LAW of thermodynamicsFor a closed system/non-flow system undergoing a process, (1 - 2)Q1-2 = ΔE + W1-2 …1)E = Stored energy of a systemThis stored energy can be viewed as the sum of microscopic and macroscopic energies.⇒ Q1-2 = Δ (U + KE + PE) + W1-2⇒ Q1-2 = ΔU + ΔKE + ΔPE + W1-2 …2)For a non-flow or closed system at equilibrium, ΔKE and ΔPE are negligible,So, these 2 terms can be neglected.⇒ Q1-2 = ΔU + W1-2 …3)Also, for a perfect gas, the internal energy is a function of temperature only.i.e. dU = mCνdT …4)Calculation: Given equation isQ1-2 = W1-2 …5)But first law states that; Q1-2 = ΔU + W1-2 …6)Comparing 5) and 6)⇒ ΔU = 0 …7)But for perfect gas; dU = mCνdT, integrating both sides\(\mathop \smallint \nolimits_1^2 dU = \mathop \smallint \nolimits_1^2 m{C_\nu }dT\)⇒ U2 – U1 = mCν(T2 – T1) {Assuming constant m, Cν}ΔU = mCν(ΔT) …8)Comparing 7) and 8)⇒ ΔT = 0 {∵ m ≠ 0, Cν ≠ 0}⇒ T2 = T1 = Constant = Isothermal process Remember the PROPERTIES of perfect gases and apply these directly instead of writing first law.Study all the basic PROCESSES in DETAIL like the adiabatic process, Isobaric, isochoric etc.

"> LAW of thermodynamicsFor a closed system/non-flow system undergoing a process, (1 - 2)Q1-2 = ΔE + W1-2 …1)E = Stored energy of a systemThis stored energy can be viewed as the sum of microscopic and macroscopic energies.⇒ Q1-2 = Δ (U + KE + PE) + W1-2⇒ Q1-2 = ΔU + ΔKE + ΔPE + W1-2 …2)For a non-flow or closed system at equilibrium, ΔKE and ΔPE are negligible,So, these 2 terms can be neglected.⇒ Q1-2 = ΔU + W1-2 …3)Also, for a perfect gas, the internal energy is a function of temperature only.i.e. dU = mCνdT …4)Calculation: Given equation isQ1-2 = W1-2 …5)But first law states that; Q1-2 = ΔU + W1-2 …6)Comparing 5) and 6)⇒ ΔU = 0 …7)But for perfect gas; dU = mCνdT, integrating both sides\(\mathop \smallint \nolimits_1^2 dU = \mathop \smallint \nolimits_1^2 m{C_\nu }dT\)⇒ U2 – U1 = mCν(T2 – T1) {Assuming constant m, Cν}ΔU = mCν(ΔT) …8)Comparing 7) and 8)⇒ ΔT = 0 {∵ m ≠ 0, Cν ≠ 0}⇒ T2 = T1 = Constant = Isothermal process Remember the PROPERTIES of perfect gases and apply these directly instead of writing first law.Study all the basic PROCESSES in DETAIL like the adiabatic process, Isobaric, isochoric etc.

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During a non-flow thermodynamic process (1-2) executed by a perfect gas, the heat interaction is equal to the work interaction (Q1-2 = W1-2) when the process is

Fluid Mechanics First Law Thermodynamics in Fluid Mechanics 7 months ago

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Concept: The first LAW of thermodynamicsFor a closed system/non-flow system undergoing a process, (1 - 2)Q1-2 = ΔE + W1-2 …1)E = Stored energy of a systemThis stored energy can be viewed as the sum of microscopic and macroscopic energies.⇒ Q1-2 = Δ (U + KE + PE) + W1-2⇒ Q1-2 = ΔU + ΔKE + ΔPE + W1-2 …2)For a non-flow or closed system at equilibrium, ΔKE and ΔPE are negligible,So, these 2 terms can be neglected.⇒ Q1-2 = ΔU + W1-2 …3)Also, for a perfect gas, the internal energy is a function of temperature only.i.e. dU = mCνdT …4)Calculation: Given equation isQ1-2 = W1-2 …5)But first law states that; Q1-2 = ΔU + W1-2 …6)Comparing 5) and 6)⇒ ΔU = 0 …7)But for perfect gas; dU = mCνdT, integrating both sides\(\mathop \smallint \nolimits_1^2 dU = \mathop \smallint \nolimits_1^2 m{C_\nu }dT\)⇒ U2 – U1 = mCν(T2 – T1) {Assuming constant m, Cν}ΔU = mCν(ΔT) …8)Comparing 7) and 8)⇒ ΔT = 0 {∵ m ≠ 0, Cν ≠ 0}⇒ T2 = T1 = Constant = Isothermal process Remember the PROPERTIES of perfect gases and apply these directly instead of writing first law.Study all the basic PROCESSES in DETAIL like the adiabatic process, Isobaric, isochoric etc.

Posted on 25 Nov 2024, this text provides information on Fluid Mechanics related to First Law Thermodynamics in Fluid Mechanics. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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