A double-pipe heat exchanger is used to cool a hot fluid (cp = 3800 J/kg·K) entering the heat exchanger at 200°C with a flow rate of 0.4 kg/s. In the cold side, cooling fluid (cp = 4200 J/kg·K) enters the heat exchanger at 10°C with a mass flow rate of 0.5 kg/s. The double-pipe heat exchanger has a thin-walled inner tube, with convection heat transfer coefficients inside and outside of the inner tube estimated to be 1400 W/m2 ·K and 1100 W/m2 ·K, respectively. The heat exchanger has a heat transfer surface area of 2.5 m2 , and the estimated fouling factor caused by the accumulation of deposit on the surfaces is 0.0002 m2 ·K/W. (a) Determine the effectiveness values for the parallel- and counter-flow configurations. (b) Determine outlet temperatures of the hot fluid for the parallel- and counter-flow configurations. (c) Determine outlet temperatures of the cold fluid for the parallel- and counter-flow configurations.

Answer:

See explaination

Explanation:

thermal conductivity is A measure of the ability of a material to transfer heat. Given two surfaces on either side of the material with a temperature difference between them, the thermal conductivity is the heat energy transferred per unit time and per unit surface area, divided by the temperature difference.

Please kindly check attachment for the step by step solution of the given problem.

Answer:

Check the explanation

Explanation:

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

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

Answer:

Check the explanation

Explanation:

For ideal regeneration heat loss in cooling aqual to heat gain in compression so temperature Tb=Td as  can be seen in the step by step solution in the attached images below.

Answer:

Check the attached images below.

Explanation:

It Is required to develop an approximate relation between . Reynolds and Grashor numbers such that the heat-transfer coefficients for pure forced convection and pure .e convection are equal, assuming laminar flow, by comparing the heat-transfer coefficients for forced or free convection over vertical hat plates.  

write the equation or heat transfer coefficient Mr (arced comedian.

Kindly check the attached images below.

Answer:

(a) 35% (b) 80% (c) Control signal is sent to a resource for activation of it's usage.

The operations are performed even in areas where it's not needed, it is ignored, because it is not used in cycles.

Explanation:

Solution

The computation of fraction  is defined below:

(a) The data memory used in lw and sw instructions

Now,

The fraction value is = sw + lw

=10 + 25

= 35

therefore the fraction value is 35%

(b) The needed sign is extended for all other instructions  other than ADD

which is shown below

The Fraction value  = addi + beq + lw + sw

=20 +25+25+ 10 =80 %

The fraction value here is 80%

(c) Now, a control signal is been sent to resource for activation of it's usage.

The operations are carried out even in case where it's not needed, it is ignored, because it is not used in cycles.

Note: The complete question to this exercise is attached below.

The fraction of all cycles that is the data memory used is 35%.

A circuit is known to be a kind of closed loop via which  electricity often pass through.

To solve for (a) which is the data memory that has been used in lw and sw instructions, you then add the value together:

That is: fraction value is = sw + lw

=10 + 25

So question A fraction value is 35%

To solve for me the required sign-extend circuit  is depicted as:

The Fraction value  = addi + beq + lw + sw

=20 +25+25+ 10 =80 %

The question (b) fraction value here is 80%

So therefore, the fraction value are 35 percent and 80 percent respectively

Learn more about circuit  from

https://brainly.com/question/2969220

Answer:

24.78 mg/L

Explanation:

The step-to-step explanation is written legibly with clear explanation in the diagram attached below.

Answer:

Answer: Input rate = 2.288 x 10⁶mg/s

             Output rate =78 x 10³Cmg/s

            Decay rate = 28.94 x 10 ⁵C mg/s

Explanation:

Assuming that complete and instantaneous mixing occurs in the lake, this implies that the concentration in the lake C is the same as the concentration of the mix leaving the lake Cm

   Input rate = Output rate  + KCV

Input rate = Q₁C₁ + QwCw

                = (75.0m³/s x 25.5mg/L + 3.0m³/s x 125.0mg/L) x 10³L/m³

                = 2.288 x 10⁶mg/s

Output rate = QmCm = (Q₁ +Qw)C

                   =(75 + 3.0)m³/s x Cmg/L x 10³L/m³ = 78 x 10³Cmg/s

Decay rate = KCV = 5/d x Cmg/L x 50 x 10⁶ x 10³L/m³  

                                           24 hr/d x 3600s/hr

                     = 28.94 x 10 ⁵C mg/s

So,

                      =2.288 x 10⁶ = 78 x 10³C + 28.94 x 10 ⁵C  = 29.72 X 10⁵C

                      = 2.288 x 10⁶          

                        29.72 X 10⁵         = 0.77 mg/l

                   C =

Answer:

Zx = 176In³

Explanation:

See attached image file

Answer:

2.83 years

Explanation:

Please kindly see the attached file at thw attachment area for a detailed and step by step solution to the given problem.

