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Correct option is (a) 1164 KJ

The BEST I can explain: Here, we need to convert grams of N2 to moles of N2 and use the CONVERSION factor 180.6 kJ/1 mol N2.

180.6 kJ × (1 mol N2/28.01 g N2) × (180.6 kJ/1 mol N2) = 1164 kJ.

The correct choice is (a) 69390.84 kJ/h

For EXPLANATION I would SAY: Heat lost by gluconic acid = heat gained by water

[2000(0.2)×(0.35×4.184×1000/1000)×(90-6) + 2000×(0.8)×(376.92-25.2)]+Q

= 2700×(209.33-8.37)×611955.84+Q=542565

Q = -69390.84 kJ/h

Negative sign MEANS heat loss by the SYSTEM to surrounding.

Hence, heat loss to the surroundings is 69390.84 kJ/h.

Correct option is (B) COMBUSTION

Easy EXPLANATION: It is burning (combustion). The combustion PROCESS generates heat that’s transformed into energy. In the combustion process biomass is burned and CONVERTED into energy.

The correct option is (a) 119.544 J/kg

The EXPLANATION is: PE INCREASE PER LB = Ep / m = gh = (9.806 m/s^2) (40 m/3.2808) = 119.54 m^2/s^2 = 119.544 J/kg.

Right choice is (b) Pyrolysis

Best explanation: Pyrolysis is defined as thermal conversion (destruction) of ORGANICS in the absence of OXYGEN. In the BIOMASS community, this commonly refers to lower temperature thermal PROCESSES producing liquids as the primary product. Possibility of chemical and food BYPRODUCTS.

The correct choice is (a) -144 kJ/KG

The explanation is: BASIS: 1 kg steam

For superheated steam:

Inlet: (60 bar, 500°C) h1 = 3422 kJ/kg

Outlet: (1 bar, 400°C) h2 = 3278 kJ/kg

min = mout = 1 kg

DECV + Sni(Ek + Ep + H)i = Q + WS (General form of the energy balance)

Now we make the following simplifying assumptions for this turbine:

▪ Adiabatic: Q = 0

▪ The inlet and outlet lines are not tremendously different in height: Ep = 0

▪ There is little difference in the velocity of the FLUID in the inlet and outlet lines: Ek = 0

▪ There are no accumulation terms (steady-state operation with steady flow): DECV = 0

The simplified form of the energy balance is therefore:

Hout – Hin = Ws = mouthout – minhin = m(hout – hin)

Substituting in the values for the specific enthalpies gives:

Ws = m (hout – hin) or Ws/m = (3278 – 3422) kJ/kg = -144 kJ/kg.

Correct option is (B) 5368 kg/hr

For EXPLANATION I would say: Mass in: Total mass = 35000/6 = 5833 kg.

Fat = 5833 x 0.04 = 233 kg.

And so Water plus solids-not-fat = 5600 kg.

Mass out: Let the mass of cream be x kg then its total fat content is 0.45x. The mass of skim MILK is (5833 – x) and its total fat content is 0.0045 (5833 – x).

Material BALANCE on fat: Fat in = Fat out

5833 x 0.04 = 0.0045(5833 – x) + 0.45x. And so x = 465 kg.

So that the flow of cream is 465 kg / hr and skim milk (5833 – 465) = 5368 kg/hr.

Correct choice is (b) 457.2 mol/hr

Easiest explanation: SOLVING for ξ ̇and substituting the corresponding quantities into this equation yields:

The rate in enthalpy change was considered to be negative since the problem is stating that the REACTION is releasing ENERGY (exothermic reaction).

Now we can ENTER the calculated extent of reaction into the equation PREVIOUSLY solved for the molar production rate of methanol, to get:

Correct option is (b) – 2001 J

To explain: Since the heat gained by the calorimeter is EQUAL to the heat lost by the system, then the substance inside must have lost the negative of +2001 J, which is – 2001 J.

The correct choice is (a) +360.139 KJ

Easy EXPLANATION: ΔH° = ƩΔvpΔH°f (products) – ƩΔvrΔH°f (REACTANTS) so this means that you add up the SUM of the ΔH’s of the products and subtract away the ΔH of the products:

(- 275.2 KJ) – (-393.509 KJ + -241.83KJ) = (-275.2) – (-635.339) = + 360.139 KJ.

Right OPTION is (c) 25°C, 1 atm

Explanation: The standard HEAT of combustion Δh°c is the specific ENTHALPY CHANGE associated with this reaction at standard conditions, usually 25°C and 1 atm PRESSURE. The standard heat of reaction is the difference between the heats of combustion of reactants and products.

Correct answer is (c) Combustion

Easy explanation: Combustion is DEFINED as thermal conversion of organic matter with an oxidant (NORMALLY oxygen) to produce primarily carbon dioxide and WATER. The oxidant is in stoichiometric EXCESS, i.e., complete oxidation.

