The flow rate of liquid metal into the downsprue of a mold is 500 cm3/s. The cross-sectional area at the top of the sprue is 8 cm2, and the height of the sprue is 15 cm. What should the cross-sectional area be at the base of the sprue in order to avoid separation during pouring of the molten metal

Answer:

6.4 m/s

Explanation:

From the equation of continuity

A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively

[tex]A1=\pi (r1)^{2}[/tex]

[tex]A2=\pi (r2)^{2}[/tex]

where r1 and r2 are radius of inlet and outlet respectively

v1 is given as 1.6 m/s

Therefore

[tex]\pi (0.01)^{2}\times 1.6 = \pi (0.005)^{2}v2[/tex]

[tex]V2=\frac {\pi (0.01)^{2}\times 1.6}{\pi (0.005)^{2}}=6.4 m/s[/tex]

Answer:

magnitude of the maximum stress is 3263 MPa

Explanation:

given data

radius of curvature = 5.5 × [tex]10^{-4}[/tex] mm

crack length = 5 × [tex]10^{-2}[/tex] mm

tensile stress = 220 MPa

to find out

magnitude of the maximum stress

solution

we know that magnitude of the maximum stress is express as

magnitude of the maximum stress = [tex]2\sigma_o ( \frac{\alpha }{2 \rho} )^{0.5}[/tex]      ..........................1

here σo is tensile stress and α is crack length and ρ is radius of curvature

so put all value in equation 1 we get

magnitude of the maximum stress = [tex]2*220 ( \frac{5.5*10^{-2}}{2 *5*10^{-4}} )^{0.5}[/tex]

solve it we get

magnitude of the maximum stress = 3263 MPa

so magnitude of the maximum stress is 3263 MPa

Answer:The awnser is 5

Explanation:Just divide all of it

Answer:

5778.86W

Explanation:

Hi!

To solve this problem follow the steps below, the procedure is attached in an image

1. Draw the complete outline of the problem.

2. find the specific weights of the water and its density using T = 20C, using thermodynamic tables

note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

3. use the Bernoulli equation to find the height of the pump's power, taking into account that the potential and kinetic energy changes are insignificant

4. find the ideal pump power

5. find the real power of the pump using the efficiency equation

Answer:

import java.util.HashMap;

import java.util.Map;

import java.util.Scanner;

public class PhoneBook {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       Map<String, String> map = new HashMap<>();

       String name, number, choice;

       do {

           System.out.print("Enter name: ");

           name = in.next();

           System.out.print("Enter number: ");

           number = in.next();

           map.put(name, number);

           System.out.print("Do you want to try again(y or n): ");

           choice = in.next();

       } while (!choice.equalsIgnoreCase("n"));

       System.out.print("Enter name to search for: ");

       name = in.next();

       if (map.containsKey(name)) {

           System.out.println(map.get(name));

       } else {

           System.out.println(name + " is not in the phone book");

       }

   }

}

Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

[tex]V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}[/tex]

V = 7.07  m/s

from bernou;ii's theorem we have

[tex]\frac{p_1}{\gamma}  +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma}  +\frac{V_2^2}{2g} + z_2 + h_l[/tex]

as we know pipe is horizontal and with constant velocity so we have

[tex]\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}[/tex]

[tex]P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma[/tex]

[tex]135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81[/tex]

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have [tex]\epsilon  = 0.15 mm[/tex]

[tex]\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025[/tex]

reynold number is

[tex]Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}[/tex]

Re = 378750

from moody chart

[tex]For Re = 378750 and \frac{\epsilon}{d} = 0.0025[/tex]

[tex]f_{new} = 0.025[/tex]

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

from moody chart

we have [tex]\frac{\epsilon}{d} =0.006[/tex]

[tex]\epsilon = 0.006 \times 60[/tex]

[tex]\epsilon = 0.36 mm[/tex]

thus pipe is old one with increased roughness

Answer:

L=0.0074 mm

Explanation:

Given that

d= 6.5 micron meter

d= 6.5 x 10⁻³ mm

Yield strength ,Sy = 73 MPa

Bond strength ,σ = 32 MPa

The fiber length given as

[tex]L=\dfrac{S_y \times d}{2 \times \sigma }[/tex]

By putting the values

[tex]L=\dfrac{S_y \times d}{2 \times \sigma }[/tex]

[tex]L=\dfrac{73 \times 6.5\times 10^{-3}}{2\times 32 }\ mm[/tex]

L=0.0074 mm

Therefore the minimum fiber length is 0.0074 mm.

