Documentation:CHBE Exam Wiki/Module 5 - Non-reactive Energy Balances

CHBE 241 Exam resources wiki
Chemical and Biological Engineering
Welcome to the CHBE Exam Resources Wiki! This wiki is intended to host past exams with fully worked-out hints and solutions
Past Exams
Final Exam 2016W
Midterm Exam 1 2016W
Midterm Exam 2 2016W
Problem Sets
Module 1 - Process Basics
Module 2 - Reactors
Module 3 - Separations 1
Module 4 - Separations 2
Module 5 - Non-reactive Energy Balances
Module 6 - Reactive Energy Balances

Question 1

At 150K and 41.64 atm, oxygen has a tabulated specific volume of 4.684cm³ /g and a specific internal energy of 1706 J/mol.
Determine the specific enthalpy of O₂ in this state.

showSolution

${\begin{aligned}{\hat {V}}&={\frac {32g}{mol}}\cdot {\frac {4.684cm^{3}}{g}}\cdot {\frac {10^{3}L}{10^{6}cm^{3}}}\\&=0.1499L/mol\\\end{aligned}}$

${\begin{aligned}{\hat {H}}&={\hat {U}}+P{\hat {V}}\\&=1706+41.64atm\cdot {\frac {0.1499L}{mol}}\cdot {\frac {8.314J/molK}{0.08206Latm/molK}}\\&=2388J/mol\end{aligned}}$

Question 2

Prior to entering a furnace, air is heated from 25°C to 150°C and the change in specific enthalpy for this process is 3640 J/mol. The flow rate of air at the outlet of the heater is 1.5 m /min and the air pressure at this point is 150 kPa absolute.

Calculate the heat needed for the process in kW. Assume ideal gas behavior and that kinetic and potential energy changes from the heater inlet to the outlet are negligible.

showSolution

Determine the molar flow rate:
${\begin{aligned}{\dot {n}}&={\frac {1.5m^{3}}{min}}\cdot {\frac {273K}{150+273K}}\cdot {\frac {150kPa}{101.3kPa}}\cdot {\frac {1mol}{22.4L}}\cdot {\frac {10^{3}L}{1m^{3}}}\\&=64.0mol/min\\\end{aligned}}$

Since the changes in kinetic and potential zero are negligible
${\begin{aligned}{\dot {Q}}&=\Delta {\dot {H}}\\&={\dot {n}}\Delta {\hat {H}}\\&={\frac {64.0mol}{min}}\cdot {\frac {1min}{60s}}\cdot {\frac {3640J}{mol}}\cdot {\frac {kW}{10^{3}J/s}}\\&=3.88kW\\\end{aligned}}$

Question 3

Superheated steam at 40 bar absolute and 500°C flowing a rate of 200 kg/min is sent to an adiabatic turbine which expands to 5 bar. The turbine outputs 1250kW. The expanded steam is then sent to a heat exchanger where isobaric heating occurs, resulting in the stream being reheated to its initial temperature. Assume no changes in kinetic energy.

Question 3a

Write the energy balance for the turbine and determine the outlet stream temperature.

showSolution

For water vapor at 500°C and 40 bar, the specific enthalpy is 3445 kJ/kg.
For water vapor at 500°C and 5 bar, the specific enthalpy is 3484 kJ/kg.

The changes in potential and kinetic energy are zero and there is no heat transfer. Hence,
${\begin{aligned}\Delta {\dot {H}}&=-{\dot {W_{s}}}\\&={\dot {m}}\cdot ({\hat {H_{2}}}-{\hat {H_{1}}})\\{\hat {H_{2}}}&={\hat {H_{1}}}-{\frac {\dot {W_{s}}}{\dot {m}}}\\&={\frac {3445kJ}{kg}}-{\frac {1250kJ}{s}}\cdot {\frac {min}{200kg}}\cdot {\frac {60s}{1min}}\\&=3070kJ/kg\end{aligned}}$

At 5 bar and specific enthalpy of 3070kJ/kg, the interpolated temperature of the outlet stream is 302°C

Question 3b

Write an energy balance on the heat exchanger and use it to determine the required input (kW) to the steam.

