3241.18, k2 = 93.28 and 2725.47.

Therefore, m1 + m2= m3.

10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.

=> 1.96 kg/s.

Answer:

The required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm²  and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2  so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ  

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Answer:

diameter is 1 cm

Explanation:

diameter is 1 cm

Answer:

1. 0.494

2. = 352.299 million years

3. It is between 0 years to 352.299 million years

4. greater than the 352.299 million years.

Explanation:

Given

Uranium 235 ———— lead 207( zircon sample)

t= 0 100%. 0%

t= t. 71%. 29%

1)- half lives elapsed = n

(1/2)n = 0.71

By taking log and solving n = 0.494

No of half lives = 0.494

2) Calculating the age of the larva

- age of lava flow = half life of uranium 235 x n

= 713 x 0.494

= 352.299 million years

3)- The rock layers was created above this lava flow is of later occurrence so the age will be lower than that of the age of the lava flow so the age of rock over the lava flow will lower than the 352.299 million years it will be in between 0 years to 352.299 million years.

4)- The rock layers created underneath this lava flow is of earlier occurrence so its age is more than that of the age of the lava flow so the age of rock will more than the lava flow is greater than the 352.299 million years.

B. The astronomers think the earth was created as all of the other rock materials in our solar system, including the oldest meteorites. The oldest meteorites ever found on Earth contain nearly equal amounts of both Uranium-238 and Lead-206.

The number of half-lives of the uranium-235to lead-207 decay pair have elapsed in the zircon crystals is; 0.494

We are given;

The atoms of Uranium-235 and lead-207 which made up the zircon sample.

At t = 0; Uranium-235 is 100% while lead-207 is 0%

At t = t;  Uranium-235 is 71% while lead-207 is 29%

1) Let the half lives that elapsed be n. Thus;

(¹/₂)ⁿ = 0.71

n*log0.5 = log 0.71

n = (log 0.71)/(log 0.5)

n = 0.494

Thus;

Number of half lives = 0.494

2) Formula to get the absolute age of the larva is;

Absolute age age of lava flow = half life of uranium-235 * n

The half life of uranium-235 is 713 million years. Thus;

Absolute age age of lava flow  = 713 * 0.494

Absolute age of lava flow = 352.22 million years

3) The rock layers above the lava flow were created after the lava flow and so the age will be lower than that of the age of the lava flow. Thus, it's age will be between 0 years and 352.299 million years.

4) A: The rock layers beneath the lava flow were in existence earlier than the lava flow and as such, the age of rock layers beneath the lava flow will be greater than 352.22 million years.

B; The astronomers think the earth was created as all of the other rock materials in our solar system.

Read more about Half life at; https://brainly.com/question/26148784

Answer:

Explanation:

Given that:

f = 250 Hz

[tex]\delta[/tex]= 2%

[tex]f_n[/tex]= 600 Hz

[tex]\zeta[/tex] = 0.5 to 1.5  increment by 0.05

[tex]F = A sin (Xt)[/tex]

For 250 Hz = 250 cycle/sec

[tex]X = 2 \pi t 250[/tex]

[tex]X = 500 \pi t[/tex]

[tex]X = Asin (500 \pi t)[/tex]

[tex]\omega = 250 \\ \\ \omega_n = 600[/tex]

M = 0.98 , 1.02

[tex]M_{(w)}} = \sqrt{[1-(\frac{w}{w_n})^2 + ( 2\zeta \frac{w}{w_n})^2}[/tex]

[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]

[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]

[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{0.98^2}-(1-(\frac{250}{600})^2)^2}[/tex]

[tex]\zeta = 0.7183[/tex]

At 0.7183 value of damping ratio the error value was 2% at 0.98 value of M

[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]

[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]

[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{1.02^2}-(1-(\frac{250}{600})^2)^2}[/tex]

[tex]\zeta = 0.6330[/tex]

At  0.6330 value of damping ratio the error value was 2% at 1.02 value of M.

Hence, the damping ratio [tex]\zeta[/tex] of the transducer must be placed between 0.6330 to 0.7183

Answer: Firebrick is referred to as "common brick" because it is the most commonly used  type of brick --- False

Explanation:

 Firebrick is not the most commonly used,   it is  a  refractory ceramic brick used in lining of  furnaces and  kilns, which is built basically  to withstand or resist high temperature.The most commonly used type of brick, is the Building Brick called "common brick"   because it is versatile and used for  applications where appearance is not an  important factor.