Answer:

full load current = 7.891151 A

so correct option is b. Ir= 7.89

Explanation:

given data

Energy E = V = 240 V

frequency =  60 Hz

full-load slip = 0.02

full-load speed = 1764 rpm

winding resistance = 0.6 Ω

winding reactance = 5 Ω

to find out

rotor current at full-load

solution

we will apply here full load current formula that is express as

full load current = [tex]\frac{S*E}{\sqrt{R^2+ (S*X)^2}}[/tex]    ...................1

here S is full-load slip and E is energy given and R is winding resistance and    X is winding reactance

put here value we get

full load current = [tex]\frac{0.02*240}{\sqrt{0.6^2+ (0.02*5)^2}}[/tex]  

full load current = 7.891151 A

so correct option is b. Ir= 7.89

Answer:

0.3

Explanation:

Please see attachment.

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

[tex]W = \dot{m}c_p \Delta T[/tex]

Where

[tex]C_p =[/tex] Specific heat

[tex]\dot{m}[/tex]= Mass flow rate

[tex]\Delta T[/tex] = Change in Temperature

Our values are given as

[tex]C_p = 1.005kJ/kgK \rightarrow[/tex] Specific heat of air

[tex]T_1 = 50\°C[/tex]

[tex]\dot{m} = 2kg/s[/tex]

[tex]W = 8kW[/tex]

From our equation we have that

[tex]W = \dot{m}c_p \Delta T[/tex]

[tex]W = \dot{m}c_p (T_2-T_1)[/tex]

Rearrange to find [tex]T_2[/tex]

[tex]T_2 = \frac{W}{\dot{m}c_p}+T_1[/tex]

Replacing

[tex]T_2 = \frac{8}{2*1.005}+(50+273)[/tex]

[tex]T_2 = 326.98K \approx 53.98\°C[/tex]

Therefore the exit temperature of air is 53.98°C

Answer:

Stories that discuss how he dealt with a crisis

Stories that illustrate how he went above and beyond expectations

Stories that discuss how he handled a tough interpersonal situation

Explanation:

For effective story-telling during an interview, Jamar needs to highlight stories by discussing how he dealt with certain crisis which relates to the job and tell the panel why he opted for a particular solution. The solution should provide a long-term solution to the crisis. He also needs to illustrate stories on how he went above and beyond expectations  to prove that he's creative. Moreover, it's important that he highlights how he handled a tough interpersonal situation to show how he deals with situations at a personal level.

Answer:

books = []

   fp = open("bookTitles.txt")

   for line in fp.readlines():

       title = line.strip()

       if title not in books:

           books.append(title)

   fp.close()

   fout = open("noDuplicates.txt", "w")

   for title in books:

       print(tile, file=fout)

   fout.close()

except FileNotFoundError:

   print("Unable to open bookTitles.txt")

Answer:

a) [tex]-7.4\frac{lb}{ft^3}[/tex]

b) [tex]-69.8\frac{lb}{ft^3}[/tex]

c) [tex]55 \frac{lb}{ft^3}[/tex]

Explanation:

1) Notation

[tex]\tau=1.85\frac{lb}{ft^2}[/tex] represent the shear stress  defined as "the external force acting on an object or surface parallel to the slope or plane in which it lies"

R represent the radial distance

L the longitude

[tex]\theta=0\degree[/tex] since at the begin we have a horizontal pipe, but for parts b and c the angle would change.