showSolution

The changes in potential and kinetic energy are zero and there is no shaft work. Hence,
${\begin{aligned}{\dot {Q}}&=\Delta {\dot {H}}\\&={\dot {m}}\cdot ({\hat {H_{3}}}-{\hat {H_{2}}})\\&={\frac {200kg}{min}}\cdot {\frac {(3484-3070)kJ}{kg}}\cdot {\frac {1min}{60s}}\cdot {\frac {1kW}{1kJ/s}}\\&=1380kW\end{aligned}}$

Question 3c

Justify that an overall energy balance on the two-unit process is satisfied.

showSolution

The changes in potential and kinetic energy are zero. Hence,
${\begin{aligned}\Delta {\dot {H}}&={\dot {Q}}-{\dot {W_{s}}}\\&={\dot {m}}\cdot ({\hat {H_{3}}}-{\hat {H_{1}}})\\{\dot {Q}}&={\dot {m}}\cdot ({\hat {H_{3}}}-{\hat {H_{1}}})+\Delta {\dot {W_{s}}}\\&={\frac {200kg}{min}}\cdot {\frac {(3484-3445)kJ}{kg}}\cdot {\frac {1min}{60s}}\cdot {\frac {1kW}{1kJ/s}}+{\frac {1250kJ}{s}}\cdot {\frac {1kW}{1kJ/s}}\\&=1380kW\\\end{aligned}}$

Question 3d

If the inlet and outlet pipes of the turbine are 0.5m, show that it is reasonable to neglect the change in kinetic energy for this unit.

showSolution

For water vapor at 500°C and 40 bar, the specific volume is 0.0864 m³/kg.
For water vapor at 302°C and 5 bar, the specific enthalpy is 0.524 m³/kg.

${\begin{aligned}u_{1}&={\frac {200kg}{min}}\cdot {\frac {1min}{60s}}\cdot {\frac {0.0864m^{3}}{kg}}\cdot {\frac {1}{0.5^{2}\cdot (\pi /4)m^{2}}}\\&=1.47m^{2}/s\\u_{2}&={\frac {200kg}{min}}\cdot {\frac {1min}{60s}}\cdot {\frac {0.524m^{3}}{kg}}\cdot {\frac {1}{0.5^{2}\cdot (\pi /4)m^{2}}}\\&=8.90m^{2}/s\\\end{aligned}}$

${\begin{aligned}\Delta {\dot {E_{K}}}&={\frac {\dot {m}}{2}}\cdot [u_{2}^{2}-u_{1}^{2}]\\&=0.5\cdot {\frac {200kg}{min}}\cdot {\frac {1min}{60s}}\cdot {\frac {[8.9^{2}-1.83^{3}]m^{2}}{s^{2}}}\cdot {\frac {1N}{1kg\cdot m/s^{2}}}\cdot {\frac {1kW\cdot s}{10^{3}N\cdot m}}\\&=0.13kW<<<<1250kW\end{aligned}}$

Question 4

A 15 m³ tank contains steam at 280°C and 15 bar. The tank and its contents are cooled until the pressure drops to 1.3 bar. Some of the steam condenses in the process.

Question 4a

What is the final temperature of the tank contents?

showSolution

At 1.3bar, the saturated stream would be at 107.1 °C

Question 4b

How much steam condensed (kg)?

showSolution

The total mass of water is
${\begin{aligned}m_{i}&={\frac {V_{i}}{\hat {V_{i}}}}\\&=15m^{3}\cdot {\frac {1kg}{0.1837m^{3}}}\\&=81.65kg\\\end{aligned}}$
See part c for the value of specific volume of the inlet stream

Total mass and volume :
${\begin{aligned}m_{v}+m_{l}&=81.65kg\\V_{v}+V_{l}&=15m^{3}\\&=m_{v}{\hat {V_{v}}}+m_{l}{\hat {V_{l}}}\\&=m_{v}(1.325m^{3}/kg)+m_{l}(0.001049m^{3}/kg)\\\end{aligned}}$

