Question
Download Solution PDFA pipe carrying 0.05 m3/s of water suddenly contracts from 30 cm to 15 cm in diameter. Then the coefficient of contraction is _________ when the loss of head is 0.5 m.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Loss of head due to sudden contraction:
h = \(KV^2\over 2g\)
h = \((V_c - V_2)^2\over 2g\) = \({V^2\over 2g}({1\over C_c} -1)^2\)
K = \(({1\over C_c} -1)^2\)
where
Cc = Coefficient of contraction
Vc = Velocity at the contracted portion
V2 = velocity after contraction
Calculation:
Q = 0.05 m3/s
h = 0.5 m
A = (π/4)× 0.152 = 0.01767
V = Q/A = 0.05/0.01767 = 2.8296 m/sec
K = h(2g)/V2 = (0.5× 2× 9.81)/(2.8296)2 = 1.2252
Now
K = \(({1\over C_c} -1)^2\)
\(({1\over C_c} -1)^2\) = K
\(({1\over C_c} -1)^2\) = 1.2252
\(({1\over C_c} -1)\) = 1.10688
CC = 0.474
Last updated on Jan 29, 2024
-> JKPSC AE Recruitment Provisional Selection List has been released.
-> Candidates can check their names and marks obtained in the written examination.
-> The selected candidates will be called for document verification.
-> Earlier, the JKPSC AE Final Answer key was released! The Jammu & Kashmir Public Service Commission released the notification for JKPSC AE (Civil) Recruitment 2024 for a total of 36 vacancies under the Jal Shakti Department.
-> The selection process comprises three stages namely Written test, Interview, and Document Verification.
-> With a JKPSC AE Salary range between Rs. 50,700 to Rs. 1,60,600, it is a great chance for aspirants who want a government job in J&K. The candidates who are preparing for the exam can check the JKPSC AE Previous Year Papers for the effective preparation.