D represent the diameter for the pipe

[tex]\gamma=62.4\frac{lb}{ft^3}[/tex] is the specific weight for the water  

2) Part a

For this case we can use the shear stress and the radial distance to find the pressure difference per unit of lenght, with the following formula

[tex]\frac{2\tau}{r}=\frac{\Delta p -\gamma Lsin\theta}{L}[/tex]

[tex]\frac{2\tau}{r}=\frac{\Delta p}{L}-\gamma sin\theta[/tex]

If we convert the difference's into differentials we have this:

[tex]-\frac{dp}{dx}=\frac{2\tau}{r}+\gamma sin\theta[/tex]

We can replace [tex]r=\frac{D}{2}[/tex] and we have this:

[tex]\frac{dp}{dx}=-[\frac{4\tau}{D}+\gamma sin\theta][/tex]

Replacing the values given we have:

[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin0]=-7.4\frac{lb}{ft^3}[/tex]

3) Part b

When the pipe is on vertical upward position the new angle would be [tex]\theta=\pi/2[/tex], and replacing into the formula we got this:

[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin90]=-69.8\frac{lb}{ft^3}[/tex]

4) Part c

When the pipe is on vertical downward position the new angle would be [tex]\theta=-\pi/2[/tex], and replacing into the formula we got this:

[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin(-90)]=55 \frac{lb}{ft^3}[/tex]

Answer:

Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for.  Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.

Explanation:

Every individual key is a big character  so to compare every keys, it is required to conduct a quite time consuming string reference procedure at every node. Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for.  Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.

The question pertains to the use of functions and pointers in programming. The task involves writing a function call that manipulates the tens and ones digit of an integer by passing the first two arguments as pointers. An example solution involves creating a function named ExtractDigits to perform this task and using pointers to directly modify specified variables.

The question refers to the concept of functions and pointers in programming, which is an advanced topic typically covered at the college level in computer science or engineering courses. The task is to write a function call that uses pointers for the first two arguments, tensPlace and onesPlace, and a regular argument for userInt. This function could be crafted to manipulate or assess the tens and ones place of a given integer, userInt.

Suppose we have a function named ExtractDigits designed to extract the tens and ones place of an integer. This function might look like:

void ExtractDigits(int* tensPlace, int* onesPlace, int userInt){
 *tensPlace = (userInt / 10) % 10;
 *onesPlace = userInt % 10;
}

To call this function with the variables tensPlace, onesPlace, and a specific integer (for example, 41), you can use the following code:

int main() {
 int tens, ones;
 ExtractDigits(&tens, &ones, 41);
 printf("tensPlace = %d, onesPlace = %d", tens, ones);
 return 0;
}

This function call passes the addresses of tens and ones to the function so that their values can be modified directly, which is a fundamental use of pointers in C and C++ programming.

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2  =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV  =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f  = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

Answer:

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes

Explanation:

In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:

[tex]Efficiency=\frac{work}{heat transfer to power plant}[/tex]

[tex]Heat transfer=\frac{work}{Efficiency\\} \\\\Heat transfer=\frac{800}{0.40}\\\\Heat transfer=2000MW[/tex]

From First law of thermodynamics:

Rate of heat transfer to river=heat transfer to power plant-work done

Rate of heat transfer to river=2000-800

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.

To solve this problem it is necessary to apply the concepts related to the normal effort and the shear effort due to failure. The shear stress can be defined based on normal stress and effective friction angle. Mathematically it can be defined as

[tex]\tau_f = \sigma_n tan\phi[/tex]

Where

[tex]\tau_f =[/tex] Shear stress at failure

[tex]\sigma_n =[/tex] Normal stress

[tex]\phi =[/tex] Effective friction angle

PART A) Our values are given as ,

[tex]\sigma_n = 192kPa[/tex]

[tex]\tau_f  = 120kPa[/tex]

Replacing at the previous equation we have,

[tex]\tau_f = \sigma_n tan\phi[/tex]

[tex]120 = 192tan\phi[/tex]

[tex]\phi = tan^{-1}(\frac{120}{192})[/tex]

[tex]\phi = 32.27°[/tex]

Therefore the effective friction angle of the sand is 32.27°

B) Using the maximum possible effort and the angle previously given we can calculate the maximum shear force, from which from its definition it is possible to find the force.