Solving the 2 simultaneous equations:
${\begin{aligned}m_{v}&=11.265kg\\m_{l}&=70.385kg\\\end{aligned}}$

Question 4c

How much heat was transferred from the tank?

showSolution

Based on the values from the tables,

INLET: steam at 280°C and 15 bar (interpolation)
${\begin{aligned}{\hat {U_{i}}}&=2748.2kJ/kg\\{\hat {V_{i}}}&=0.1837m^{3}/kg\\\end{aligned}}$

OUTLET: vapor(v) and liquid(l) at 107.1°C and 1.3 bar
${\begin{aligned}{\hat {U_{v}}}&=2514.7kJ/kg\\{\hat {V_{v}}}&=1.325m^{3}/kg\\{\hat {U_{l}}}&=449.1kJ/kg\\{\hat {V_{l}}}&=0.001049m^{3}/kg\\\end{aligned}}$

The energy balance:
${\begin{aligned}Q&=\Delta U\\&=m_{v}{\hat {U_{v}}}+m_{l}{\hat {U_{l}}}-m_{i}{\hat {U_{i}}}\\&=11.265kg(2514.7kJ/kg)+70.385kg(449.1kJ/kg)-81.65kg(2748.2kJ/kg)\\&=-1.64\cdot 10^{5}kJ\\\end{aligned}}$

Question 5

The maximum pressure that a 250L water tank can withstand is 20bar (absolute). At time x, the tank contains 180kg of liquid water and the inlet and outlet valves are closed, and the absolute pressure in the space of above the liquid is 3 bar (assume that the space only contains water vapor). Out of carelessness, the tank heater was turned on and the tank was left unmonitored. Let t₁ be the time when the heater is turned on and t₂ be the moment before the tank explodes. Use the steam tables for the following calculations.

Question 5a

Determine the water temperature, the liquid and head-space volumes (L), and the mass of water vapor in the head space (kg) at time t₂ .

showSolution

At 3 bar, the saturated stream would be at 133.5 °C

Also from the steam table at 3 bar,
${\begin{aligned}{\hat {V_{l}}}&=0.001074m^{3}/kg\\{\hat {V_{v}}}&=0.606m^{3}/kg\\\end{aligned}}$

${\begin{aligned}V_{l}&=0.001074m^{3}/kg\cdot 1000L/m^{3}\cdot 180kg\\&=193.3L\\V_{space}&=V_{tank}-V_{l}\\&=56.7L\\m_{v}&=56.7L\cdot {\frac {1m^{3}}{1000L}}\cdot {\frac {1kg}{0.606m^{3}}}\\&=0.0935kg\\\end{aligned}}$

Question 5b

Determine the water temperature, the liquid and head-space volumes (L), and the mass of water vapor (kg) that evaporates between t₁ and t₂.

showSolution

${\begin{aligned}m_{total}&=m_{l}+m_{v}\\&=180+0.0935\\&=180.09kg\\\end{aligned}}$

Since the maximum pressure is 20 bar, the final pressure is also 20 bar.
From the steam table at 20 bar,
${\begin{aligned}T_{1}&=212.4^{\circ }C\\{\hat {V_{l}}}&=0.001177m^{3}/kg\\{\hat {V_{v}}}&=0.0995m^{3}/kg\\\end{aligned}}$

${\begin{aligned}V_{total}&=m_{l}\cdot {\hat {V_{l}}}+m_{v}\cdot {\hat {V_{v}}}\\&=m_{l}\cdot {\hat {V_{l}}}+(m_{total}-m_{l})\cdot {\hat {V_{v}}}\\250L\cdot {\frac {1m^{3}}{1000L}}&=m_{l}\cdot (0.001177m^{3}/kg)+(180.09-m_{l})\cdot (0.0995m^{3}/kg)\\m_{l}&=179.72kg\\m_{v}&=0.38kg\\\end{aligned}}$

${\begin{aligned}V_{l}&=0.001177m^{3}/kg\cdot 1000L/m^{3}\cdot 179.72kg\\&=211.5L\\V_{space}&=250-211.5\\&=38.5L\\m_{evaporated}&=(0.38-0.0935)kg\cdot \\&=0.284kg\\\end{aligned}}$

Question 5c

Calculate the amount of heat (kJ) transferred to the tank contents between t₁ and t₂. Give two reasons why the actual heat input to the tank must have been greater than the calculated value.

showSolution

We know that the changes in kinetic and potential energy along with work done are zero.