Answer:

For a 1 kWh, heat supplied for the cost of furnace is = $1.66 *1 0^-5 /kJ * 3600 =$0.06/kWh.

Therefore, supplied heat for natural gas furnace is $0.06/kWh which is lower than $0.12/kWh cost of electricity of refrigerator.

Explanation:

Solution

Given that:

The thermal efficiency of furnace = 0.8 or 80%

The cop of refrigerant COP = 3.5

The unit cost of electricity for refrigerant is =$ 0.12kWh

The unit cost of natural gas of furnace is = $1.40/therm

1 therm = 105,500 kJ

Now,

In the summer the usage or utilization of lightening in the house will be lesser than winter, hence the total cost of energy used in the household will be decreased.

Thus,

In winter, the utilization of lightening will be higher with regards to electricity

For a natural gas furnace it is given below:

The unit cost =$1.40/ηth (I therm/105,500)

= 1.40/0.8 (I therm/105,500)

= $1.66 *1 0^-5 /kJ

For a 1 kWh, heat supplied for the cost of furnace is = $1.66 *1 0^-5 /kJ * 3600 =$0.06/kWh.

So,for the heat supplied for natural gas furnace is $0.06/kWh which is less er than $0.12/kWh cost of electricity of refrigerator.

Therefore, the efficiency of energy lightning will reduce the total energy cost of the building both in winter and summer.

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem

Answer:

4.2ms

Explanation:

Calibrated time= 0.3+0.2+0.5+1.4= 2.4

Measured time= 6.6ms

Accuracy is closeness of measurement to an observed or true value

Accuracy= 6.6-2.4= 4.2ms

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

[tex]\tau_{ave} = \frac{T}{2tA_{m}}[/tex]

[tex]A_m[/tex] = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, [tex]A_m[/tex] = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

[tex]\tau_{ave} = \frac{T}{2tA_{m}}[/tex]

Plugging in then values, gives;

[tex]\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa[/tex]

The shear stress will be 80,000,000 Pa or 80 MPa.

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 =  0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 =  0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97  = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

Answer:

atomic percentage = 143 %

Explanation:

Let  x be the number of tin atoms and there are 4 atoms / cell in the FCC structure , then 4 -x  be the number of copper atoms . Therefore, the value of x can be determined by using the density equation as shown below:

[tex]\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }[/tex]

where;

the lattice parameter is given as : 4.7589 × 10⁻⁸ cm

The atomic mass of tin is 118.69 g/mol

The atomic mass of copper is 63.54 g/mol

The density is 8.772 g/cm³

[tex]\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }[/tex]

569.32 = 118.69x + 254.16-63.54x

569.32 - 254.16 = 118.69x - 63.54 x

315.16 = 55.15x

x = 315.16/55.15

x = 5.72 atoms/cell

As there are four atoms per cell in FCC structure for the metal, thus, the atomic percentage of the tin is  calculated as follows :

atomic % = [tex]\frac{no \ of \ atoms \ per \ cell \ in \ tin }{no \ of \ atoms \ per \ cell \ in \ the \ metal}*100[/tex]

atomic % = [tex]\frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100[/tex]

atomic % = 143 %

Answer:

A one -two pages menu was written with questions directed to the CSP which is stated below in the explanation section

Explanation:

Solution

If CSP has no team or limited staff, you will need to ask the following questions to understand how the CSP is set up:

To access evidence in the cloud :

Answer:

$151094

Explanation:

Solution

Recall that:

Acme Chemical in 2002 purchased a large pump worth of = 112,000

The estimation for the pump to the industrial pump index is =100

The index in 2002= 212

The current index is = 286

k = is the  reference year for which cost or price is known.

n =  the year for which cost or price is to be estimated (n>k).

Cn = the estimated cost or price of item in year n.

Ck =  the cost or price of item in reference year k.

Cn = Ck * (In / Ik )

Now,

We find the estimated cost of the new pump which is stated as follows:

Cn = (112,000 * 286) /212

=32032000/212

=$151094

Therefore, the estimated cost of the new pump is $151094

Answer:

Explanation:

The image that is supposed to be attached to the question is displayed in the diagram below.

