Consider a 2.4-kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.10/kWh and $1.20/therm (1 therm = 105,500 kJ), respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.

Answer:

a) x = 5.7791 m ( this a turbulent flow )

b) x = 0.04 m ( this is a Laminar flow )

Explanation:

For Air, take

p = 1.2 kg/m³ and u = 1.8×10⁻⁵ kg/m³

So for A , x = 10 cm

Guess turbulent flow;

(Sturb / x) = (0.16 / (p∪x / u)^1/7)

we substitute

(0.1/x = 0.16) / ( 1.2 × 15 × x / 1.8×10⁻⁵)^1/7

x = 5.7791 m

CHECK

Re = (1.2 × 15 × 5.7791) / 1.8×10⁻⁵ = 5.779×10⁶

Therefore this a turbulent flow

for B, x = 1mm

Guess Laminar flow

(S-laminar / x) = ( 5 / (Reₓ)^1/2)

x = (Stutb)²p∪ / 23u

x = ((0.001)² × 1.2 × 15) / (25 × 1.8×10⁻⁵)

x = 0.04 m

CHECK

Re = (1.2 × 15 × 0.04) / (1.8×10⁻⁵)

= 40000

Therefore  this is a Laminar flow

Answer:

technicians

Explanation:

Just answer this question with this answer...

I got it right so you should too...

Answer:

Technicians

Explanation:

Answer:

The answer ix below

Explanation:

There are 12 apple pies left.

Given that:

n = number of apple pies left = 12

x = number of food banks = 6

1) For the 12 apple pies to be distributed among 6 food banks. The number of ways in which this can be done is:

C(n + x - 1, x - 1) = C(12 + 6 -1, 6 - 1) = C(17, 5) = [tex]\frac{17!}{(17-5)!5!} =\frac{17!}{12!5!}=6188 \ ways[/tex]

12 apple pies can be distributed among 6 food banks in 6188 ways

2) For the 12 apple pies to be distributed among 6 food banks if each food bank must receive one pie, 6 pies would be remaining. The number of ways in which this can be done is:

C((n - x) + x - 1, n - x) = C(12 - 6 + 6 - 1, 12 - 6) = C(11, 6) = [tex]\frac{11!}{(11-6)!6!} =\frac{11!}{6!5!}=462 \ ways[/tex]

12 apple pies can be distributed among 6 food banks if each food bank must receive one pie in 462 ways

Answer:

Problem definition

Explanation:

The engineering design is a series of step that the engineer gradually take in other to create a functioning product or process. There are many different models of the engineering design process , but they can be shrunk into four basic steps which are

The students are within the first step, and the problem here is 'an appropriate means for the students to keep their drinks cold, for 3 hours for the ball game.'

Answer:

some parts of your question is missing attached below is the missing part

Answer :  Fa = 4.46 Ib

Explanation:

use the equation

Z = 0.1 sin2∅

next we will differentiate the equation to get the locus of the velocity

z = 0.2 cos2∅∅

differentiate the equation furthermore to get the locus of acceleration in the horizontal axis

z = -0.4sin2∅(∅)^2 + 0.2cos2∅∅

note : express ∅ as 6 rad.[tex]s^{-1}[/tex] for angular velocity and ∅ = 0 for angular acceleration

equation above becomes :

Z = - 0.4 sin 2∅ ( 6)^2 + 0.2 cos 2∅(0)

  = - 14.4 sin 2∅ ( acceleration of the follower in horizontal direction )

next calculate The force at the end of A of the follower

Fa - Kx = mz  

note: m = w / g hence : Fa - Kx = w/g z  ------- (2)

w = weight of the spring-held follower = 0.75 Ib

x = compression of the spring = 0.4

k = spring stiffness = 12 Ib/ft

∅ = 45⁰

g = 32.2 ft/s^2

input these values into equation 2

hence : Fa = 4.46 Ib ( force at the end A of the follower )