${\begin{aligned}Q&=\Delta U\\&=0.38kg\cdot (2598.2kJ/kg)+179.72kg\cdot (906.2kJ/kg)-0.0935kg\cdot (2543kJ/kg)-180kg\cdot (561.1kJ/kg)\\&=6.26\cdot 10^{4}kJ\\\end{aligned}}$

where,
From the steam table at 20 bar,
${\begin{aligned}{\hat {U_{l}}}&=906.2kJ/kg\\{\hat {U_{v}}}&=2598.2kJ/kg\\\end{aligned}}$

From the steam table at 3 bar,

${\begin{aligned}{\hat {U_{l}}}&=561.1kJ/kg\\{\hat {U_{v}}}&=2543kJ/kg\\\end{aligned}}$

Two reasons: Heat absorbed by the tank walls and heat loss to the surroundings

Question 5d

List three different factors responsible for the increase in pressure resulting from the transfer of heat to the tank.

showSolution

• The formation of vapor from liquid vaporization results in pressure increase
• The specific volume of water increases with temperature, which results in the increase in occupied space with the same mass of water
• The decrease in the head space area

Question 6

Two beakers of aqueous sulfuric acid with different concentrations were mixed to form a 70 wt% product solution. The first solution at 20 wt% H₂SO₄ was taken from the bench top at 25°C while the second solution with 80wt% sulfuric acid was taken from the cooling bath at 15°C. Assume that there was no temperature loss between the mixing process and the initial temperature of the solutions from the respective locations.

Question 6a

If the mass of the 20wt% solution is 5 lb, what is the mass of the 80 wt% solution ?

showSolution

Overall mass balance,
${\begin{aligned}m_{1}+m_{2}&=m_{final}\\5+m_{2}&=m_{final}\\\end{aligned}}$

Sulfuric acid mass balance,
${\begin{aligned}w_{1}\cdot m_{1}+w_{2}\cdot m_{2}&=w_{final}\cdot m_{final}\\(0.2)\cdot 5+0.8\cdot m_{2}&=0.7\cdot m_{final}\\\end{aligned}}$

Solving the simultaneous equations,
${\begin{aligned}m_{2}&=25lb\\m_{final}&=30lb\\\end{aligned}}$

Question 6b

Use the enthalpy-concentration chart for H₂SO₄/Water to estimate the product solution temperature if the mixing is adiabatic.

showSolution

The enthalpy for the 20wt% H₂SO₄ @ 25C and 80wt% H₂SO₄ @ 15C are -28 Btu/lb and -120 Btu/lb respectively from the chart.
Energy balance,
${\begin{aligned}Q&=\Delta H\\&=0\\0&={\hat {H}}_{final}\cdot m_{final}-{\hat {H}}_{1}\cdot m_{1}-{\hat {H}}_{2}\cdot m_{2}\\&=30\cdot {\hat {H}}_{final}-5\cdot (-28)-25\cdot (-120)\\{\hat {H}}_{final}&=-104Btu/lb\\\end{aligned}}$

Based on the diagram, the estimated temperature is 120°F (≈49°C)

Question 6c

If the product solution drops to 25°C, how much heat is lost from the solution to the surrounding under a constant atmospheric pressure.

showSolution

The enthalpy for the 70wt% H₂SO₄ @ 25°C is approximately -124 Btu/lb respectively from the chart.
Energy balance,
${\begin{aligned}Q&=m_{final}\cdot [{\hat {H}}_{final,2}-{\hat {H}}_{final,1}]\\&=30\cdot [-124-(-104)]\\&=-600Btu\\\end{aligned}}$

Question 6d

In terms of safety, is it safer to add a concentrated acid to a dilute acid or vice versa? Use the chart to explain your response.

showSolution

The temperature rise is much slower when the concentration of the product solution increases from low to high (left to right on the chart).