Question

A glass tube has several different cross-sectional areas with the values indicated in the figure. A piston at the left end of the tube exerts pressure so that mercury within the tube flows from the right end with a speed of 8.0 m/s. Three points within the tube are labeled A, B, and C.
http://edugen.wileyplus.com/edugen/courses/crs3976/art/qb/qu/c11/r5-1.png

Notes: The drawing is not drawn to scale.
Atmospheric pressure is 1.01 × 105 N/m2; and the density of mercury is 13 600 kg/m3.

What is the total pressure at point B?

Answer 1

Answer on Question #37652 - Physics - Other

A glass tube has several different cross-sectional areas with the values indicated in the figure. A piston at the left end of the tube exerts pressure so that mercury within the tube flows from the right end with a speed of $8.0 \, \text{m/s}$ . Three points within the tube are labeled A, B, and C.

http://edugen.wileyplus.com/edugen/courses/crs3976/art/qb/qu/c11/r5-1.png

Notes: The drawing is not drawn to scale.

Atmospheric pressure is $1.01 \times 105 \, \text{N/m}^2$ ; and the density of mercury is 13 600 kg/m3.

What is the total pressure at point B?

Solution:

First, let's find speed at which mercury flowing past point B:

$$S _ {c} V _ {c} = S _ {B} V _ {B}$$ $$V _ {B} = \left(\frac {S _ {c}}{S _ {B}}\right) V _ {c} = \left(\frac {6 . 0 \mathrm {c m} ^ {2}}{5 . 6 \mathrm {c m} ^ {2}}\right) \cdot 8 \frac {\mathrm {m}}{\mathrm {s}} = 8. 6 \frac {\mathrm {m}}{\mathrm {s}}$$

Now, let's find the total pressure at point B (pressure equation):

$$\frac {1}{2} \rho V _ {B} ^ {2} + p _ {B} = \frac {1}{2} \rho V _ {C} ^ {2} + p _ {C}$$ $$p _ {C} = p _ {\text {a t m}} = 1. 0 1 \times 1 0 ^ {5} \frac {\mathrm {N}}{\mathrm {m} ^ {2}}$$

(2)in(1):

$$\begin{array}{l} p _ {B} = \frac {1}{2} \rho \left(V _ {C} ^ {2} - V _ {B} ^ {2}\right) + p _ {\text {a t m}} = 1. 0 1 \times 1 0 ^ {5} \frac {\mathrm {N}}{\mathrm {m} ^ {2}} + \frac {1}{2} \cdot 1 3 6 0 0 \frac {\mathrm {k g}}{\mathrm {m} ^ {3}} \cdot \left(\left(8 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2} - \left(8. 6 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}\right) \\ = 3. 1 \times 1 0 ^ {4} \mathrm {P a} \\ \end{array}$$

Answer: total pressure at point B is $3.1 \times 10^{4} \mathrm{~Pa}$ .