Bioprocess Engineering Questions and Answers – Solving Unsteady – State Energy Balances

This set of Bioprocess Engineering Questions and Answers for Aptitude test focuses on “Solving Unsteady – State Energy Balances”.

1. “Chemical reactions or phase changes occur” is this statement applicable for unsteady state energy balance?
a) True
b) False
View Answer

Answer: b
Explanation: No chemical reactions or phase changes occur as the system is well mixed with uniform temperature and composition. Properties of the outlet stream are therefore the same as within the system and mixtures and solutions are ideal.

2. A boat has a thermo-server attached to it. While floating with the fluid, it measures the temperature of the fluid. Both flow and temperature are under unsteady-state condition:
T (x, y, z, t) = x2 + yz + t
\(\vec{V}(x, y, z, t) = 2xi + 2 t^2y\hat{j} + \hat{k}\)
Determine the rate of change of the temperature recorded by the sensor at t = 1, when the boat flows past the location, whose spatial co-ordinates are 2i + j + k.
DT/ dt = \(\frac{\partial T}{\partial t} + (V.\nabla) T\)
\(\frac{\partial T}{\partial t} = 1\)
\(\frac{\partial T}{\partial x} = 2x; \frac{\partial T}{\partial y} = z; \frac{\partial T}{\partial z} = y\)
Here, (x, y, z) = (2, 1, 1)
a) 30
b) 25
c) 20
d) 35
View Answer

Answer: c
Explanation: Substituting in the equation given in the question:
\(\frac{DT}{dt}\)|t=1 = 1+(2×2)(2×2)
(x, y, z) = (2, 1, 1)\(V_x \frac{\partial T}{\partial x}\)
+ (2 × 12 x 1) × 1 + 1×1
\(V_y\frac{\partial T}{\partial y}   V_z\frac{\partial T}{\partial z}\)
= 1 + 16 + 2 + 1 = 20
(Note: Vx = 2x, Vy = 2t2y, Vz = 1)

3. The exothermic elementary liquid-phase reaction is as below:
The exothermic elementary liquid-phase reaction with cooling coil to reactor isothermal
is carried out in a batch reactor with a cooling coil to keep the reactor isothermal at 27°C. The reactor is initially charged with equal concentrations of A and B and no C, cAo = cBo = 2.0 mol/L, cCo = 0.
Additional data:
Rate constant, k = 0.01725 L/mol. min, at 27°C
Heat of reaction, ΔHR = -10 kcal/mol A, at 27°C
Partial molar heat capacities, \(\bar{C}_{PA} = \bar{C}_{PB} \) = 20 cal/mol. K, \(\bar{C}_{PC}\) = 40 cal/mol K
Reactor volume, VR = 1200 L
How long does it take to reach 95% conversion?
a) 550 min
b) 552 min
c) 553 min
d) 551 min
View Answer

Answer: d
Explanation: Assuming constant density, the material balance for component A is
The material balance for component A at constant density

advertisement
advertisement

The stoichiometry of the reaction, and the material balance for B gives
The stoichiometry of the reaction, & the material balance for B

or cA = cB. Substitution into the material balance for species A gives

The material balance for species A for cA = cB

Separation of variables and integration gives

advertisement

By integrating the values for k & cAo gives t 551 min

Substituting cA=0.05cAo and the values for k and cAo gives
t = 551 min.

advertisement

4. Refer to Q3, and estimate what is the total amount of heat (kcal) that must be removed by the cooling coil when this conversion is reached?
Using the following equation:
Q ̇= ΔHR r VR
a) -2.3 × 104 kcal
b) -2.2 × 104 kcal
c) -2.4 × 104 kcal
d) -2.6 × 104 kcal
View Answer

Answer: a
Explanation: We assume the incompressible-fluid energy balance is accurate for this liquidphase reactor. If the heat removal is manipulated to maintain constant reactor temperature, the time derivative in Equation vanishes leaving
Q ̇ = ΔHR r VR
Substituting dcA/ dt = -r and multiplying through by dt gives
dQ = – ΔHR VR dcA
Integrating both sides gives
Q = – ΔHR VR (cA– cAO) = -2.3 × 104 kcal.