Answer:

torque to raise the load = 16.411 Nm

torque to lower the load = 8.40 Nm

overall efficiency = 0.24

Explanation:

Given:

max load on vertical direction of the screw = Force = F = 5kN

frictional  diameter of the collar = 45 mm

Diameter = 25 mm

length of pitch = 5 mm

coefficient of friction for thread µ  = 0.09

coefficient of friction for collar µ[tex]_{c}[/tex] = 0.06

To find:

torque to "raise" the load

torque to and "lower"

overall efficiency

Solution:

Compute torque to raise the load:

[tex]T_{R} = \frac{ Fd_{m}}{2} (\frac{L+(\pi ud_{m}) }{\pi d_{m}-uL }) +\frac{Fu_{c} d_{c} }{2}[/tex]

where

[tex]T_{R}[/tex] is the torque

F is the load

[tex]d_{m}[/tex] is diameter of thread

[tex]d_{c}[/tex] is diameter of collar

L is the thread pitch distance

µ is coefficient of friction for thread

µ[tex]_{c}[/tex]  is coefficient of friction for collar

Putting the values in above formula:

[tex]T_{R}[/tex] = 5(25) / 2 [5+ (π(0.09)(25) / π(25)-0.09(5)] + 5(0.06)(45) / 2

    = 125/2 [5 + (3.14)(0.09)(25)/ 3.14(25)-0.45] + 13.5/2

    = 62.5 [(5 + 7.065) / 78.5 - 0.45] + 6.75

    = 62.5 [12.065 / 78.05 ] + 6.75

    = 62.5 (0.15458) + 6.75

    = 9.66125 + 6.75

    = 16.41125

[tex]T_{R}[/tex] = 16.411 Nm

Compute torque to lower the load:

[tex]T_{L} = \frac{ Fd_{m}}{2} (\frac{(\pi ud_{m}) - L }{\pi d_{m}-uL }) +\frac{Fu_{c} d_{c} }{2}[/tex]

     = 5(25) / 2 [ (π(0.09)(25) - L / π(25)-0.09(5) ] + 5(0.06)(45) / 2

     = 125/2 [ ((3.14)(0.09)(25) - 5) / 3.14(25)-0.45 ] + 13.5/2

     = 62.5 [ (7.065 - 5) / 78.5 - 0.45 ] + 6.75

    = 62.5 [ 2.065 / 78.05 ] + 6.75

     = 62.5 (0.026457) + 6.75

     = 1.6535625 + 6.75

     = 8.40 Nm

Since the torque required to lower the the load is positive indicating that an effort is applied to lower the load, Hence the thread is self locking.

Compute overall efficiency:

overall efficiency = F(L) / 2π [tex]T_{R}[/tex]

                             = 5(5) / 2(3.14)( 16.411)

                             = 25/ 103.06108

overall efficiency = 0.24

Answer:

the base-emitter junction is open and the emitter resistor is open

Explanation:

Because here there will be no base current only when the the base emitter junction is kept open and. Also when emitter resistor is kept open or with thus there will be no voltage drop across the Resistor meaning the base voltage Will be equal to that in the voltage divider circuitry

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

h₁ = 763.088 kJ/kg

s₁ = 4.287 kJ/(kg·K)

Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

h₂ = 79.572 kJ/kg

The saturation temperature at the given

T₂ = 79°C

The rate of work output [tex]\dot W[/tex] = [tex]\dot m[/tex]×[tex]c_p[/tex]×(T₂ - T₁)

Where;

[tex]c_p[/tex] = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)

[tex]\dot m[/tex] =  The mass flow rate = 2.0 kg/s

Substituting the values, we have;

[tex]\dot W[/tex] = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

[tex]\dot W[/tex] = -396.17 kJ/s

Answer:

One cycle for a sine wave is 360o or 2 radians. a) A sine wave with maximum amplitude at time t=0. The amplitude of a sine wave is maximum at the peak of the wave. Case 1: assuming that the wave is starting its cycle at t=0 then there is no phase shift for the wave at time t=0 without considering the amplitude...