Applying Nodal Analysis at node 1;

[tex]\dfrac{V_o -50}{12.5*10^{-3}} + \dfrac{V_o}{50*10^3}+\dfrac{V_o-7500 \ in}{10*10^3}=0[/tex]

where;

[tex]in = \dfrac{V_o}{50*10^3}[/tex]   (from the circuit)

= [tex]\dfrac{V_o-50}{12.5}+\dfrac{V_o}{50} + V_o -\dfrac{7500 *V_o }{\frac{50*10^3}{10}}=0[/tex]

= [tex]V_o [ \dfrac{1}{12.5}+\dfrac{1}{50}+\dfrac{1}{10}-\dfrac{75}{500}] = \dfrac{50}{12.5}[/tex]

= [tex]V_o[ \dfrac{500*500+12.5*5000+12.5*5000*5-75*12.5*500}{12.5*50*10*500}]= \dfrac{50}{12.5}[/tex]

= [tex]V_o = 80 \ volts[/tex]

[tex]in = \frac{80}{50*10^3}= 1.6 mA \\ \\ 7500*in = 120 volts \\ \\ I = \frac{120-80}{10(10^3} =4*10^{-3} Amps \\ \\ \\ \\ P_{generated} = 75000*in*I \\ \\ P_{generated} = 120*4*10^{-3} \\ \\ P_{generated} = 480 \ MW[/tex]

Answer:

Find explanation below.

Explanation:

Seed drills make it possible for seeds to be planted at equal spacing or distances.

Seed drills are important because they allow seeds to be planted at specific depths and distances. This allows for increased crop yield. They also help to curb the growth of weeds.

Answer:

0.09

Explanation:

Packet switching involves breaking a message into packets and sending them independently. Since the user only needs to transmit 10 percent of the time, the probability that a given (specific) user is transmitting = 10% = 0.1

The  probability that a user is not transmitting = 100% - 10% = 90% = 0.9

Therefore, the probability that a given (specific) user is transmitting, and the remaining users are not transmitting = 0.1 * 0.9 = 0.09

The image attached that is supposed to be attached to the question is shown in the first file below.

Answer:

t = 2.19 seconds

Explanation:

The free body diagram showing the center of mass A and B is attached in the second diagram below.

NOTE : that from the second diagram; Mass A and B do not have any acceleration

Taking the moment about wheel A:

[tex]\sum M_A = I_A \alpha _A[/tex]

[tex]-f(r_A) = I_A \alpha _A ----- (1)[/tex]

The equilibrium forces in the y-direction is 0

i.e

[tex]F_y = 0[/tex]

So;

[tex]N +T sin 30^0 -W_A = 0 ----- (2)[/tex]

The equilibrium forces in the x-direction is as follows:

[tex]\sum F_x = 0[/tex]

[tex]Tcos 30^0 + f= 0 -----(3)[/tex]

The kinetic friction f can be expressed as :

[tex]f = \mu _k N[/tex]

From above equation (2) and equation (3);

[tex]N + [\dfrac{-f}{cos 30^0}]sin 30^0 -150 =0[/tex]

[tex]N - \mu _k N \ tan 30^0 -150 =0[/tex]

[tex]N = \dfrac{150}{1-0.3 \ tan 30^0}[/tex]

N = 181.423 lb

Similarly; from equation(1)

[tex]\alpha_A = - \dfrac{f(r_A)}{I_A}[/tex]

[tex]\alpha _A = \dfrac{-\mu_k N(r_A)}{I_A}[/tex]

[tex]\alpha _A = \dfrac{-0.3*181.423*1.25}{\frac{150}{32.2}*I^2}[/tex]

[tex]\alpha _A =-14.6045 \ rad/s^2[/tex]

However; from the kinematics ; as moments are constant ; so is the angular acceleration is constant )

Thus;

[tex]\omega _A - \omega_o^A = \alpha_A t[/tex]

[tex]\omega _A = \omega_o^A + \alpha_A t[/tex]

[tex]\omega _A = 100 -14.6045 \ t ---- (4)[/tex]

Let's take a look at wheel B now;

Taking the moment about wheel B from the equation of motion:

[tex]\sum M_B = I_B \alpha _B[/tex]

