A piston–cylinder device contains 0.8 kg of steam at 300°C and at pressure of 800 kPa. Steam is cooled at constant pressure until one-half of the mass condenses. The final temperature is

a.
178°C

b.
184°C

c.
195°C

d.
None of these

e.
167°C

f.
170°C

Answers

Answer 1

Answer:

d

Explanation:

d.

None of these


Related Questions

A private plane pilot is what kind of individual transportation position? professional level mid-level entry-level EPA-certified

Answers

Answer:

A private plane pilot is a mid-level position.

Explanation:

The six pilot certifications in the US are as follows:

Sport PilotRecreational PilotPrivate Plane PilotCommercial Plane PilotFlight InstructorAirline Transport Pilot

From this listing, it is evident that the Private Plane Pilot is in the mid of the line so it is a mid-level position.

Illinois furniture , Inc produces all types of coffee furniture the executive secretary is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130 there are 480 minutes available during the day and the average daily demand has been 50 chairs there are eight tasks

Answers

This question is incomplete, the complete question is;

Illinois furniture , Inc produces all types of office furniture. The "Executive Secretary" is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130. There are 480 minutes available during the day and the average daily demand has been 50 chairs. There are eight tasks.

TASK     PERFORMANCE TIME( MIN )     TASK MUST FOLLOW TASK LISTED

A                            4                                                     ---------

B                            7                                                     ----------

C                            6                                                        A,B

D                            5                                                          C

E                            6                                                          D

F                            7                                                           E

G                            8                                                          E

H                            8                                                          F,G

1) What is the cycle time for operation?

2) What is the theoretical minimum number of workstation?

Answer:

1) the cycle time for operation is 9.6 min

2) the theoretical minimum number of workstation is 5

Explanation:

Given the data in the question;

production time per day = 480 minutes

average daily demand = 50

Given the data in the question;

1) cycle time for operation

this is simply referred to as the total time for the process from start to finish.

cycle time = production time per day / units demand per day

we substitute

cycle time = 480 min / 50

cycle time = 9.6 min

Therefore, the cycle time for operation is 9.6 min

2) theoretical minimum number of workstation.

theoretical minimum number of workstation = total task time / cycle time

Total task time = ( 4 + 7 + 6 + 5 + 6 + 7 + 8 + 6 ) = 49 min

∴  theoretical minimum number of workstation = 49 min / 9.6 min

theoretical minimum number of workstation = 5.104 ≈ 5

Therefore, the theoretical minimum number of workstation is 5

In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor principal stresses.

Answers

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

[tex]\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}[/tex]

[tex]\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}[/tex]

Here

[tex]\sigma_x[/tex] is the normal stress which is 1750 psf[tex]\sigma_y[/tex] is 0[tex]\tau_{xy}[/tex] is the shear stress which is 800 psf

So the formula becomes

[tex]\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}[/tex]

Similarly, the minimum normal stress is given as

[tex]\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}[/tex]

The maximum shear stress is given as

[tex]\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}[/tex]

Which location sharing service offers items for users as a gaming component and also allows them to collectively link their check ins to publish a trip

Answers

Answer:

Gowalla

Explanation:

Gowalla is a Social networking service operating as a gaming component in which it includes the registration of users' cities. Such information is then saved and utilized during the fun activities on the platform such as meeting up with people nearby and discuss interesting things relating to them.

Hence, considering there are no available options, GOWALLA is the location-sharing service that offers items for users as a gaming component and also allows them to collectively link their check-ins to publish a trip

Which of the following are the same as 1545.5347
Select one:
a. 1.545534e: 3
b. 1545534.0e 4
c0.15455340-5
d. 154553.4e 3

Answers

Explanation:

D the answer is D and anyone can answer this Question

Based on the provided options, the scientific expression that is the same as 1545.5347 is 154553.4e3. Therefore, the correct option is option D.

A scientific expression, often known as scientific notation, is a method of representing extremely big or extremely small integers. It is often used to simplify the representation of such numbers in scientific and mathematical operations. A number is expressed in scientific notation as a product of two parts: a coefficient and a power of ten.