The correct option is (b) 1.00 J/kg

To explain I would say: Kinetic energy Ek =mv^2, tube dia. D = 3.0 CM, m = 1 kg.

Cross-section area of tube A = \(\FRAC{1}{4}\) πD^2 = \(\frac{1}{4}\) π(3.0/100 m)^2 = 7.0686×10^-4m^2

Average velocity of water v = Q/A = 0.001 m^3 /(7.0686×10^-4m^2) = 1.415 m/s

KE PER kg = Ek/m = 1/2 v^2 = 1/2 (1.415 m/s)^2 = 1.00 m^2/s^2 = 1.00 J/kg.

Correct OPTION is (B) 23.2 kg-AIR/kg-fuel

For explanation I WOULD say: 50% EXCESS air ( 150% Theoretical air):

Right CHOICE is (C) 0.0514 btu/lb

For EXPLANATION: PE increase PER lb = Ep / m = GH = (9.806 m/s^2) (40 m/3.2808) = 119.54 m^2/s^2 = 119.544 J/kg = (119.544 btu/1055.6)/2.2046 lb = 0.0514 btu/lb.

The correct choice is (c) 44.5 tons of refrigeration

For EXPLANATION I would say: With motors outside, the motor INEFFICIENCY = (1 – 0.86) does not impose a LOAD on the refrigeration

Total heat load = (104 + [0.86 x 53.7] + 6.3)

= 156 kW

= 44.5 tons of refrigeration.

The CORRECT CHOICE is (c) 7.4%

The best explanation: NCO2 = 3, Ntotal = 3+4+3.05+ 30.27 = 40.32

The correct OPTION is (a) 45.8°C

For explanation I would SAY: NH2O = 4, \(\frac{P_{H_{2}O}}{P} = \frac{N_{H_2O}}{N_{total}} = \frac{4}{40.32} => P_{H_2O} = (101.32 kPa)[\frac{4}{40.32}]\) = 10.05 kPa

Tdew-point = [email protected] = 45.8°C.

Correct option is (a) – 18.61 kJ

The best I can explain: The REACTION is NH4NO3(s) → N2O(g) + 2 H2O(g)

Given 2 mol of NH4NO3 decomposes, hence multiplying equation by 2

2 NH4NO3(s) → 2 N2O(g) + 4 H2O(g)

Δn = nproduct (g) – nreactant (g) = (2 + 4) -(0) = 6

Work done in chemical reaction is given by

∴ W = – Δn RT = – (6) mol × 8.314 J K^-1 mol^-1 × 373 K = – 18607 J

∴ W = – 18.61 kJ

Negative sign indicates that work is done by the system on the surroundings

Work done by the surroundings on the system in the reaction is – 18.61 kJ.

Right choice is (a) FAT = 4.5%, Water = 86.5%, protein = 3.3%, carbohydrate = 4.9%, ASH = 0.8%

EASY explanation: Basis: 100 kg of skim milk.

This contains, therefore, 0.1kg of fat. Let the fat which was removed from it to make skim milk be x kg.

Total original fat = (x + 0.1) kg

Total original mass = (100 + x) kg

and as it is shown that the original fat content was 4.5% so,

(x+0.1)/(100+x) = 0.045

where = x+0.1 = 0.045(100+x)

x = 4.6 kg

So the COMPOSITION of the whole milk is then fat = 4.5%, water = 90.5/104.6 = 86.5%, protein = 3.5/104.6 = 3.3%, carbohydrate = 5.1/104.6 = 4.9% and ash = 0.8%.

The CORRECT option is (d) Condensation

The best EXPLANATION: Condensation is the change of the PHYSICAL STATE of matter from gas PHASE into liquid phase, and is the reverse of evaporation.

The correct choice is (b) False

Easy explanation: Heat transferred to raise or lower the temperature of a material is called sensible heat; change in the enthalpy of a SYSTEM due to VARIATION in temperature is called sensible heat change, WHEREAS the TERM specific heat refers to heat capacity expressed on a per-unit-mass BASIS.

The correct choice is (c) 3.6 MJ

For explanation I would say: ENERGY = power x time

1 kW is 1,000 watts

Since 1 WATT is 1 joule PER sec

1,000 watts is 1,000 joules per sec

In one hour there are 3,600 seconds

Substituting in the equation energy = 1,000 x 3,600 joules

= 3,600,000 joules

= 3.6 MJ

Therefore a 1 kW heater converts 3.6 MJ of electrical energy into heat per hour.

The CORRECT choice is (b) 100 W

For explanation: A man doing no or very little physical work needs about 2,000 kcal (or less) of energy in his daily food. The body converts this energy almost entirely into heat.

