Thermal and Hydraulic Machines part1

Name: Thermal and Hydraulic Machines part1
Brand: Skuola.net
Price: 4.99 EUR
Availability: InStock
Rating: 3.0 (1 reviews)
Author: Em4nuel3

Aggiornato il 21/04/2026

di Em4nuel3

Publisher

Vota 3,0/5 (1)

Contenuto verificato e approvato dal Team di Esperti di Skuola.net

Appunti utili per la preparazione dell'esame di Thermal and Hydraulic Machines del professor Claudio Don Giovanni al Politecnico di Torino, Prima Facoltà di Ingegneria, corso di laurea in …

Esame Thermal and Hydraulic Machines

Facoltà Ingegneria i

Dal corso del Prof. Don Giovanni Claudio

Università Politecnico di Torino

A.A. 2013-2014

153 pagine

1 download

Appunto

Scarica

Estratto del documento

Nozzles and diffusors

Stagnation properties

h₀, T₀, p₀, v₀, i₀, s₀ (Total)

State that the fluid reaches when the stream is slowing down up to rest by means of a reversible and adiabatic process.

We apply the first law of thermodynamics (considering adiabatic process)

Q̇_E + Ċ_I = 0

Δh + ΔE_K,g,Ψ

Technical equation conservation law

di = ν dp + dE_K,g,Ψ + dΨ_w = 0

Ideal O = ν dp + dE_K

We integrate this relation between initial condition and total condition:

O = ∫_p^p₀ ν dp + ∫_c^c₀ dE_K

We follow an ideal process and an adiabatic process

Ψ_w = 0

Q_E = 0

Tds = dQ_e + dΨ = 0

dS = 0 (Isentropic Process)

Now:

∫_p^p₀ ν dp = ∫_p^k^=const dp = const^1/kp^k+1/_k |_p^p₀ = (p^v)^1/k = (p⁰)^k - p^k+1= 1/_1/k ν = K/_k-1 [ ( p₀/^{P )^k-1 - 1 ]}

And also, about energy:

∫_c^c₀ dE_K = ∫_c^c₀ c²/₂ dc = c²/₂ |_c^c₀ = -c²/₂

Nozzles and diffusors

h, T, p, v, i, sStagnation Properties

h⁰, T⁰, p⁰, v⁰, i⁰, s⁰ (total)

State that the fluid reaches when the stream is slowing down or to rest by means of reversible and adiabatic process.

We apply the first law of thermodynamics (considering adiabatic process)

∑Q̇_e + ∑Ẇ_i = Δh + ΔE_{k, p, g, w}

Technical Equation Conservation Law

dẋ_i = vdp + dE_{k, p, g, w} + dẆ_w = 0

ideal 0 = vdp + dE_k

We integrate this relation between initial condition and total condition:

0 = ∫_p^p⁰ vdp + ∫_c^c⁰ dE_k

We follow an ideal process and an adiabatic process

Ẇ_w : 0

∑Tds = dq_c + dẋ_w -> dS = 0 (isentropic process)

Now:

∫_p^p⁰ vdp = ∫_p^p⁰ const/ p^1/k dp = const/kP^1/k ∫_p^p⁰ p^k+1 dp = (p^v)^1/k (k/k-1)^k-k-1 [p^k]= [p^k]^1/k - :=[p^1/k v^k/k-1 P^k-1/k P^k-1/k k/k-1 ]

And also, about energy:

∫_c^c⁰ dE_k = ∫_c^c⁰ d^c²/2 = -c²/2

PV_k^k-1 = c²/2

p^o = p[1 + ^k-1/₂ M²]^k/k-1

For Definition

c_s² = (^∂p/_∂ρ)_s = k const ρ^k = const p/ρρ^k = const => P/ρ_k = const

SO k PV = c_s²^c/_{c_s} = M (MACH NUMBER)

M < 1 → Subsonic
M > 1 → Hypersonic

Substituting above we hold:

p^o = p[1 + ^k-1/₂ M²]^k/k-1

To compute the total density we consider ^p/_{p_k} = ^{p^o}/_{p_{o_k}} (Adiabatic transformation)

ρ^o = ρ(^{p^o}/_p)^1/k => ρ^o = ρ[1 + ^k-1/₂ M²]^1/k-1

For Ideal Gas

PV = RT

dh = C_pdT => h° - h = C_p(T° - T)

P°V° PV = T°T => T°T = ^P°V/_PV(1 + ^{k - 1}/₂Π²)k - ¹/_{k - 1} = 1

Hence for Ideal Gas:

T° = T(1 + ^{k - 1}/₂Π²)

We introduce the meaning of nozzles

In general, these are in machines. In a nozzle, the velocity increases => dc > 0, while if we have a diffuser, dc < 0.

We consider the inlet section and outlet section:

=> { } => 0 2

We don't know what there is in the middle. To study it in a simple way, we introduce some hypotheses:

Ideal process (dL = 0)
Adiabatic process (dQ = 0)
One-dimensional fluid flow
Steady state condition G = const

This is the meaning of steady state condition.

dG = 0 (remember that G = ρAC)

=> ^dG/_G = ^dP/_ρ + ^dA/_A + ^dC/_C = 0

Anteprima

Vedrai una selezione di 10 pagine su 153