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Question: Water is flowing through a horizontal pipe of variable cross-section. The rate of volume flow is \[0...

Water is flowing through a horizontal pipe of variable cross-section. The rate of volume flow is 0.2liters/sec0.2{\text{liters/sec}}. The pressure is 30,000Pa.30,000Pa. at a point where the area of the cross-section is 5×10−5m25 \times {10^{ - 5}}{m^2}. The pressure at the point where the area of cross-section is 10−4{10^{ - 4}}:
A. 6000Pa6000Pa
B. 36000Pa36000Pa
C. 24000Pa24000Pa
D. 30000Pa30000Pa

Explanation

Solution

Bernoulli theorem states that when an incompressible and non-viscous liquid or gas flows in streamline motion from one place to another, then at every point of its path the total energy per unit volume is constant i.e., pressure energy, kinetic energy, and potential energy remains constant.

Complete step by step solution:
Given-Rate of volume flow = 0.2Litre/sec⁡=0.2×10−3m3/sec⁡0.2Litre/\sec = 0.2 \times {10^{ - 3}}{m^3}/\sec , P1=30,000Pa{P_1} = 30,000Pa, A1=5×10−5m2{A_1} = 5 \times {10^{ - 5}}{m^2},A2=10−4m2{A_2} = {10^{ - 4}}{m^2}, find P2{P_2}.
We can write Bernoulli’s theorem mathematically as follows.
P+12ρv2+ρgh=constantP + \dfrac{1}{2}\rho {v^2} + \rho gh = {\text{constant}}
Now, because the pipe is horizontal, its potential energy will be zero. So, the above formula reduces to the following.
P+12ρv2=constantP + \dfrac{1}{2}\rho {v^2} = {\text{constant}}
Now, let us use this formula for any two ends of the pipe having area A1{A_1} and A2{A_2}, Pressure P1{P_1} and P2{P_2}.
P1+12ρv12=P2+12ρv22{P_1} + \dfrac{1}{2}\rho {v_1}^2 = {P_2} + \dfrac{1}{2}\rho {v_2}^2…….(1)
Also, we know, rate of flow is equal to the area multiplied by velocity.
Q=A1v1=A2v2Q = {A_1}{v_1} = {A_2}{v_2}
Let us now substitute the values in equation (1).
P1+12×ρ(QA1)2=P2+12×ρ(QA2)2{P_1} + \dfrac{1}{2} \times \rho {\left( {\dfrac{Q}{{{A_1}}}} \right)^2} = {P_2} + \dfrac{1}{2} \times \rho {\left( {\dfrac{Q}{{{A_2}}}} \right)^2}
Let us rearrange the above equation.
P2=P1+12×Q2ρ(1A12−1A22){P_2} = {P_1} + \dfrac{1}{2} \times {Q^2}\rho \left( {\dfrac{1}{{A_1^2}} - \dfrac{1}{{A_2^2}}} \right)
Now, substituting the values and solving them.
P2=30000+12×(0.2×10−3)2(1000)(1(5×10−5)2−1(10−4)2){P_2} = 30000 + \dfrac{1}{2} \times {\left( {0.2 \times {{10}^{ - 3}}} \right)^2}\left( {1000} \right)\left( {\dfrac{1}{{{{\left( {5 \times {{10}^{ - 5}}} \right)}^2}}} - \dfrac{1}{{{{\left( {{{10}^{ - 4}}} \right)}^2}}}} \right)
P2=30000+(0.02×10−3)(1(25×10−10)−1(10−8))=30000+6000{P_2} = 30000 + \left( {0.02 \times {{10}^{ - 3}}} \right)\left( {\dfrac{1}{{\left( {25 \times {{10}^{ - 10}}} \right)}} - \dfrac{1}{{\left( {{{10}^{ - 8}}} \right)}}} \right) = 30000 + 6000
P2=36000Pa{P_2} = 36000 Pa
Hence, option (B) 36000Pa36000 Pa is correct.

Note:
Bernoulli's theorem is based on the principle of conservation of energy for a flowing fluid.
According to the equation of continuity of fluid flow, if a liquid flows in streamlined motion through a tube of non-uniform cross-section, then the product of the area of cross-section and the velocity of flow is the same at every point in the tube.
Hence, we can say the velocity of the liquid is smaller in the wider part and larger in the narrow parts.