[tex]f(r_B) = I_B \alpha _B[/tex]

[tex]\mu_k N (r_B) = I_B \alpha_B[/tex]

[tex]\mu_k N (r_B) = \dfrac{W_B}{g}* k^2_B \alpha_B[/tex]

[tex]\alpha_B = \dfrac{0.3*181.423*1}{\frac{100}{32.2}*0.75^2}[/tex]

[tex]\alpha = 31.1563 \ rad/s^2[/tex]

Again; from the kinematics; as the moments are constant which lead to the angular accleration;

[tex]\omega _B = \omega _o^B + \alpha _B \ t[/tex]

[tex]\omega _B =0 + 31.156 \ t-----(5)[/tex]

From equation 4 and 5 which attain the same angular velocity; we have;

[tex]\omega^A = \omega^B[/tex]

100 - 14.6045 t = 31.1563 t

100 = 31.1563 t + 14.6045 t

100 = 45.761 t

t = 100/45.761

t = 2.19 seconds

Answer:

Explanation:

Base on the scenario been described in the question, this solution, I have assumed that the river has the exact flow all through and therefore, dispersion characteristics will is also the same

This implies that the river doesn't separate or any other water body does not mixed with Ohio river

For us to begin we can assume that all phenol was injected at a time at upstream from Cincinnati.

Please check the image below

Answer:

Explanation:

Find the temperature at exit of compressor

[tex]T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k[/tex]

Find the work done by the compressor

[tex]\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg[/tex]

Find the actual workdone by the compressor

[tex]\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg[/tex]

Find the temperature at exit of the turbine

[tex]T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k[/tex]

Find the actual workdone by the turbine

[tex]1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg[/tex]

Find the temperature of the regeneration

[tex]\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k[/tex]

Find the heat supplied

[tex]Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg[/tex]

Find the thermal efficiency

[tex]n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4[/tex]

60.4%

Find the mass flow rate

[tex]m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42[/tex]

Find the actual workdone by the compressor

[tex]\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg[/tex]

Find the actual workdone by the turbine

[tex]\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg[/tex]

Find the temperature of the compressor exit

[tex]\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k[/tex]

Find the temperature at the turbine exit

[tex]4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k[/tex]

Find the temperature of regeneration

[tex]\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k[/tex]

Answer:

a) 60.4%;  18.42 kg/s

b) 37.8% ;    35.4 kg/s

Explanation:

a) at an isentropic efficiency of 100%.

Let's first find the exit temperature of the compressor T2, using the formula:

[tex](r_p) ^k^-^1^/^k = \frac{T_2}{T_1}[/tex]

Solving for T2, we have:

[tex] T_2 = 300 * (8)^1^.^6^6^7^-^1^/^1^.^6^6^7 = 689.3 K [/tex]

Let's now find the work dine by the compressor.

[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]

The actual work done by the compressor =

[tex] W_c = 1 * 2020.4 = 2020.4 KJ/kg [/tex]

Let's find the temperature at the exit of the turbine, T4

[tex](r_p) ^k^-^1^/^k = \frac{T_3}{T_4}[/tex]

Solving for T4, we have:

[tex]T_4 = \frac{1800}{(8)^1^.^6^6^7^-^1^/^1^.^6^6^7} = 783.3 K[/tex]

Let's find the work done by the turbine.

[tex]\frac{W_t}{m} = c_p(T_3 - T_4)[/tex]

[tex]\frac{W_t}{m} = 5.19(1800 - 783.3) = 5276.6 KJ/kg[/tex]

The actual work done by the turbine:

= 1 * 5276.6 = 5276.6 KJ/kg

Let's find the regeneration temperature, using the formula:

[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]

Substituting figures, we have:

[tex] 0.75 = \frac{T_r - 689.3}{783.3 - 689.3} [/tex]

[tex] T_r = [0.75(783.3 - 689.3)] + 689.3 = 759.8 [/tex]

Let's calculate the heat supplied.

