A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The other side is evacuated . The membrane ruptures, filling the entire volume. The finial pressure is 100bar. Determine the final temperature of the steam and the volume of the vessel.

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Answer 1
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1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/sspeed zone. A police car observed the automobile. At the instant that the twovehicles are abreast of each other, the police car starts to pursue the automobile ata constant acceleration of 1.96 m/s². The motorist noticed the police car in his rearview mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)a) Find the total time required for the police car to overtake the automobile.b) Find the total distance travelled by the police car while overtaking theautomobile.c) Find the speed of the police car at the time it overtakes the automobile.d) Find the speed of the automobile at the time it was overtaken by the police car.

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

Answers

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

(d_(2) )/(d_(1) ) =(1)/(2) (\sqrt{1+8F^(2) } -1)

10*2=\sqrt{1+8F^(2) } -1

Solving for F:

F = 7.4162

v=F\sqrt{gd_(1) }

Here, g = gravity = 32.2 ft/s²

v=7.4162*√(32.2*1) =42.0833ft/s

b) The flow rate:

q=v*L*d_(1) =42.0833*80*1=3366.664ft^(3) /s

c) The Froude number:

v_(2) =(q)/(L*d_(2) ) =(3366.664)/(80*10) =4.2083ft/s

F=\frac{v_(2)}{\sqrt{gd_(2) } } =(4.2083)/(√(32.2*10) ) =0.2345

d) The flow energy dissipated:

E=((d_(2)-d_(1))^(3) )/(4d_(1)d_(2)) =((10-1)^(3) )/(4*1*10) =18.225ft

e) The critical depth:

d_(c) =((((q)/(L))^(2)  )/(g) )^(1/3) =((((3366.664)/(80))^(2)  )/(32.2) )^(1/3) =3.8030ft

(d) Arches NP is known for its spectacular arches that develop in the jointed areas of the park. Placemark Problem 2d flies you to Landscape Arch, the arch with the largest span in Arches NP. If the stresses that stretched the rock to form the joints were oriented perpendicular to the joint surfaces and the rock fins that contain the arches, what was the direction that the rocks were stretched? ☐ N-S
☐ E-W
☐ NW-SE
☐ NE-SW

Answers

Answer:

☐ NE-SW

Explanation:

Based on the description, the rock direction is North East - South West (NE-SW). Rocks generally can expand or compress depending on the type and magnitude of stress applied on the rocks. However, if the applied stress is sufficiently high, cracks and fractures will be created on the rock and it can ultimately lead to the formation of particles.

Can anyone tell me all the corrects answers to these? I’m sorry if this is the wrong subject I’m not sure what to put it under but I really need help!

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Answer:

Crankshaft position sensor - F     I can't quite make out the letter but it's the thing at the bottom almost touching the notched wheel.

Coil Module - B  

Knock Sensor - D

Coil Pack -E

Fuse Block - A

Powertrain Control Module - C

Compute the longitudinal tensile strength of an aligned glass fiber-epoxy matrix composite in which the average fiber diameter and length are 0.010 mm and 2.5 mm, respectively, and the volume fraction of fibers is 0.40. Assume that thefiber-matrixbondstrengthis75MPa, the fracture strength of the fibers is 3500 MPa and the matrix stress at fiber failure is 8.0 MPa.

Answers

Answer:

Explanation:

Given that,

Average fiber diameter is 0.01mm

d = 0.01mm = 1 × 10^-5m

The average fiber length is 2.5mm

L = 2.5mm = 0.0025m

Volume of the fraction of fibers is 0.40

Vf = 0.40

Fiber matrix bond strengths is 75MPa

τ = 75 MPa

The fraction strength of the fibers is 3500 Mpa

σf = 3500 MPa

The matrix street fiber is 8 MPa

σm = 8 MPa

We need to find the critical fiber length and compare it to original fiber length

Ic = σf•d / 2τ

Ic = 3500 × 0.01 / 75 × 2

Ic = 0.233 mm

Since the critical fiber length of 0.233 mm is much less than the provided length of the fiber (2.5mm) , so we can use the following equation to find the longitudinal tensile strength

σcd = σf•Vf(1—Ic / 2L) + σm(1—Vf)

σcd = 3500×0.4[1—0.233/(2 × 2.5)] + 8(1—0.4)

σcd = 1400(1—0.0467) + (8 × 0.6)

σcd = 1334.67 + 4.8

σcd = 1339.47 MPa

The longitudinal tensile strength of the aligned glass fiber-epoxy matrix composite is 1339.47 MPa

Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 seconds. In Eager Mode, the app utilizes a faster timer resolution for its computations, so the execution time in Eager Mode is 2 seconds (i.e., Eager Mode execution time is 60% of Lazy Mode execution time).After finishing computation, the app sends some data to the cloud, regardless of the mode it’s in. The data size sent to the cloud is 600 MB. The bandwidth of communication is 15 MBps for WiFi and 5 MBps for 4G. Assume that the communication radio is idle during the computation time. Assume that the communication radio for WiFi has a power consumption of 75 mW when active and 15 mW when idle. Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization. Assume that the power consumption of the CPU is a linear function of its utilization. In other words: P = (Idle Power) + (Utilization)*(Power per unit Utilization). A configuration of the mobile app involves choosing a timer resolution (Lazy or Eager) and choosing a type of radio (WiFi or 4G). For example, faster timer resolution (Eager) and 4G network is a configuration, while slower resolution (Lazy) and WiFi is another. There are four possible configurations in all.

Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?

Answers

The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Why reducing leads to increasing wages?

Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.

Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.

Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.

Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Learn more about average power on:

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N didn’t do it for you toroeriot everyone wwas wowowowoww

You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

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The solution is in the attachment

Answer:

please find attached.

Explanation:

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