Answer:

B.  1.75 microfarads

Explanation:

When capacitors are connected in parallel, the total capacitance is equivalent to the sum of each individual capacitor's capacitance.

With this in mind, we have the values, 0.1, 0.6, 1.0, and 0.05, all microfarads.  So simply, we need to sum these values.

0.1 + 0.6 + 1.0 + 0.05

= 0.7 + 1.0 + 0.05

= 1.7 + 0.05

= 1.75

So the total capacitance for this circuit would be 1.75 microfarads.

Cheers.

Answer:

Water Hammer is a knocking sound in an water pipe which occurs when the tap is turned off briskly

Explanation:

Answer:

  5 microhenries

Explanation:

The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.

  10 uH ║ 10 uH = 5 uH

The effective inductance is 5 uH.

Answer:

the turbine work in MW is   [tex]\mathbf{W_t = 10.47299 \ kW}[/tex]

Explanation:

Given that:

the inlet temperature = 500°C

pressure = 600 kPa

exit temperature = 45°С

mass flow rate = 11 kg/s

The objective is to find the turbine work  in MW given that the mass flow rate of the steam is at 11 kg/s.

From the steam tables, the data obtained for the enthalpies at the inlet temperature of  500°C and a pressure of  600 kPa is:

[tex]\mathtt{h_1 : 3482.7 \ kJ/kg}[/tex]

The turbine expansion process is also the isentropic process, as such:

Inlet entropy [tex]\mathbf{s_1}[/tex] = Exit entropy [tex]\mathbf{s_2}[/tex]

The data obtained from the steam table for [tex]\mathbf{s_1}[/tex] = 8.003 kJ/kg.K

[tex]s_2 = s_{f2}+ x_2 *s_{fg2}[/tex]

The data obtained from the steam tables for this entities are as follows:

[tex]\mathsf{s_{f2} = 0.638 \ kJ/Kg/K }[/tex]

[tex]\mathtt{s_{fg2}=7.528 \ kJ/kg/K}[/tex]

since  [tex]\mathbf{s_1}[/tex] =  [tex]\mathbf{s_2}[/tex] = 8.003 kJ/kg.K

Therefore;

[tex]8.003 = 0.638 + x_2 * 7.528[/tex]

[tex]8.003 -0.638 = 7.528x_2[/tex]

[tex]7.365= 7.528x_2[/tex]

[tex]x_ 2= \dfrac{ 7.365}{7.528}[/tex]

[tex]x_2 = 0.978[/tex]

From the steam tables, the data obtained for the enthalpies at the exit temperature of  45°C and a pressure of  600 kPa is:

[tex]h_{f2}[/tex] = 188.4

[tex]h_{fg2[/tex] =2394.9

Thus;

[tex]h_ 2= 188.4 +0.978*2394.9[/tex]

[tex]h_ 2=2530.61[/tex] kJ/kg

The workdone for the turbine process can be computed as:

[tex]\mathsf{W_t = m(h1-h_2)}[/tex]

[tex]\mathsf{W_t = 11(3482.7-2530.61)}[/tex]

[tex]\mathsf{W_t = 11(952.09)}[/tex]

[tex]\mathsf{W_t = 10472.99}[/tex] kW

To MW, we have

[tex]\mathbf{W_t = 10.47299 \ kW}[/tex]

Explanation:

a) Given B(n) is the number of the length 'n' bit sequences which have no three consecutive 0s(i.e., they does not contain substring “000”)

Any bit string which has no 000 should have a 1 in at least one of the 1st three positions. Then, the n we will break all the bit strings by avoiding the 000 by when the 1st 1 occurs. i.e., each of the bit of string of the length of n will avoid 000 falls into the exactly any one of these cases:

1 is followed by the bit string of the length of (n-1) avoiding the 000.