5. Refer to Q3 and Q4 and estimate what is the maximum rate at which heat must be removed by the cooling coil (kcal/min) and at what time does this maximum occur?
Using the following equation:
Q ̇ = ΔHR kc2Ao VR
a) −828 kcal/min
b) –880 kcal/min
c) -825 kcal/min
d) – 850 kcal/min
View Answer

Answer: a
Explanation: The right-hand side is a maximum in absolute value (note it is a negative quantity) when cA is a maximum, which occurs for cA = cAo, giving

Qmax = ΔHR kc2Ao VR = (−10 kcal/mol A) (0.01725 L/mol•min) (2.0 mol/L )2 (1200 L)
= −828 kcal/min.

6. Refer to Q3, Q4 and Q5 estimate what is the adiabatic temperature rise for this reactor and what is its significance?
Using the equation:
The adiabatic temperature rise for the reactor 250 K
a) 230 K
b) 200 K
c) 250 K
d) 260 K
View Answer

Answer: c
Explanation: The maximum temperature rise corresponds to complete conversion of the reactants and can be computed from the given data

The maximum temperature rise corresponds to complete conversion of the reactants

The adiabatic temperature rise indicates the potential danger of a coolant system failure. In this case the reactants contain enough internal energy to raise the reactor temperature by 250 K.

7. “All of the coolant is at a uniform temperature, Tc”, is this statement correct according to the assumptions for the Unsteady-state energy balance?
a) True
b) False
View Answer

Answer: a
Explanation: All of the coolant is at a uniform temperature, Tc as the increase in coolant temperature as the coolant passes through the coil is neglected.

8. Fruit juice is fed to a heat exchanges at the rate of 15 kg/h. Saturated steam at 145 kPa pressure is used to heat the juice from 70C to 900C. Assuming the heat capacity of juice 5kJ/kg°C, find out the quantity of steam required for the operation.
Quantity of steam required = 2.97 kg/h if latent heat of condensation is 2229

Using the equation:
FHF + SHS = PHP + CHC
λs (Latent heat of condensation) = 2229 at 145 kPa.
a) 2.90 kg/h
b) 2.97 kg/h
c) 2.99 kg/h
d) 2.95 kg/h
View Answer

Answer: b
Explanation: Equation given in the question can be rewritten as :
Amount of steam required = 2.97 kg/h for raising temperature of 15 kg/h fruit juice
The negative sign indicates the heat is given from the steam to juice.
Hence, the amount of steam required is 2.97 kg/h for raising the temperature of 15 kg/h fruit juice from 7°C to 900°C.

9. The evaporator economy is dependent on the ________
a) Heat transfer rate
b) Mass transfer rate
c) Energy balance considerations
d) Mass balance considerations
View Answer

Answer: a
Explanation: There are three main measures of evaporator performance:
Capacity (kg vaporized/time), Economy (kg vaporized/ kg steam input), Steam consumption (kg/hr).
The performance of a evaporator is evaluated by the capacity and the economy.
The rate of heat transfer q through the heating surface of an evaporator, by the definition of overall heat transfer coefficient, is product of three factors
i) The area of heat transfer surface A
ii) The overall heat transfer coefficient U
iii) The overall temperature drop ΔT
Q = U * A * ΔT.

10. The fouling factor is _________
a) Is a safety factor
b) Accounts for all resistances due to mass transfer
c) Is a dimensionless number
d) Accounts for all resistances due to heat transfer
View Answer

Answer: d
Explanation: The fouling factor represents the theoretical resistance to heat flow due to a build-up of a layer of dirt or other fouling substance on the tube surfaces of the heat exchanger, but they are often overstated by the end user in an attempt to minimise the frequency of cleaning. In reality, if the wrong fouling factor is used, cleaning may actually be required more frequently.

11. The units for the log mean temperature difference are __________
a) 1 / °C
b) °C
c) 1/2 °C
d) Dimensionless
View Answer

Answer: b
Explanation: In the particular system, whatever unit is assigned to temperature, the same is the unit of LMTD. It is just a special kind of temperature difference. Logarithmic mean temperature difference. LMTD is a logarithmic average of the temperature difference between the hot and cold streams in heat exchanger. Its unit is whatever is the unit of the temperature. °F or °C or Kelvin.

Sanfoundry Global Education & Learning Series – Bioprocess Engineering.

To practice all areas of Bioprocess Engineering for Aptitude test, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.