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Blababa [14]
3 years ago
14

A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the

inlet temperature if the gas is argon, helium, or nitrogen.
Engineering
1 answer:
Margaret [11]3 years ago
5 0

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$ T_{in}  = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p}  + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

C_p = 520 \:\: J/kg.K

So, the inlet temperature of argon is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520}  + 300$

$ T_{in}  = \frac{1}{2} \times 119  + 300$

$ T_{in}  = 360K $

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

C_p = 5193 \:\: J/kg.K

So, the inlet temperature of helium is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193}  + 300$

$ T_{in}  = \frac{1}{2} \times 12  + 300$

$ T_{in}  = 306K $

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

C_p = 1039 \:\: J/kg.K

So, the inlet temperature of nitrogen is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039}  + 300$

$ T_{in}  = \frac{1}{2} \times 60  + 300$

$ T_{in}  = 330K $

Note: Answers are rounded to the nearest whole numbers.

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a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: 4.3\cdot 10^{-15} J

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

V(r)=\frac{kq}{r}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

q_1=+2.00\mu C=+2.00\cdot 10^{-6}C

and is located at the origin (x=0, y=0)

Charge 2 is

q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

r_{1A}=0.40 m

So the potential due to charge 2 is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V

The distance of point A from charge 2 is

r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m

So the potential due to charge 1 is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V

Therefore, the net potential at point A is

V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m

So the potential due to charge 1 at point B is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V

The distance of charge 2 from point B is

r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m

So the potential due to charge 2 at point B is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V

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V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V

So the potential difference is

V_B-V_A=-29,500 V-(-2700 V)=-26,800 V

c)

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W=q\Delta V

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q is the charge of the particle

\Delta V is the potential difference

In this problem, we have:

q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-26,800 V is the potential difference

Therefore, the work required on the electron is

W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

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