01 is followed by some bit string of the length of (n-2) avoiding a 000.

001 which is followed by any bit string of the length(n-3) avoiding the 000.

Therefore, recurrence is

B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(0)=1, B(1)=2, B(2)=4

Or

B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(1)=2, B(2)=4, B(3)=7

b) Let the S(n)={1,2,3,…,n}. We will say that the subset A of S(n) is very good if A does not have any of the two integers that are consecutive.

For any k, let a(k) be a number of any good subsets of a S(k).

There are two types of good subsets of S(n).

Type 1 of  good subsets of S(n) which  contain the element n,

Type 2 of good subsets of S(n)  which do not contain n.

We will first get an expression for a number of Type 1 of good subsets of the S(n) , where n≥2. This a subset does not contain n-1. So any Type 1 of the good subset of the S(n)  is obtainable when adding n to good subset of the S(n-2) . Also, on adding n to a good subset of the S(n-2) , we always get a Type 1 good subset of the S(n). Thus there are exactly as many good Type 1 of subsets of S(n) as there is good subsets of S(n-2) . By the definition there are a(n-2) good subsets of S(n-2) .

A good subset of a S(n)  is either Type 1 or Type 2. Thus the number of the good subsets of S(n)  is a(n-2)+a(n-1)

We have therefore shown that a(n) = a(n-2)+a(n-1)

c). Here firstly see that if the n is not the multiple of a 3, then there will be no chance to tile the rectangle. And also if n is a multiple of a 3, then there may be two ways to tile the 1st three columns:

And the rest of tilling is the tilling of  2x(n-3) rectangle for which there are T(n-3).

So, the recurrence is

T(n )= {2T(n-3), with T(0)=1     if n=0(mod 3)   or 0 else

We could use the base case T(3)=2

So, the following recurrence can be aslo equivalent to T(n)= 2T(n-3), with T(0)=1, T(1)=0,T(2)=0

Or

T(n)= 2T(n-3), with T(1)=0, T(2)=0,T(3)=2

d)A ternary string may be defined as a sequence of some digits, where each digit has either 0,1,or 2.

According to the problem given, we do not have a 2 anywhere after  0, and the dot which represents the binary string of length(n-1) with the property which we cannot use 2 anywhere after we use the 0.

Now for base case,  we should note that any of the ternary string which has a length of 1 satisfies the given required property. Hence the recurrence is

T(1)=3

T(n)=2T(n-1)+2^(n-1)

Answer:

class helloSarah {

public static void main(String [] args) {

Name = "Hello Sarah";

System.out.print(Name);

}

}

Explanation:

The question seem incomplete; However, since the program is incorrect, I'll assume the question is to correct the given code;

The program segment was written in Java and the correct program is in the answer section

See bold texts for line by line explanation

This line declares the class name

class helloSarah {

This line declares the main method of the program

public static void main(String [] args) {

This line initializes string variable Name

String Name = "Hello Sarah";

This line prints the initialized variable

System.out.print(Name);

}

}

Answer:

class helloSarah {

public static void main(String [] args) {

Name = "Hello Sarah";

System.out.print(Name);

]

Explanation:

Answer:

Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L

Explanation:

Given Data:

Amount of sewage received=500 m^3/d

Surface Area= 10 hectares=10*10^4 m^2

Depth=1 m

Pollutant concentration=200 mg/L

Decay coefficient=0.75 d-1

Required:

Steady-state concentration of the pollutant in the effluent= ?

