Chapter 28, Problem 2TS

A particular stream widens as it progresses downstream. Using your answers for part (a) and (b), briefly describe the changes in discharge.

(a) If the cross-sectional area of the stream is $1{\text{m}}^{\text{2}}$ and the stream speed is $0.5\text{m}/\text{s}$, what is the stream’s discharge?

(b) If the cross-sectional area of the stream is $2{\text{m}}^{\text{2}}$ and the stream speed is $0.5\text{m}/\text{s}$, what is the stream’s discharge?

To determine

The stream discharge.

Answer to Problem 2TS

The steam discharge is $0.5{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given Info: The area of the stream is $1{\text{m}}^2$ and speed of the stream is $0.5\text{m}/\text{s}$.

Write the expression for the discharge of a stream.

$d=A\times v$

Here,

$A$ is the cross-sectional area of stream

$v$ is the speed of the stream

Substitute $1{\text{m}}^2$ for $A$ and $0.5\text{m}/\text{s}$ for $v$ in the above expression to get the value of $d$.

$\begin{array}{c}{d}_a=1{\text{m}}^{\text{2}}\text{ × 0}\text{.5}\text{m}/\text{s}\\ =\text{0}\text{.5 }{\text{m}}^3/\text{s}\end{array}$

Conclusion:

Therefore, the steam discharge is $0.5{\text{m}}^{\text{3}}/\text{s}$.

To determine

The steam discharge.

Answer to Problem 2TS

The steam discharge is $1.0{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Write the expression for the discharge of a stream.

$d=A\times v$

Here,

$A$ is the cross-sectional area of stream

$v$ is the speed of the stream

Substitute $2{\text{m}}^2$ for $A$ and $0.5\text{m}/\text{s}$ for $v$ to get the value of $d$.

$\begin{array}{c}{d}_b=2{\text{m}}^{\text{2}}\text{ × 0}\text{.5}\text{m}/\text{s}\\ =1.0{\text{m}}^3/\text{s}\end{array}$

Thus from both the parts, it is clear that as the cross-section area increases while the speed of stream is constant, the discharge rate increases. The discharge of stream in part (b) is double to that of part (a).

Conclusion:

Therefore, the steam discharge is $1.0{\text{m}}^{\text{3}}/\text{s}$.