---
title: "Mass and energy equations for CV; nozzle, condenser"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:mass-energy-cv-nozzle-condenser"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/mass-energy-cv-nozzle-condenser
site: BCT Engineering
---

# Mass and energy equations for CV; nozzle, condenser

Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics

- Asked in: 2069 Bhadra, 6 marks
- Video: [Steady Flow Systems - Nozzles and Diffusers | Thermodynamics | (Solved examples)](https://www.youtube.com/watch?v=u5ysyw0mc1g) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/mass-energy-cv-nozzle-condenser
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

Write down general expressions for mass and energy conservation for a control volume. Reduce these equations for an adiabatic nozzle and condenser.

## Short answer

Mass conservation: dm_cv/dt = Σṁ_i - Σṁ_e. Energy: dE_cv/dt = Q̇ - Ẇ + Σṁ_i(h_i + V_i²/2 + gz_i) - Σṁ_e(h_e + V_e²/2 + gz_e). Adiabatic nozzle: h₁ + V₁²/2 = h₂ + V₂²/2. Condenser: ṁ_s(h₁ - h₂) = ṁ_w(h₄ - h₃).

## Answer

#### General expressions for a control volume
**Mass:** dm_cv/dt = Σṁ_i − Σṁ_e, with ṁ = ρAV = AV/v

**Energy:** dE_cv/dt = Q̇ − Ẇ + Σṁ_i (h_i + V_i²/2 + g z_i) − Σṁ_e (h_e + V_e²/2 + g z_e)

For steady state (dm_cv/dt = 0, dE_cv/dt = 0): Σṁ_i = Σṁ_e, and  
Q̇ − Ẇ = Σṁ_e (h + V²/2 + gz)_e − Σṁ_i (h + V²/2 + gz)_i

#### Adiabatic nozzle
One inlet (1), one exit (2); Q̇ = 0, Ẇ = 0, ΔPE = 0.

Mass: ṁ₁ = ṁ₂ ⇒ A₁V₁/v₁ = A₂V₂/v₂

Energy: h₁ + V₁²/2 = h₂ + V₂²/2

**V₂ = √[ V₁² + 2 (h₁ − h₂) ]**

#### Condenser
A steam stream (1 → 2) and a cooling-water stream (3 → 4) exchange heat inside; with the whole condenser as CV, Q̇ = 0 (insulated), Ẇ = 0, ΔKE = ΔPE = 0.

Mass (each stream): ṁ_s1 = ṁ_s2 = ṁ_s, ṁ_w3 = ṁ_w4 = ṁ_w

Energy: ṁ_s h₁ + ṁ_w h₃ = ṁ_s h₂ + ṁ_w h₄

**ṁ_s (h₁ − h₂) = ṁ_w (h₄ − h₃)**

The heat given up by condensing steam equals the heat gained by the cooling water. Cooling water flow needed: ṁ_w = ṁ_s (h₁ − h₂)/(c_w (T₄ − T₃)).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a control volume at steady state, what does the mass conservation equation reduce to?
   A. dm_cv/dt = Σṁ_i + Σṁ_e
   B. Σṁ_i = Σṁ_e
   C. Σṁ_i = 0
   D. Σṁ_i V_i = Σṁ_e V_e
   Answer: B. At steady state dm_cv/dt = 0, so total mass inflow rate equals total mass outflow rate. (Key point: Steady-state mass balance)

2. Which assumptions reduce the energy equation of an adiabatic nozzle to h1 + V1^2/2 = h2 + V2^2/2?
   A. Q = 0, W = 0, ΔPE = 0 (steady flow, one inlet, one exit)
   B. Q = 0 only
   C. ΔKE = 0 and W = 0
   D. W = 0 and constant pressure
   Answer: A. Adiabatic gives Q = 0, a nozzle has no shaft work, and the elevation change is negligible; KE change is retained because it is the purpose of a nozzle. (Key point: Nozzle energy equation)

3. From the nozzle energy equation, the exit velocity is given by:
   A. V2 = sqrt[V1^2 + (h1 − h2)]
   B. V2 = sqrt[V1^2 − 2(h1 − h2)]
   C. V2 = sqrt[2(h2 − h1)]
   D. V2 = sqrt[V1^2 + 2(h1 − h2)]
   Answer: D. Rearranging h1 + V1^2/2 = h2 + V2^2/2 gives V2^2 = V1^2 + 2(h1 − h2); the enthalpy drop becomes kinetic energy. (Key point: Nozzle exit velocity)

4. Taking the whole condenser (steam 1→2, cooling water 3→4) as the control volume, the energy balance gives:
   A. ṁ_s (h1 − h2) = ṁ_w (h4 − h3)
   B. ṁ_s h1 = ṁ_w h4
   C. ṁ_s (h2 − h1) = ṁ_w (h4 − h3)
   D. ṁ_s = ṁ_w
   Answer: A. With Q = W = 0, the heat given up by condensing steam equals the heat gained by the cooling water. (Key point: Condenser energy balance)

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