Question
Download Solution PDFA fully insulated and completely empty tank is being filled with some fluid with negligible fluid velocity. Which of the following is correct?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
This is the case of an unsteady state.
Energy associated with entering fluid = \((m_2-m_1)(h_1+\frac{V_1^2}{2})~\)
Q = heat transfer of the control volume, then
\((m_2-m_1)(h_1+\frac{V_1^2}{2})+Q=m_2u_2-m_1u_1~\)
The tank is fully insulated so Q = 0, then
\((m_2-m_1)(h_1+\frac{V_1^2}{2})=m_2u_2-m_1u_1~\)
Also, the tank is initially empty so \(m_1=0~\)
\(m_2(h_1+\frac{V_1^2}{2})=m_2u_2~\)
The velocity of entering fluid is negligible, so \(V_1=0\)
The final result is,
\(h_1=u_2~\)
So, the specific enthalpy of entering fluid is equal to final specific internal energy of fluid in the charged tank.
Last updated on May 28, 2025
-> SSC JE ME Notification 2025 will be released on June 30.
-> The SSC JE Mechanical engineering application form will be available from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 2 to 31.
-> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.