Question
A lawnsprinkler has 20 holes of cross sectional area 2×10^-2cm^2,the sprinkler is connected to a hose pipe of cross sectiomal area. If the speed of the water in the hose pipe is 1.5ms^-1 estimate the speed of the water as it emerges from the hole
Answers
GPT-4o mini
To estimate the speed of the water as it emerges from the holes of the lawn sprinkler, we can use the principle of conservation of mass, which states that the mass flow rate must be the same in the hose pipe and the total of all the holes of the sprinkler.
The mass flow rate can be expressed as:
\[
A_1 v_1 = A_2 v_2
\]
where:
- \( A_1 \) = cross-sectional area of the hose pipe
- \( v_1 \) = speed of water in the hose pipe (1.5 m/s)
- \( A_2 \) = total cross-sectional area of the holes in the sprinkler
- \( v_2 \) = speed of water as it emerges from the holes
1. **Calculate the total cross-sectional area of the holes:**
Since there are 20 holes, each with a cross-sectional area of \( 2 \times 10^{-2} \text{ cm}^2 \):
\[
A_2 = 20 \times 2 \times 10^{-2} \text{ cm}^2 = 40 \times 10^{-2} \text{ cm}^2 = 0.40 \text{ cm}^2
\]
To convert this into square meters, we use the conversion \( 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \):
\[
A_2 = 0.40 \times 10^{-4} \text{ m}^2 = 4.0 \times 10^{-3} \text{ m}^2
\]
2. **Define the cross-sectional area of the hose pipe \( A_1 \)**:
Assuming we know or have the cross-sectional area of the hose pipe. Let's denote it as \( A_1 \), which should be provided. However, if it isn't, we can express it generally.
3. **Using the conservation of mass**:
Rearranging our equation:
\[
v_2 = \frac{A_1 v_1}{A_2}
\]
4. **If we set \( A_1 = A_h \) (the area of the hose pipe)** and plug in the known values:
Substituting values, we can find \( v_2 \):
\[
v_2 = \frac{A_h \times 1.5 \text{ m/s}}{4.0 \times 10^{-3} \text{ m}^2}
\]
Since we do not have \( A_h \), we can conclude generally rather than providing a numerical answer without specific area values. Therefore, if you have the value for \( A_h \), substitute it as follows to find \( v_2 \).
Assuming \( A_h = 1 \times 10^{-3} \text{ m}^2 \) (example only):
\[
v_2 = \frac{1 \times 10^{-3} \times 1.5}{4.0 \times 10^{-3}} = \frac{1.5 \times 10^{-3}}{4.0 \times 10^{-3}} = 0.375 \text{ m/s}
\]
**So, you will need the area of the hose pipe to find the exact speed of water emerging from the holes.**
The mass flow rate can be expressed as:
\[
A_1 v_1 = A_2 v_2
\]
where:
- \( A_1 \) = cross-sectional area of the hose pipe
- \( v_1 \) = speed of water in the hose pipe (1.5 m/s)
- \( A_2 \) = total cross-sectional area of the holes in the sprinkler
- \( v_2 \) = speed of water as it emerges from the holes
1. **Calculate the total cross-sectional area of the holes:**
Since there are 20 holes, each with a cross-sectional area of \( 2 \times 10^{-2} \text{ cm}^2 \):
\[
A_2 = 20 \times 2 \times 10^{-2} \text{ cm}^2 = 40 \times 10^{-2} \text{ cm}^2 = 0.40 \text{ cm}^2
\]
To convert this into square meters, we use the conversion \( 1 \text{ cm}^2 = 10^{-4} \text{ m}^2 \):
\[
A_2 = 0.40 \times 10^{-4} \text{ m}^2 = 4.0 \times 10^{-3} \text{ m}^2
\]
2. **Define the cross-sectional area of the hose pipe \( A_1 \)**:
Assuming we know or have the cross-sectional area of the hose pipe. Let's denote it as \( A_1 \), which should be provided. However, if it isn't, we can express it generally.
3. **Using the conservation of mass**:
Rearranging our equation:
\[
v_2 = \frac{A_1 v_1}{A_2}
\]
4. **If we set \( A_1 = A_h \) (the area of the hose pipe)** and plug in the known values:
Substituting values, we can find \( v_2 \):
\[
v_2 = \frac{A_h \times 1.5 \text{ m/s}}{4.0 \times 10^{-3} \text{ m}^2}
\]
Since we do not have \( A_h \), we can conclude generally rather than providing a numerical answer without specific area values. Therefore, if you have the value for \( A_h \), substitute it as follows to find \( v_2 \).
Assuming \( A_h = 1 \times 10^{-3} \text{ m}^2 \) (example only):
\[
v_2 = \frac{1 \times 10^{-3} \times 1.5}{4.0 \times 10^{-3}} = \frac{1.5 \times 10^{-3}}{4.0 \times 10^{-3}} = 0.375 \text{ m/s}
\]
**So, you will need the area of the hose pipe to find the exact speed of water emerging from the holes.**