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harkovskaia [24]
3 years ago
5

A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,

700 R. The valve is opened and the two tanks come to a uniform state at 580 R.
What is the final pressure?

Engineering
1 answer:
gulaghasi [49]3 years ago
8 0

Answer:

final pressure = 200KPa or 29.138psia

Explanation:

The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.

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Answer:

fluid nozzle that is too large

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Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin
Andre45 [30]

Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

5 0
3 years ago
A fisherman tries to escape a flash flood by walking up the 10 degree hill. If the water is rising at 0.11 m/s, how fast must he
FromTheMoon [43]

10.5 mph I’m pretty sure so so sorry if I’m wrong

6 0
3 years ago
Read 2 more answers
The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s
Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

lA=\sqrt{0.2^{2}+0.2^{2}  }=0.2\sqrt{2}m

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B} (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+      \frac{1}{2}k(l_{A}-l_{ul})  ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul})  ^{2}

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

\frac{m_{C}v_{B}^{2}   }{R}-N_{B}-k(l_{B}-l_{ul})=0

Replacing values and clearing NB:

NB = 694 N

6 0
3 years ago
Read 2 more answers
List, in ascending order, the cutoff frequencies for the first ten modes of a rectangular waveguide, normalized to the cutoff fr
Alisiya [41]

Answer:

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

Explanation:

Assuming a = 2b

Attached below is the required steps to the solution

The cutoff frequencies for the first ten modes of a rectangular waveguide listed in ascending order  is :

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

5 0
3 years ago
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