Answer:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

Answer:

A schedule is strict if it satisfies the following conditions:

Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.

Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.

Explanation:

See attached image

Answer:

Find the given attachment

Answer: S = 0.284

Explanation:

Data: d = 80 mm;

l =40 mm;

c = 0.06 mm;

F = 9kN;

n = 3600rpm = 60 rps

SAE 40 oil

T= 65°C

Therefore:

p = F / ld

= 9 x1000 /40 x 80

= 2.813 MPa

μ = 30 cp at 65°C for SAE 40 oil

S = r^2 x μ x n / c^2 x p

S= ( 40 )^2 x 30*10^-3 x 60 / (0.06)^2 x 2.813*10^6

S = 2880 / 10,126.6

S = 0.284

l/d = ½,

h o /c = 0.38

ε = e /c = 0.62

h o = 0.38 x C

= 0.382 x 0.06

=0.023mm

= 23µm

e = 0.62 x C

= 0.62 x 0.06

= 0.037 mm

Viscosity temperature curves of SAE graded oils

(r /c) f = 7.5,

S = 0.284

l /d = ½

f = 7.5 x (c / r)

= 7.5x (0.06/40)

= 0.0113

Answer:

[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]

[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]

And we can find this probability with this difference:

[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

[tex]\mu = 4.18,\sigma =1.19[/tex]

We can assume that the distribution for X is normal

[tex] X \sim N(\mu = 4.18 , \sigma =1.19)[/tex]

And we want to find this probability:

[tex] P(3 <X<5)[/tex]

And we can use the z score formula given by:

[tex] z=\frac[X- \mu}{\sigma}[/tex]

And we can find the z score for each limit and we got:

[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]

[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]

And we can find this probability with this difference:

[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

The number of poles required for a 3-phase stepping motor to achieve 0.05 mm spatial control depends on the increments per revolution, which was not provided. To move the axis at 90 mm/sec, the pulses per second from the controller will be calculated based on the screw pitch and the number of increments per revolution. Additional information on the stepper motor is needed for precise calculations.

To answer the student's questions regarding a 3-phase stepping motor for spatial control in a robotics application:

Without the specific number of increments per revolution of the stepping motor, it's impossible to provide an exact answer. However, typically stepper motors with more poles can provide finer control, and thus would be preferred in this application to meet the 0.05 mm spatial control requirement.

Answer:

See attached images

Answer:

Explanation:

the solution to the problem is given in the pictures attached. (b) is answered first then (a). I hope the explanation helps you.Thank you

Answer: The feed rate is

17,020kg/he and the rate is 13,520kg/h

Check the attachment for step by step explanation

Answer:

[tex]x = 0.58\,GW[/tex]

Explanation:

The conversion is:

[tex]x = (580000\,kW)\cdot \left(\frac{1\,GW}{1000000\,kW} \right)[/tex]

[tex]x = 0.58\,GW[/tex]

The function RunningLate returns true for the variable onTime when noTraffic is true and gasEmpty is false, which is achieved by the logical expression noTraffic && ~gasEmpty in MATLAB.

The task is to complete the function RunningLate in MATLAB, which returns a logical variable onTime that is true if noTraffic is true and gasEmpty is false. The finished MATLAB code inside the function should be:

This code uses the logical AND operator (&&) to check that there is no traffic, and the NOT operator (~) to check that the gas tank is not empty.

Answer: (a).power developed = 776.1 kW

(b). Rate of entropy production = 1.023 kW/K

(c). efficiency = 63%

Explanation:

Let us carry a step by step process to solve this problem;

from the question we have that

P₁ = 5 bar

T₁ = 320°C

where V₁ = 0.5416 m³/Kg, S₁ = 7.5308 KJ/Kg-K and R₁ = 0.3105.6 KJ/Kg

the volumetric flow rate is given as (φ) = 1.825 m³/s

Remember that φ = ṁ V

where ṁ is the mass flowrate, and V is the volume

ṁ = φ/V = 1.825/0.5416 = 3.37 Kg/s

Also given for the Exit state;

P₂ = 1 bar

T₂ = 200°C

where V₂ = 0.5416 m³/Kg, S₂ = 7.5308 KJ/Kg-K and R₂ = 0.3105.6 KJ/Kg

(a). we are asked to determine the power developed in the Kw.

using the Flow energy equation to turbine we have;

ṁ(R₁ + V₁²/2 + gZ₁) + φ = ṁ(R₂ + V₂²/2 + gZ₂₂) + ш

canceling out terms from both steps we have that

ш = 3.37 (3105-2815.3) = 776.1 kW

Therefore the Power output is 776.1 kW

(b). The rate of entropy production in Kw/K.

Rate(en) = ṁ (S₂-S₁) = 3.37 (7.8343 - 7.5308)

Rate(en) = 1.023 kW/K

(c). The percent isentropic  turbine efficiency.