The coefficient is a value between 1 and 10, and the power of 10 determines how far the decimal point should be moved. The "e3" signifies multiplying the integer by 10 raised to the power of 3, which is similar to moving the decimal point three places to the right in this formula. As a result, 154553.4e3 equals 154553.4103, which simplifies to 154553400.

Therefore, the correct option is option D.

To know more about scientific expression, here:

https://brainly.com/question/15361382

#SPJ4

In a production facility, 3-cm-thick large brass plates (k 5 110 W/m·K, r 5 8530 kg/m3, cp 5 380 J/kg·K, and a 5 33.9 3 1026 m2/s) that are initially at a uniform temperature of 25°C are heated by passing them through an oven maintained at 700°C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h 5 80 W/m2·K, determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be

Answers

Answer:

the surface temperature of the plates when they come out of the oven is approximately 445 °C

Explanation:

Given the data in the question;

thickness t = 3 cm = 0.03 m

so half of the thickness L = 0.015 m

thermal conductivity of brass k = 110 W/m°C

Density p = 8530 kg/m³

specific heat [tex]C_p[/tex] = 380 J/kg°C

thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s

Temperature of oven T₀₀ = 700°C

initial temperature T[tex]_i[/tex] = 25°C

time t = 10 min = 600 s

convection heat transfer coefficient h = 80 W/m².K

Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.

So, using analytical one-term approximation method, the Fourier number > 0.2.

now, we determine the Biot number for the process

we know that; Biot number Bi =  hL / k

so we substitute

Bi =  hL / k

Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109

Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )

The interpolation method used to find the

λ₁ = 0.1039 and A₁ = 1.0018

so

The Fourier number т = ∝t/L²

we substitute

Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²

т = 0.02034 / 0.000225

т = 90.4

As we can see; 90.4 > 0.2

So,  analytical one-term approximation can be used.

∴ Temperature at the surface will be;

θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation

θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )

so we substitute

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )

θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998

θ(L,t)[tex]_{wall[/tex] = 0.3775

so we substitute into equation 1

θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)

0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )

0.3775 = ( T(L,t) - 700 ) / ( - 675 )

0.3775 × ( - 675 ) = ( T(L,t) - 700 )

- 254.8125 = T(L,t) - 700

T(L,t) = 700 - 254.8125

T(L,t) = 445.1875 °C ≈ 445 °C

Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C

el protozoos es del reino protista?

Answers

Answer: Si (Yes)

Explanation:

Answer:

Protozoario o protozoo es un organismo unicelular y eucariota (con núcleo celular definido) perteneciente al Reino protista. Los protozoarios se encuentran junto con los protófitos o algas simples, generalmente acuáticas, dentro del Reino protista o también denominado Reino protoctista.

Explanation:

please help with my economics problem

Answers

Answer:

You first get a new job, and make a new company and then by amazon to traumatize Jeff Bezos after his divorce

Explanation:

Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.

Answers

Answer:

3  kmol/m^3

Explanation:

Determine the concentration of C in the stream leaving the reactor

Given that the CTSR reaction ; A (k1) → B (k2) → C

K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2

attached below is the detailed solution

concentration of C leaving the reactor= 3 kmol/mol^3

Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3

Consider a normal population distribution with the value of known.

Answers

Answer:

it is alba kk

Explanation:

rbeacuse wrong

Write a program that can take a tree as input and travers it in 3 different format (pre-order, in-order and post-order). Write a report, you can follow the attached file. You can modify the report as you want.

Answers

Answer:

Let us see different corner cases.

Complexity function T(n) — for all problem where tree traversal is involved — can be defined as:

T(n) = T(k) + T(n – k – 1) + c

Where k is the number of nodes on one side of root and n-k-1 on the other side.

Let’s do an analysis of boundary conditions

Case 1: Skewed tree (One of the subtrees is empty and other subtree is non-empty )

k is 0 in this case.