Solution:

Volume=Surface Area * Depth

[tex]Volume=10*10^4 *1\\Volume=10*10^4\ m^3[/tex]

Time to fill the lagoon=[tex]\frac{Volume}{Amount\ received\ per\ day}[/tex]

[tex]Time\ to\ fill\ the\ lagoon=\frac{10*10^4\ m^3}{500\ m^3}\\ Time\ to\ fill\ the\ lagoon= 200\ days[/tex]

Formula for steady State:

[tex]A_t=\frac{A_0}{1+kt}[/tex]

where:

A_t is the steady state concentration

A_0 is the initial concentration

k is the decay constant

t is the time

[tex]A_t=\frac{200\ mg/L}{1+0.75*200}\\ A_t=1.3245033\ mg/L[/tex]

Steady-state concentration of the pollutant in the effluent= 1.3245033 mg/L

Answer:

In order to logically identify alternative solutions to problems

Explanation:

Electricians are specialized in electrical wiring of buildings, transmission lines, stationary machines, and related equipment. They are either employed in the installations of new electrical components, or to maintain an already installed component. The job of an electrician can be mentally tasking, especially in troubleshooting for fault, and methods of fixing of faults. Some problems might require an out-of-norm approach to solve, and the electrician has to be able to logically identify alternative solutions to problems.

Answer:

in order to logically identify alternative solutions to problems

Explanation:

it makes the most sense

Answer:

A centrifugal pump is a rotating machine that pumps liquid by forcing it through a paddle wheel or propeller called an impeller. It is a most common type of industrial pump. By the effect of the rotation of the impeller, the pumped fluid is drawn axially into the pump, then accelerated radially, and finally discharged tangentially.

Explanation:

Suppliers generally offer charts in the plan, which present the various things at the nominal operating point using two main arrangements: the inductors and the balancing pants.

Answer: provided in the explanation section

Explanation:

The question says;

In a 1-phase UPS, Vd = 350 V, vo(t) = 170 sin(2π * 60t) V, and io(t) = 10 sin(2π * 60t - 30ᴼ) A.Calculate and plot da(t), db(t), vaN(t), vbN(t), Id, id2(t) and id(t). Switching frequency fs = 20 kHz.

Answer

From this we have come to this;

da = Ṽan/Vd = 0.5 + (0.5 * 0.4857) sin w,t

db = Ṽbn/Vd = 0.5 - (0.5 * 0.4857) sin w,t

Ṽan(t) = da * 350 V

Ṽbn(t) = db * 350 V

Id = 0.5 * Ṽo/ Vd * Îo cosΦ1 = 0.5 * (170/350) * 10 * cos 30ᴼ = 2.429 A

Id2 = -0.5 * 170/350 * 10 * sin(2 w,t - 30ᴼ) = -2.429 sin(2 w,t - 30ᴼ)

īd = Id + id2 = 2.429 A + -2.429 sin(2 w,t - 30ᴼ).

Note: Attached is a copy of an image showing and explaining thr plots.

cheers i hope this has been helpful !!!!

Answer:

The amount of shaft work the turbine cam do per second is 3660.29 kJ

Explanation:

The given parameters are;

Pressure at entry p₁ = 300 kPa

The mass flow rate,  [tex]\dot {m}[/tex] = 5.0 kg/sec

The initial temperature, T₁ = 175°C

Therefore;

The enthalpy at 300 kPa and 175°C, h₁ = 2,804 kJ/kg

At the turbine exit, we have;

The pressure at exit, p₂ = 100 kPa

The quality of the steam at exit, in percentage, x₂ = 60%

Therefore, for the enthalpy, h₂ for saturated steam at 100 kPa and quality 60%, we have;

h₂ = 417.436 + 0.6 × 2257.51 = 1771.942 kJ/kg

Heat loss from the turbine, [tex]h_l[/tex]= 1,500 kW

By energy conservation principle we have;

dE/dt = [tex]\dot Q[/tex] - [tex]\dot W[/tex] + ∑[tex]m_i \cdot (h_i[/tex]+ [tex]ke_i[/tex] +[tex]pe_i[/tex]) - ∑[tex]m_e \cdot (h_e[/tex] + [tex]ke_e[/tex] +[tex]pe_e[/tex] )

0 = -[tex]h_l[/tex]  -  [tex]\dot W[/tex] + [tex]m_i \cdot h_i[/tex] - [tex]m_e \cdot h_e[/tex]  

[tex]\dot W[/tex] = [tex]\dot {m}[/tex] × (h₁ - h₂) - [tex]h_l[/tex] = 5.0×(2,804 - 1771.942) - 1500 = 3660.29 kJ/s

The rate of work of the shaft = 3660.29 kJ/s

The amount of shaft work the turbine cam do per second = 3660.29 kJ.