[tex]\tau_f = \sigma_n tan\phi[/tex]

[tex]\tau_f = 200tan(32\°)[/tex]

[tex]\tau_f = 124.97kPa[/tex]

With the value of the shear stress from its basic definition we can find the force. By definition the shear stress is given by

[tex]\tau_f = \frac{F}{A}[/tex]

Re-arrante to find the Force

[tex]F = A\tau_f[/tex]

The shear stress is a function of the Force and the Area, therefore, the area would be the square of the sides (50mm)

Replacing in the equation we have to,

[tex]F = 124.97*10^{-3}*(50*50)[/tex]

[tex]F = 312.425N[/tex]

Therefore the shear force required to cause failure of the specimen is 312.425N

Answer:

a) RC = 1.03 mseg.

b) Qmax = CV = 85.2 μC

c) Q = 53.9 μC

Explanation:

a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.

This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.

In this case , it can be calculated as follows:

ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.

b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:

Q = C V = 11.5 μF . 7.41 V = 85.2 μC

c) During the charging process, the charge increases following this equation:

Q = CV (1 - e⁻t/RC)

When t = RC, the expression for Q is as follows:

Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is    [tex]= 1.751 \times 10^{-5} \% per hr[/tex]

Explanation:

we know Arrhenius expression is given as

[tex]\dot \epsilon =Ce^{\frac{-Q}{RT}[/tex]

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is[tex] \dot \epsilon = 5.5\times 10^{-2} [/tex]% per hr

At 800 degree C  creep rate is[tex] \dot \epsilon = 1 [/tex]% per hr

activation energy for creep is [tex]\frac{\epsilon_{800}}{\epsilon_{700}}[/tex] = [tex]= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}[/tex]

[tex]\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}[/tex]

[tex]\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}][/tex]

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

[tex]C = \epsilon e^{\frac{Q}{RT}}[/tex]

    [tex]=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr[/tex]

[tex]\epsilon_{500} = C e^{\frac{Q}{RT}}[/tex]

                         [tex]= 1.804 \times 10^{12}  e{\frac{251758}{8.314(500+273}}[/tex]

                         [tex]= 1.751 \times 10^{-5} \% per hr[/tex]

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

Answer:

[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]

Explanation:

Lets take

Length of pipe B = L

Length of pipe A = 5 L

Discharge in pipe A = Q₁

Discharge in pipe B = Q₂

We know that head loss in the pipe given as

[tex]h_f=\dfrac{FLQ^2}{12.1d^5}[/tex]

F=Friction factor, Q=Discharge,L=length

d=Diameter of pipe

here all only Q and L is varying and all other quantity is constant

So we can say that

LQ²= Constant

L₁Q₁²=L₂Q²₂

By putting the values

5LQ₁²=LQ²₂

[tex]\dfrac{Q_2}{Q_1}=\sqrt{5}[/tex]

Therefore

[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]

Answer:

a) [tex]h_L=-3.331ft[/tex]

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

[tex]p_a =31.1psi=4478.4\frac{lb}{ft^2}[/tex]

[tex]p_b =27.3psi=3931.2\frac{lb}{ft^2}[/tex]

For above conversions we use the conversion factor [tex]1psi=144\frac{lb}{ft^2}[/tex]

[tex]z_a =56.7ft[/tex]

[tex]z_a =68.8ft[/tex]

[tex]h_L =?[/tex] head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

[tex]\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L[/tex]

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

[tex]\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L[/tex]

3)Part a

And on this case we have all the values in order to replace and solve for [tex]h_L[/tex]

[tex]\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L[/tex]

[tex]h_L=(71.769+56.7-63-68.8)ft=-3.331ft[/tex]

4)Part b

Analyzing the value obtained for [tex]\h_L[/tex] is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

Answer:

Explanation:

1. The problems in the above pseudocode include (but not limited to) the following

1. The end year is not properly represented.

2. Though, the factor of 5 is declared, it's not implemented as increment in the pseudocode

3. Incorrect use of control structures. (While ...... And .......Endif)

2.