T(n) = T(0) + T(n-1) + c

T(n) = 2T(0) + T(n-2) + 2c

T(n) = 3T(0) + T(n-3) + 3c

T(n) = 4T(0) + T(n-4) + 4c

A 1.00 liter solution contains 0.46 M hydrocyanic acid and 0.35 M potassium cyanide If 25.0 mL of water are added to this system, indicate whether the following statements are true or false. (Note the the volume MUST CHANGE upon the addition of water.)
A. The concentration of HCN will increase.
B. The concentration of CN" will decrease.
C. The equilibrium concentration of Hy0 will remain the same 4
D. The pH will remain the same.
E. The ratio of [HCN]/[CN] will decrease.

Answers

I won leader solution contain 0.46 mL of hydronic I said of 0.3 potassium

hmmmmmmmm i already put the photo as attachment its

Answers

Answer:

letse see

Explanation:

Route Choice The cost of roadway improvements to the developer is a function of the amount of traffic being generated by the theater as well as the routes that these vehicles use in getting to/from the theater. Thus, you need to determine traffic demand on the available routes between the housing area (apartment complexes) and the proposed theater site. There are two potential routes from this housing area to the proposed theater site. The total flow on these two routes (origin-to-destination demand plus other traffic) is 5500 vehicles. These routes have the following speed and length characteristics:

Route Free Flow Speed(mi/h) Length(mi)
1 45 5.5
2 40 3.5
3 35 3.75

It is known that the individual route travel times increase (in units of minutes) according to the following functions (with x in units of 1000 vehicles per hour):

Route Travel time increase as a function of traffic volume
1 0.5x1
2 x2
3 0.25

Required:
Determine user equilibrium flows and travel times if the total flow on these routes (origin-to-destination demand plus other traffic) is 5500 veh/h.

Answers

Row choice the cost of roadway improvements to the developer and functional the amount of trafficBeing generated by the theater as well as the Ralph’s ladies

What happens to the speed of light if the IOR increases?

Answers

Because index of reflection is defined as the radio between the spray speed of light in a vacuum and the speed of light in a medium as a light traveling to roll the medium increases in speed at index of refraction decreases

You are designing a package for 200 g of snack food that is sensitive to oxygen, and fails when it absorbs 120 ppm of oxygen (by weight). Marketing tells you it wants the snack to be in a plastic pouch measuring 6 inches by 6 inches (ignore seems), so it will have a total surface area available for permeation of 72 in(6" x 6" x 2 sides). You need to recommend an appropriate plastic material for this product, to provide a minimum of 70 days shelf life. Follow these steps:
a. Calculate the allowable oxygen gain, in cm at STP. (5 pts)
b. At this point, you do not know the material you will use, so you do not know the permeability coefficient or the thickness. The better the barrier the plastic you choose, the thinner the material can be to provide the appropriate barrier. Rather than simple trial and error, a sensible approach is to solve for the ratio of P/L that is required. We can solve the basic permeability equation for this ratio: P = 9 At Ap L Use the information you have to determine the required value for P/L, expressing your answer in cm/(100 in? d atm). (5 pts)
c. Use the information in the textbook (chapters 4 and 14 or in another reliable source; provide reference if you have used chapter 4, 14 or any other source) on oxygen permeability coefficients for various polymers to select a polymer that would be suitable, and calculate the required thickness. (Be sure this is reasonable; for example, if the required thickness is more than 20 mils, you need to choose a different polymer!) Note that chapter 4 presents these values in the units you used in (b) while chapter 14 presents values with different units, so unit conversion would be required. In your answer, state the material you have chosen, its oxygen permeability coefficient, and the minimum thickness you recommend. (Be sure to express the thickness with no more than one decimal place.) Obviously, there is more than one solution to this problem, but you only need one. (10 pts)

Answers

Snack food is the sensitivity to auction and avails win the George 120

The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.

Answers

Answer:

the liquid woulriekwvhrnsshsnekwb ndrhwmoadi

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