Answer:

(b) NAND

(d) NOR

(f) XNOR

Explanation:

This is best explained using a truth table.

A truth table containing 3 hypothetical inputs and their corresponding outputs using the AND, NAND, OR, NOR, XOR, and XNOR gates, has been attached to this response.

From the table, it can be seen that the NAND, NOR and XNOR gates produce a false value (0) when all inputs are true.

Note:

On the table,

For the AND column, a 1 is being output only when all three inputs are 1. Otherwise, a 0 results.

For the NAND column (which is the negation of the AND), a 0 is being output only when all three inputs are 0. Otherwise, a 1 results.

For the OR column, a 1 is being output when at least one of the three inputs is 1. Otherwise, a 0 results.

For the NOR column (which is the negation of the OR), a 0 is being output when at least one of the three inputs is 0. Otherwise, a 1 results.

For the XOR column, a 1 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 0 results.

For the XNOR column (which is the negation of the XOR), a 0 is being output when the number of inputs that are true (1) is odd. i.e when the number of 1s is 1 or 3. Otherwise, a 1 results.

Answer:

0.067wc

Explanation:

The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction

We substitute values

actual static pressure = 0.1

Total equivalent length = 150 ft

0.1 = (150ft/100) multiplied by Rate of friction

Friction rate at 100ft = 0.067

So we have that the required friction needed is 0.067wc

Answer:

Attachment...?

Explanation:

Answer:

Discharge = area of the pipe or channel × velocity of the liquid

Q = Av

Explanation:

The flow rate of a liquid is a measure of the volume of liquid that moves in a certain amount of time. The flow rate depends on the area of the pipe or channel that the liquid is moving through, and the velocity of the liquid. If the liquid is flowing through a pipe, the area is A = πr2, where r is the radius of the pipe. For a rectangle, the area is A = wh where w is the width, and h is the height. The flow rate can be measured in meters cubed per second (m3/s), or in liters per second (L/s). Liters are more common for measures of liquid volume, and 1 m3/s = 1000 L/s.

Answer:

C

Explanation:

Answer:

B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv

    sending end current : Is = 1.241 ∠ 15.5⁰ KA

    real power = 730.5 Mw

C) percent voltage regulation = 8.7%

D) Transmission line efficiency  = 95.8%

Explanation:

attached is the detailed solution to the problem

Given data:

l = 200 km

z = 0.032 + j0.35 Ω/km

y = j4.2 * 10^-6 S/km

A) find the total series impedance and shunt  admittance

B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv

    sending end current : Is = 1.241 ∠ 15.5⁰ KA

    real power = 730.5 Mw

C) percent voltage regulation = 8.7%

D) Transmission line efficiency  = 95.8%

Answer:

Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.

When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.

Answer:

Explanation:

hola friend!!!!!

there is only one crankshaft in V8 engine

Hope this helps

plz mark as brainliest!!!!!!!!

Answer:

The correct option is A

Explanation:

Thermoplastics are polymers that can be recycled. They can be melted with heat and moulded into any shape which becomes hard when cooled. The heat applied to it in the process of recycling does not affect its chemical composition.

There are two main classes of thermoplastics; semi-crystalline and amorphous.

An amorphous thermoplastic has a disorderly arranged molecular structure which makes it difficult to have a definite melting temperature (Tm). However, a semi-crystalline thermoplastic has a well arranged molecular structure which makes it melt after a certain/definite amount of heat is absolved. Hence, the correct answer is A