
The cylindrical tube of a spray pump has a cross-section of $8c{m^2}$, one end of which has 40 fine holes each of area ${10^{ - 8}}{m^2}$. If the liquid flows inside the tube with a speed of $0.15m/\min $, the speed with which the liquid is ejected through the holes is.
A. $50m/s$
B. $5m/s$
C. $0.05m/s$
D. $0.5m/s$
Answer
558.3k+ views
Hint: Here the liquid flows out of the spray pump and the liquid passes through the cylindrical pipe and reaches the other end which has 40 holes. In the meanwhile we assume there are no evaporative losses or any external input or output to the pipe except from the spray pump to the cylindrical pipe. So the flow is continuous and we use continuity equation here.
Formula used:
${A_1}{v_1} = {A_2}{v_2}$
Complete step by step solution:
We assume one control volume. Here that volume is a cylindrical pipe. The volume of that pipe is constant. If we multiply the volume with density then we will get mass. Since we assume the liquid is incompressible the density of that liquid will be constant. Since the volume and density is constant, its product also will be constant that means mass will be constant. So the rate of change of mass will be zero. So the rate at which mass of liquid enters will be the same as the rate at which it leaves the 40 holes. So the flow is continuous and we use a continuity equation here to solve it. The continuity equation is
${A_1}{v_1} = {A_2}{v_2}$
Here
${A_1}$ is given as $8c{m^2}$
${v_1}$ is given as $0.15m/\min $
${A_2}$ will be 40 times of ${10^{ - 8}}{m^2}$
${v_2}$ is needed.
$\eqalign{
& {A_1}{v_1} = {A_2}{v_2} \cr
& \Rightarrow \left( {8c{m^2}} \right)\left( {0.15m/60\sec } \right) = \left( {40 \times {{10}^{ - 8}}{m^2}} \right){v_2} \cr
& \Rightarrow \left( {8 \times {{10}^{ - 4}}{m^2}} \right)\left( {0.15m/60\sec } \right) = \left( {40 \times {{10}^{ - 8}}{m^2}} \right){v_2} \cr
& \therefore {v_2} = 5m/s \cr} $
Hence, option B is the answer.
Note:
This continuity equation is frequently used in fluid mechanics. The other equation used is Bernoulli's equation. Bernoulli’s equation is the base for fluid mechanics and it is used to manufacture several machines. Machines like pump and diffuser are vastly used in mechanical industry in order to extract the desired characteristics from the fluid flowing through them.
Formula used:
${A_1}{v_1} = {A_2}{v_2}$
Complete step by step solution:
We assume one control volume. Here that volume is a cylindrical pipe. The volume of that pipe is constant. If we multiply the volume with density then we will get mass. Since we assume the liquid is incompressible the density of that liquid will be constant. Since the volume and density is constant, its product also will be constant that means mass will be constant. So the rate of change of mass will be zero. So the rate at which mass of liquid enters will be the same as the rate at which it leaves the 40 holes. So the flow is continuous and we use a continuity equation here to solve it. The continuity equation is
${A_1}{v_1} = {A_2}{v_2}$
Here
${A_1}$ is given as $8c{m^2}$
${v_1}$ is given as $0.15m/\min $
${A_2}$ will be 40 times of ${10^{ - 8}}{m^2}$
${v_2}$ is needed.
$\eqalign{
& {A_1}{v_1} = {A_2}{v_2} \cr
& \Rightarrow \left( {8c{m^2}} \right)\left( {0.15m/60\sec } \right) = \left( {40 \times {{10}^{ - 8}}{m^2}} \right){v_2} \cr
& \Rightarrow \left( {8 \times {{10}^{ - 4}}{m^2}} \right)\left( {0.15m/60\sec } \right) = \left( {40 \times {{10}^{ - 8}}{m^2}} \right){v_2} \cr
& \therefore {v_2} = 5m/s \cr} $
Hence, option B is the answer.
Note:
This continuity equation is frequently used in fluid mechanics. The other equation used is Bernoulli's equation. Bernoulli’s equation is the base for fluid mechanics and it is used to manufacture several machines. Machines like pump and diffuser are vastly used in mechanical industry in order to extract the desired characteristics from the fluid flowing through them.
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