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gladu [14]
3 years ago
13

In a steady two-dimensional flow, particle trajectories are given byx(t) =x0cosγt;y(t) =y0sinγta. From these trajectories determ

ine the Lagrangian particle velocity componentsu(t) =dx/dtandv(t) =dy/dt. Convert these to Eulerian velocity componentsu(x,y) andv(x,y). Note that the Eulerian velocity does not depend on time.
Engineering
1 answer:
user100 [1]3 years ago
6 0

Answer with Explanation:

Given that

x(t)=x_ocos(\gamma t)............(i)

y(t)=y_osin(\gamma t)..............(ii)

Thus by definition x component of velocity in Lagrangian system is given by

u=\frac{dx}{dt}\\\\u=\frac{d(x_ocos(\gamma t)}{dt}=-\gamma x_osin(\gamma t)

Thus by definition y component of velocity in Lagrangian system is given by

v=\frac{dy}{dt}\\\\v=\frac{d(y_osin(\gamma t)}{dt}=\gamma y_ocos(\gamma t)

Since in eulerian system we need to eliminate time from the equations

From euations 'i' and 'ii' we can write

cos(\gamma t)=\frac{x}{x_o}\\\\sin(\gamma t)=\frac{y}{y_o}

Applying these values in the velocity components as obtained in Lagrangian system we get

u=\frac{-\gamma x_o}{y_o}\cdot y\\\\v=\frac{\gamma y_o}{x_o}\cdot x

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Define extensive and intensive properties of thermodynamic system.
Flura [38]

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An intense property is a physical attribute of a system that is independent of the size of the system or the quantity of material it contains. An extensive property of a system, on the other hand, is dependent on the size of the system or the amount of material in it.

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Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss
klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

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ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

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3 0
3 years ago
An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in
Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, T_{a} = 140^{\circ}C = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, T_{i} = 60^{\circ}C = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, C_{p} = 0.949\ kJ/kgK

At room temperature

Specific heat of iron, C_{p} = 0.45\ kJ/kgK

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

MC_{p}\Delta T = mC_{p}\Delta T

28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)

Thus

T_{e} = 109.71^{\circ}C = 273 + 109.71 = 382.71 K

where

T_{e} = Equilibrium temperature

Now,

To calculate the changer in entropy:

\Delta s = \Delta s_{a} + \Delta s_{i}

Now,

For Aluminium:

\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K

For Iron:

\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K

Thus

\Delta s =-2.025 + 2.253 = 0.228\ kJ/K

6 0
3 years ago
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