Blending Steam Streams to Make 300 °C Superheated Steam: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2013, Q3

Question 3: Blending Steam Streams to Make 300 °C Superheated Steam (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Adiabatic mixing of two steam streams, all at 1.0 atm: saturated steam $\dot m_1 = 1000$ kg/h and superheated steam at 400 °C ($\dot m_2$, unknown), producing superheated steam at 300 °C. Steam-table data (1 atm ≈ 0.1 MPa):

State (1 atm)	$\hat H$ (kJ/kg)	$\hat V$ (m³/kg)
Saturated vapour (100 °C)	2676.1	—
Superheated, 300 °C	3074.3	2.639
Superheated, 400 °C	3278.2	3.103

Find. the volumetric flow rate of the 400 °C steam and the production rate of 300 °C steam.

Figure 2 — Adiabatic mixing junction: saturated steam and 400 °C steam blend to a 300 °C product, all at 1 atm.

Approach. Because the pipes are adiabatic and at constant pressure, the mixing is a steady enthalpy balance: the enthalpy carried in by the two feeds equals that of the product. Combine it with the total mass balance to get $\dot m_2$, then convert to volume with the 400 °C specific volume.

Mass and energy balances on the junction. With $\dot m_3 = \dot m_1 + \dot m_2$ and no heat loss or work, $$\dot m_1 \hat H_1 + \dot m_2 \hat H_2 = (\dot m_1+\dot m_2)\hat H_3.$$
Solve for the 400 °C stream. Rearranging for $\dot m_2$ with the tabulated enthalpies, $$\dot m_2 = \dot m_1\,\frac{\hat H_3-\hat H_1}{\hat H_2-\hat H_3} = 1000\,\frac{3074.3-2676.1}{3278.2-3074.3} = \frac{398{,}200}{203.9} = \boxed{1953\ \text{kg/h at }400\ ^\circ\text{C}}.$$
Production rate of 300 °C steam. The total mass balance gives $$\dot m_3 = 1000 + 1953 = \boxed{2953\ \text{kg/h of }300\ ^\circ\text{C steam}}.$$
Volumetric flow of the 400 °C steam. Multiply the mass rate by the specific volume at 400 °C, 1 atm: $$\dot V_2 = \dot m_2\,\hat V_2 = 1953(3.103) = \boxed{6.06\times10^{3}\ \text{m}^3/\text{h}}.$$

Quantity	Result
400 °C steam required	1953 kg/h
Volumetric flow of 400 °C steam	≈ 6060 m³/h
300 °C steam produced	2953 kg/h

Concept & Theory

Topic. Adiabatic isobaric mixing of two vapour streams — an open-system energy balance with tabulated enthalpies.

Key relations. Steady open-system balance $\Delta(\dot m \hat H)=\dot Q - \dot W_s$; here $\dot Q = \dot W_s = 0$, so enthalpy is conserved: $\sum_\text{in}\dot m\hat H = \sum_\text{out}\dot m\hat H$. Specific volume from the steam tables converts mass to volumetric flow.

Why this works. Enthalpy is the correct energy variable for a flowing, constant-pressure stream because it bundles internal energy with the flow work $P\hat V$. Reading $\hat H$ straight from the steam tables sidesteps any need for heat capacities or a reference temperature — the tables already carry a common reference (triple point).

Common pitfall. Using $\Delta H = m c_p \Delta T$ with a single average $c_p$ across the saturation line; that ignores the latent heat contained in "saturated vapour" versus superheated states. Always use the tabulated $\hat H$ values when a phase or large superheat is involved.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §2.9–2.10 (mass and energy balances for open systems; steam tables).

Question 3 of 7: Blending Steam Streams to Make 300 °C Superheated Steam

Question 3: Blending Steam Streams to Make 300 °C Superheated Steam (equal value)