Πt = (R₁-R₂) / (h₁ - h₂s)

Πt = (3105.6 - 2875.3) / (3105.6 - 2740) = 63%

Πt = 63%

cheers i hope this helped!!!!!

Answer:

Find the attachments sequence wise for complete solution.

Answer:

See attached images for the diagrams and tables

Answer:

4.8 years

Explanation:

The rate of corrosion (CPR) is defined as the rate at which a metal corrodes in a specific environment. It depends on the environmental condition and the type of metal. It is expressed in inches per year or melts per year. CPR is given by:

[tex]CPR=\frac{KW}{\rho At}[/tex]

Where K is a constant = 534, W is the weight corroded = 2.1 kg = 2.1 × 10⁶mg, A is the area = 17 in², ρ is the density = 7.9 g/cm³

From the CPR equation:

[tex]t=\frac{KW}{\rho A(CPR)}=\frac{534*2.1*10^6}{7.9*17*200}= 4.17*10^4 hrs=4.8years[/tex]

The question involves applying principles of static equilibrium and mechanical advantage in physics to calculate forces exerted by a hydraulic cylinder and at a support point in a construction telescoping arm scenario.

The question relates to the application of static equilibrium and mechanical advantage concepts in physics to solve for forces in a telescoping arm used in construction. Specifically, it involves calculating force exerted by a hydraulic cylinder and the force at a support point when a mass is placed at a specific location on the arm.

In order to solve for the force exerted at point B by the hydraulic cylinder (part a) and the force exerted on the supporting carriage at A (part b) when the arm makes an angle  heta = 25 degrees with the horizontal, one would typically use the principles of torques and static equilibrium. Summation of torques around a point, often the pivot, and summation of forces in horizontal and vertical directions are standard procedures. This requires knowledge of the geometry of the construction, the location of the center of gravity, and the mass of the workers and platform.

The information given about the forces in a wheelbarrow and cranes are analogous examples that illustrate similar principles of physics applied to different scenarios. These examples demonstrate how to calculate the mechanical advantage and forces based on lever arms and mass distribution.

The force exerted at B by the hydraulic cylinder BD and the force exerted on the supporting carriage at A can be calculated using principles of moments and equilibrium of forces, respectively.

For part (a): The force exerted at B by the hydraulic cylinder BD can be calculated using the principle of moments. By summing the moments about point A, you can find the force at B.

For part (b): The force exerted on the supporting carriage at A can be determined by considering.

Answer:

volume flow rate Q  = 53.23 in³/s

Explanation:

given data

diameter = 4.22 inches

length = 75 inches

screw rotates = 65 revolutions per minute

depth = 0.23 in

flight angle = 21.4 degrees

head pressure = 705 lb/in²

viscosity = 145 x [tex]10^{-4}[/tex] lb·sec/in²

solution

we get here volume flow rate of platstic in barrel that is express as

volume flow rate Q = volume flow rate of die - volume flow of extruder barrel   ................1

here

volume flow rate extruder barrel is

flow rate   = [tex]\frac{\pi \times 705 \times 4.22 \times 0.23\times sin21.4 }{12\times 145 \times 10^{-4}\times 75 }[/tex]    

flow rate = 60.10  

and

volume flow rate of die is express as

volume flow rate = 0.5 × π² × D² × Ndc ×  sinA × cosA   .............2

put here value and w eget

volume flow rate = 0.5 × π² × 4.22² × 0.23 × 1 × sin21.4 × cos21.4

volume flow rate = 6.866

so put value in equation 1 we get

volume flow rate Q  = 60.10 - 6.866  

volume flow rate Q  = 53.23

Given Information:  

distance = s₁ = 6050 ft

velocity = v₁ = 0 mi/hr

velocity = v₂ = 150 mi/hr

velocity = v₃ = 195  mi/hr

Acceleration = a = 2 ft/s²

Required Information:  

distance = s₂ = ?

Answer:

distance = s₂ = 14,399 ft

Explanation:

We know from the equations of motion,

v₃² = v₂² + 2a(s₂ - s₁)

We want to find out the distance s₂

2a(s₂ - s₁) = v₃² - v₂²

s₂ - s₁ = (v₃² - v₂²)/2a

s₂ = (v₃² - v₂²)/2a + s₁

First convert given velocities from mi/hr to ft/s

1 mile has 5280 feet and 1 hour has 3600 seconds

velocity = v₂ = 150*(5280/3600) = 220 ft/s

velocity = v₃ = 195*(5280/3600) = 286 ft/s

s₂ = (v₃² - v₂²)/2a + s₁

s₂ = (286² - 220²)/2*2 + 6050

s₂ = 33396/4 + 6050

s₂ = 8349 + 6050

s₂ = 14,399 ft

Therefore, the plane would have traveled a distance of 14,399 ft when it reaches a constant speed of 286 ft/s

Answer:

D

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.