1 day = 24 x 60 x 60 s = 86,400 s 1 cal = 4.2 J

Hence, 2000 kcal/day = 2000 × 4.2 kj/day = \(\frac{8.4 \,Mj}{86600s}\)

= 100 j/s = 100 W

We see that a HUMAN body doing no work is EQUIVALENT to a heat source of about 100 W – the equivalent of a GOOD bulb.

The correct choice is (a) True

Best explanation: EXTENSIVE properties are linearly dependent on the amount of substance. Examples: mass, volume, energy. TAKE two identical samples with all properties identical and combine them into a single sample. Properties that DOUBLE are extensive. Properties that remain the same are intensive. For example, if I took 1.0 liter of WATER at room temperature (25 °C) and added another 1.0 liter of water at the same temperature then I would have 2.0 liters of water at 25 C. From this example we see that Volume and Mass are extensive properties (i.e., volume and mass doubled), while Temperature is an intensive property (i.e., temperature stayed the same). You would also expect the density to remain the same, so it is also an intensive property.

Correct CHOICE is (a) 629.6 BTU/LBM

Easiest explanation: h = u + Pv

h = \(600 \frac{Btu}{lbm} + (100 \frac{lbf}{in^2}) (1.6 \frac{ft^3}{lbm}) (144 \frac{in^2}{ft^2}) (\frac{Btu}{778 \,ft-lbf})\) = 629.6 btu/lbm.

The CORRECT answer is (b) 100.4 KJ

Explanation: Enthalpy (“Water”, 20, 000,”l”)

= 6.032 kJ/mol \((∫_{20}^{100} C_{pH_2O} dT)\)

= \( (\frac{6.032 kJ}{mol}) (\frac{1g}{ml}) (\frac{mol}{18.02 G}) (\frac{300 ml}{1})\) = 100.4 kJ

Correct choice is (b) 1.17 kg/m^3, 0.00117, 140 kg

To ELABORATE: At specified conditions, air can be treated as ideal gas.

The gas constant of air is R = 0.287 K Pa-m^3/Kg K

The density of the air is determined from the ideal-gas RELATION, P = ρ R T to be

ρ = P/RT = (100 kPa)/((0.287 k Pa-m^3/Kg K) (25+273.15)K) = 1.17 kg/ m^3

Then the SPECIFIC gravity of the air becomes

SG = (ρ)/ρH2O = (1.17 kg/m^3)/(1000 kg/m^3) = 0.00117

Finally, the volume and mass of the air in the room are

V = (4m) (5m) (6m) = 120 m^3

m = ρV = (1.17 kg/m^3) (120 m^3) = 140 kg.

Right option is (d) SPECIFIC internal energy

The EXPLANATION is: The symbol “Û” refers to Specific internal energy in a close system in Enthalpy CHANGE in non-reactive PROCESSES WHEREAS in an open- system symbol “Ĥ” refers to specific enthalpy.

Right choice is (B) False

To explain: Energy input for downstream PROCESSING is minimised to avoid DAMAGING heat-labile products. Nevertheless, energy EFFECTS are important because biological catalysts are very sensitive to heat and changes in temperature. In large-scale processes, heat released during reaction can CAUSE cell death or denaturation of enzymes if it is not quickly removed.

The correct answer is (b) Dry- bulb temperature

For explanation I would say: The dry-bulb temperature (DBT) is the temperature of air measured by a THERMOMETER FREELY EXPOSED to the air but shielded from radiation and moisture. DBT is the temperature that is USUALLY thought of as air temperature, and it is the TRUE thermodynamic temperature.

Right answer is (C) Mass

For explanation I would say: TEMPERATURE, density, and mole fraction are examples of properties which are independent of the SIZE of the system; these quantities are called intensive variables. On the other hand, mass, volume and energy are extensive variables which change if mass is added to or REMOVED from the system.

Correct ANSWER is (a) 0.001341 hp

The BEST I can explain: 1 watt = 1 W = 1 J/1 s = 10^-3 kW = 10^-6 MW

= 3.412 BTU/h

= 0.001341 hp (HORSEPOWER UNITS).

The correct CHOICE is (a) 4.6 bar

The best explanation: ΔP/ρ + Δv^2/2 + gΔh + F = W

ΔP = P2 – P1 (Pa)

Δv^2 = v2^2 – v1^2

Δh = h2 – h1 (m)

F = frictional energy loss (mechanical energy loss to system) (J/kg)

W = WORK DONE on system by pump (J/kg)

ρ = 1000 kg/m^3

Volumetric flow is 20/ (1000.60) m^3/s

= 0.000333 m^3/s

v1 = 0.000333/(π(0.0025)^2) = 16.97 m/s

v2 = 0.000333/ (π(0.005)^2) = 4.24 m/s

(101325 – P1)/1000 + [(4.24)^2 – (16.97)^2]/2 + 9.81.50 = 0

P1 = 456825 Pa (4.6 bar).