[tex]Q = c_p(T_3 - T_r)[/tex]

[tex] Q = 5.19(1800 - 759.8) [/tex]

Q = 5388.2 kJ/kg

For thermal efficiency, we have:

[tex] n = \frac{W_t - W_c}{Q} [/tex]

Substituting figures, we have:

[tex] n = \frac{5276.6 - 2020.4}{5388.2} = 0.604 [/tex]

0.604 * 100 = 60.4%

For mass flow rate:

Let's use the formula:

[tex] m = \frac{W_n_e_t}{P} [/tex]

Wnet = 60MW = 60*1000

[tex] m = \frac{60*10^3}{5276.6 - 2020.4} = 18.42 [/tex]

b) at an isentropic efficiency of 80%.

Let's now find the work done by the compressor.

[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]

The actual work done by the compressor =

[tex] W_c = \frac{2020.4}{0.8}= 2525.5 KJ/kg [/tex]

Let's find the work done by the turbine.

[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]

[tex] \frac{W_t}{m} = 5.19(1800 - 787.5) = 5276.6 KJ/kg[/tex]

The actual work done by the turbine:

= 0.8 * 5276.6 = 4221.2 KJ/kg

Let's find the exit temperature of the compressor T2, using the formula:

[tex]\frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] 2525.5 = 5.19(T_2 - 300) [/tex]

Solving for T2, we have:

[tex] T_2 = \frac{2525.5 + 300}{5.19} = 787.5 [/tex]

Let's find the temperature at the exit of the turbine, T4

[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]

[tex] 4221.2 = 5.19(1800 - T_4) [/tex]

Solving for T4 we have:

[tex] T_4 = 958 K[/tex]

Let's find the regeneration temperature, using the formula:

[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]

Substituting figures, we have:

[tex] 0.75 = \frac{T_r - 787.5}{985 - 787.5} [/tex]

[tex] T_r = [0.75(958 - 787.5)] + 787.5 = 935.5 K [/tex]

Let's calculate the heat supplied.

[tex]Q = c_p(T_3 - T_r)[/tex]

[tex] Q = 5.19(1800 - 935.5) [/tex]

Q =  4486.2 kJ/kg

For thermal efficiency, we have:

[tex] n = \frac{W_t - W_c}{Q} [/tex]

Substituting figures, we have:

[tex] n = \frac{4221.2 - 2525.2}{4486.2} = 0.378 [/tex]

0.378 * 100 = 37.8%

For mass flow rate:

Let's use the formula:

[tex] m = \frac{W_n_e_t}{P} [/tex]

Wnet = 60MW = 60*1000

[tex] m = \frac{60*10^3}{4221.2 - 2525.2} = 35.4 kg/s [/tex]

Answer: Because of the role the base region play in the transistor.

Explanation:

The base region of BJT transistor - an opposite polarity charge carrier from emitter region to collector region, plays a vital role in triggering for a sufficient emiter - to - collector current.

The current received by the base region of BJT determines the effect of the continue flow of current into the collector region which will eventually determine the output current.

Answer:

A counter which counts from 0 to 255 with seven segment display

Timer Mode Control (TMOD)

Explanation:

Answer:

Using 1's complement

a)

Therefore the difference is -10001

b)

Therefore the difference is 00001

c)

Therefore the difference is 01001

d)  

Therefore the difference is 10100

e)

Therefore the difference is 00111

Explanation:

Using 1's complement

a) The 1's complement of the subtrahend 11010 = 00101.

Therefore 01001-11010 = 01001 + 00101 = 01110

Since no overflow, we take the 1's complement of the result and it is negative.

Therefore the difference is -10001

b) The 1's complement of the subtrahend 11001 = 00110.

Therefore 11010-11001 = 11010 + 00110 =1 00000

Since there is an overflow, we add the overflow to the result

Therefore the difference is 00001

c) The 1's complement of the subtrahend 01101 = 10010

Therefore 10110-01101 = 10110 + 10010 =  1 01000

Since there is an overflow, we add the overflow to the result

Therefore the difference is 01001

d)  The 1's complement of the subtrahend 00111 = 11000

Therefore 11011-00111= 11011 + 11000 =  1 10011

Since there is an overflow, we add the overflow to the result

Therefore the difference is 10100

e) The 1's complement of the subtrahend 10101 = 01010

Therefore 11100-10101= 11100 + 01010 = 1 00110

Since there is an overflow, we add the overflow to the result

Therefore the difference is 00111

Using 2's complement

a) The 2's complement of the subtrahend 11010 = 00110.