Start

Start_Year = 2017

Kount = 1

While Kount <= 30

Display Start_Year

Start_Year = Start_Year + 5

Kount = Kount + 1

End While

Stop

3. Yes, by doing the following

* Apply increment of 5 to start year within the whole loop, where necessary

* Use matching control structures

While statement ends with End While (not end if, as it is in the question)

4. Using VBA

Dim Start_Year: Start_Year = 2017 ' Declare and initialise year to 2017

Dim Counter: Counter = 1 ' Declare and Initialise Counter to 1

While Count <= 30 ' Test Value of Counter; to check if the desired number of year has gotten to 30. While the condition remains value, the following code will be executed.

msgbox Start_Year ' Display the value of start year

Start_Year = Start_Year + 5 'Increase value of start year by a factor of 5

Counter = Counter + 1 'Increment Counter

Wend

5. Yes

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

[tex]N_F=n_F\times \frac{N_S}{n_S}[/tex] here,

[tex]n_F[/tex] = the number of unseeded faults = 6

[tex]N_S[/tex] = number of seeded faults = 30

[tex]n_s[/tex] = number of seeded faults found = 15

So NF will be calculated as,

[tex]N_F=6\times \frac{30}{15}=12[/tex]

And the estimate of faults remaining is  [tex]N_F-n_F[/tex] = 12 - 6 = 6

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. Assume that the air conditioner operates steadily.

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

a) The COP of the air conditioner is 2.083.

b) Rate of heat transfer to the outside air is 18.5 kW.

Step 1

Given Data:

- Heat removal rate from the house, [tex]\( \dot{Q}_{in} = 750 \, \text{kJ/min} \)[/tex]

- Electric power input, [tex]\( \dot{W}_{in} = 6 \, \text{kW} \)[/tex]

Converting Units:

[tex]\[ \dot{Q}_{in} = 750 \, \text{kJ/min} = \frac{750 \, \text{kJ}}{60 \, \text{s}} = 12.5 \, \text{kW} \][/tex]

Calculations:

(a) Coefficient of Performance (COP):

The COP of an air conditioner is defined as the ratio of the heat removed from the house to the work input (electric power):

[tex]\[ \text{COP} = \frac{\dot{Q}_{in}}{\dot{W}_{in}} \][/tex]

Step 2

Substituting the given values:

[tex]\[ \text{COP} = \frac{12.5 \, \text{kW}}{6 \, \text{kW}} = 2.083 \][/tex]

(b) Rate of Heat Transfer to the Outside Air:

The rate of heat transfer to the outside air, [tex]\( \dot{Q}_{out} \)[/tex], can be determined using the first law of thermodynamics for a steady-state system:

[tex]\[ \dot{Q}_{out} = \dot{Q}_{in} + \dot{W}_{in} \][/tex]

Substituting the given values:

[tex]\[ \dot{Q}_{out} = 12.5 \, \text{kW} + 6 \, \text{kW} = 18.5 \, \text{kW} \][/tex]

Answer:

Q = 2.532 x 10³ W

Explanation:

The properties of air at the film temperature of

T(f) = (T(s) + T()∞)/2

T(f) = (120 + 20)/2 = 70 °C

From the table of properties of air at T = 20 °C we know that ρ = 1.028 kg/m³ , k = 0.02881W/mK, Pr = 0.7177 , v = 1.995 x 10⁻⁵ m²/s

Calculate the Reynold’s Number

Re(l) = V x L/ν, where V is the speed of air flowing, L is the length of the surface in meters and ν is the kinematic viscosity

Re(l) = 62 x 0.45/(1.995 x 10⁻⁵) = 1.398 x 10⁶

Re(l) is greater than the critical Reynold Number. Thus, we have combined laminar and turbulent flow and the average Nusselt number for the entire plate is determined by

Nu = (0.037 x (Re(l)^0.8) - 871) x Pr^1/3, where Pr is the Prandtl’s Number

Nu = (0.037 x (1.398 x 10⁶)^0.8) – 871) x (0.7177)^1/3

Nu = 1952.85

We also know that

Nu = h x L/k, where h is the heat transfer coefficient, L is the length of the surface and  k is the thermal conductivity

Rearranging to solve for h

h = (Nu x k)/l

h = 1952.85 x 0.02881/0.45 = 125.03 W/m²C

The heat transfer rate Q, may be determined by

Q = h x A x (T(s) - T(∞))

Q = 125.03 x 0.45 x 0.45 x (120 - 20)

Q = 2.532 x 10³ W

The heat transfer rate is 2.532 x 10³ W to the air.