Question

The temperature distribution in a fluid is given by T = 10x + 5y, where x and y are the horizontal and vertical coordinates in meters and T is in degrees centigrade. Determine the time rate of change of temperature of a fluid particle traveling:
(a) horizontally with u = 20 m/s, v = 0.
(b) vertically with u = 0, v = 20 m/s.
(c) diagonally with u = 20 m/s, v = 20 m/s.

Answer 1

Answer:

The rates of change of T for each case are the following:

(a) $\frac{dT}{dt}$ = 200 °C/s

(b) $\frac{dT}{dt}$ = 100 °C/s

(c) $\frac{dT}{dt}$ = 300 °C/s

Explanation:

We need to find the rate of change for temperatura of the fluid, which is the derivate of the function as a function of time (t). We know that T is defined as a function of coordinates x and y, but we also know that velocity u is the derivate of x in terms of time (t) of the coordinate x and v is the derivate of y in terms of time (t).

So, to obtain the derivate of T in terms of t, we have:

$\frac{dT}{dt} =\frac{dT}{dx} \frac{dx}{dt} +\frac{dT}{dy} \frac{dy}{dt}$

And, as we said before,

$\frac{dx}{dt} =u$     and    $\frac{dy}{dt} =v$

Then,

$\frac{dT}{dt}=\frac{dT}{dx} u+\frac{dT}{dy} v$

And, as we know the equation that defines T in terms of x and y, we can derivate this function and obtain,

$\frac{dT}{dx} =10\\\frac{dT}{dy} =5$

So,

$\frac{dT}{dt}=10u+5v$

And finally, we replace the value of u and v for each case,

(a) u=20 and v=0

$\frac{dT}{dt} =10(20)+5(0)=200°C/s$

(b) u=0 and v=20

$\frac{dT}{dt} =10(0)+5(20)=100°C/s$

(c) u=20 and v=20

$\frac{dT}{dt} =10(20)+5(20)=300°C/s$

Answer 2

T= 10x + 5y

Applying chains rule to determine the rate of change in temperature :

DT/ Dt = dT/dt + u dt/dx + v dt/dx + w dt/dz

Given that:

a). Horizontally

u=20m/s

v=0, w=0

DT/Dt= 0 + 20 d/dx(10x+5y) + 0 d/dy(10x+5y) + 0 d/dz(10x+5y)

DT/ Dt= 200°C/s

b). Vertically

DT/Dt = 0 + 0 d/dx(10x+5y) + 20 d/dy(10x+5y) + 0

DT/Dt = 100° C/s

c). Diagonally

DT/Dt = 0 ° C/s

This is because VARIABLES were given in x and y directions respectively.