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marysya [2.9K]
3 years ago
12

Solving Expressions Analytically 1 point Consider the following equation, which describes the speed of sound a in an ideal gas:

The Mach number M describes the ratio of a velocity v to the acoustic velocity or speed of sound a The Mach number can also be written in terms of the stagnation temperature To. 2 (To Combine these equations and solve symbolically for the temperature T in terms of all other quantities except M. Your answer should be stored in a variable T.result, which will be a list containing one or more syspy expressions. Prefer Te instead of T.e in this case Starter code (oick to view 1 ihport sympy Answer Type here to search

Engineering
1 answer:
WINSTONCH [101]3 years ago
6 0

Answer:

Check the explanation

Explanation:

From the question above the M describes the ratio of velocity v to the acoustic velocity or the speed sound a

M=  \frac{v}{a} this can be termed as equation 1

From the above equation squaring of both sides will be calculated in the attached image below

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The specific volume of mercury is .00007 m^3/Kg. What is its density in lbm/ft^3?
aalyn [17]

Answer:

891.027 lbm/ft³

Explanation:

See it in the pic

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3 years ago
Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
Leona [35]

Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

4 0
3 years ago
What are the height and width of scissors?
timofeeve [1]
Short ones are 4.5 inches but long ones can be up to 8 inches.
8 0
2 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
jok3333 [9.3K]

Answer:

Q=4.98\times 10^{-3}\ m^3/s.

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

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\mu =0.00164\ Pa.s

We know that

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where Q is volume flow rate

L is length of tube

d_i is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s.

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A(n) ______ is used to measure fluid flow in engineering
Arte-miy333 [17]

Answer:

A pitot tube is used to measure fluid flow in engineering

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