Therefore 01001-11010 = 01001 + 00110 = 01111

Since no overflow, we take the 2's complement of the result and it is negative.

Therefore the difference is -10001

b) The 2's complement of the subtrahend 11001 = 00111.

Therefore 11010-11001 = 11010 + 00111 =1 00001

Since there is an overflow, we drop the overflow

Therefore the difference is 00001

c) The 1's complement of the subtrahend 01101 = 10011

Therefore 10110-01101 = 10110 + 10011 =  1 01001

Since there is an overflow, we drop the overflow

Therefore the difference is 01001

d)  The 1's complement of the subtrahend 00111 = 11001

Therefore 11011-00111= 11011 + 11001 =  1 10100

Since there is an overflow, we drop the overflow

Therefore the difference is 10100

e) The 1's complement of the subtrahend 10101 = 01011

Therefore 11100-10101= 11100 + 01011 = 1 00111

Since there is an overflow, we drop the overflow

Therefore the difference is 00111

Answer:

a) I₂ = 2 mA   (The current has decreased)

b) L₂ = 2.4 cm

Explanation:

Consider the old wire as wire 1 and the new wire as wire 2. The data that we have from the question is:

Current through wire 1 = I₁ = 12.5 mA

Diameter of wire 1 = d₁ = 5 mm

Length of wire 1 = L₁ = 15 cm

Cross-Sectional Area of wire 1 = A₁ = πd₁²/4 = π(5 mm)²/4 = 19.63 mm²

Diameter of wire 2 = d₂ = 2 mm

Cross-Sectional Area of wire 2 = A₂ = πd₂²/4 = π(2 mm)²/4 = 3.14 mm²

a)

Length of wire 2 = L₂ = 15 cm

Since, the battery is same. Therefore, the voltage will be same for both wires.

V₁ = V₂

using Ohm's Law (V = IR)

I₁R₁ = I₂R₂

Since resistance of wire is given by formula:  R = ρL/A

Therefore,

I₁ρ₁L₁/A₁ = I₂ρ₂L₂/A₂

where,

ρ = resistivity, and it depends upon material of wire and the material of both wires is same and the length of wires is also same.

Hence,    ρ₁ = ρ₂

and      L₁ = L₂

and the equation becomes:

I₁/A₁ = I₂/A₂

I₂ = I₁A₂/A₁

I₂ = (12.5 mA)(3.14 mm²)/(19.63 mm²)

I₂ = 2 mA

Thus, the current has decreased.

b)

In order to have same current the resistance of both wires must be same:

R₁ = R₂

ρ₁L₁/A₁ = ρ₂L₂/A₂

Since,   ρ₁ = ρ₂

Therefore,

L₁/A₁ = L₂/A₂

L₂ = L₁A₂/A₁

L₂ = (15 cm)(3.14 mm²)/(19.63 mm²)

L₂ = 2.4 cm

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

V = 36.4 m³ = 4.86 gallons

W = 80193.88 lbf = 356720 N = 356.72 KN

Explanation:

We have the following data given in the question:

At = Total area of roof =  1967 ft²

h = Annual Rainfall = 14 inches = 1.17 ft

V = Volume of tank in m³  and gallons = ?

W = Weight of water in N and lbf = ?

So, for volume we know that the area of roof that receives rainfall is 56% of total area and 14 inches of annual rainfall means  that there is a standing height of 14 inches of rain water for a given area, for 1 year.

Area to receive rain = A = 0.56*1967 ft² = 1101.52 ft²

Now,

Volume = V = A * h = 1101.52 ft²)(1.17 ft)

V = 1285.11 ft³

Converting to m³:

V = (1285.11 ft³)(1 m³/35.3147 ft³)

V = 36.4 m³

Converting to gallons:

V = (1285.11 ft³)(1 m³/264.172 gal)

V = 4.86 gal

Now, for the weight of water, we use formula:

W = ρVg

where,

W = weight of water = ?

ρ = Density of water = 1000 kg/m³

V = Volume of tank = 36.4 m³

g = 9.8 m/s²

Therefore,

W = (1000 kg/m³)(36.4 m³)(9.8 m/s²)

W = 356720 N = 356.72 KN

Converting to lbf:

W = (356720 N)(1 lbf/4.44822 N)

W = 80193.88 lbf