### YEAR 2017

• Q.1

The curve measured during a pulse tracer experiment is shown below. In the figure, C(t) is the concentration of the tracer measured at the reactor exit in mol/liter at time t seconds.

The mean residence time in the reactor, rounded to 1 decimal place, is _________ s.

Marks:2 Explanation

Mean residence time = $$\frac{\int_0^\infty tc(t)dt}{\int_0^\infty c(t)dt}$$ .

$$\int_0^\infty tc(t)dt \space= \space \int_0^1t\times tdt \space +\space \int_1^2(2-t)\times t dt\space +\space \int_2^30\times dt$$

$$\implies \space \int_0^\infty tc(t)dt\space =\space \left|\frac{t^3}{3}\right|_0^1\space+\space \left|(t^2\space-\space \frac{t^3}{3})\right|_1^2\space =\space \frac{1}{3}\space + [(4\space-\space\frac{8}{3})\space-\space(1\space-\space\frac{1}{3})]\space=\space\frac{1}{3}\space+\space (4-\frac{8}{3}-1+\frac{1}{3})$$

$$\implies \space \int_0^\infty t c(t)dt\space =\space \frac{1}{3}\space+3\space-\frac{7}{3}\space =\space \frac{(1+9-7)}{3}\space=\space\frac{3}{3}\space=\space 1$$.

$$\int_0^\infty c(t)dt \space= \space \int_0^1 tdt \space +\space \int_1^2(2-t) dt\space +\space \int_2^30dt\space = \space \left|\frac{t^2}{2}\right|_0^1\space+\left|2t\space-\space\frac{t^2}{2}\right|_1^2\space=\space \frac{1}{2}\space+\space (4\space-\space\frac{4}{2})\space-\space(2\space-\space\frac{1}{2})\space= 1$$

Mean residence time = $$\frac{\int_0^\infty tc(t)dt}{\int_0^\infty c(t)dt}\space=\space\frac{1}{1}\space = 1$$s.

• Q.2

The Sherwood number ($$Sh_l$$) correlation for laminar flow over a flat plate of length L is given

$$\space\space\space\space\space\space\space\space\space\space\space\space\space\space Sh_L = 0.664 Re_l^{0.5}Sc^{1/3}$$

where $$Re_l$$ and Sc represent Reynolds number and Schmidt number, respectively. The correlation, expressed in the form of Chilton-Colburn $$j_D$$ factor, is

(A)

$$j_D = 0.664$$

(B)

$$j_D = 0.664\space Re_L^{-0.5}$$

(C)

$$j_D = 0.664\space Re_L$$

(D)

$$j_D = 0.664\space Re_L^{0.5}Sc^{2/3}$$

Marks:2 Explanation

.$$j_D\space=\space St_MSc^{2/3}\space=\space \frac{Sh}{Re_lSc}\times Sc^{2/3}\space=\space\frac{Sh}{Re_lSc^{1/3}}\space=\space\frac{0.664Re_l^{0.5}Sc^{1/3}}{Re_lsc^{1/3}}\space=\space0.664Re_l^{-0.5}$$

• Q.3

The vapour phase composition and the realative volatilities (with respect to n-propane) on an ideal tray of a distillation column are

 Component Methane Ethane n-propane Mole fraction in vapour 0.12 0.28 0.60 Relative volatility 10 4 1

The mole fraction of n-propane in the liquid phase, rounded to 2 decimal places, is ______________.

Marks:2 Explanation

.Given : $$y_M =0.12,y_E=0.28,y_P=0.60$$

$$\alpha_{MP}=10,\space\alpha_{EP}=4,\space\alpha_{PP}=1$$

$$\alpha_{MP}\space=\space \frac{(\frac{y_M}{x_M})}{(\frac{y_P}{x_P})}\space=\space \frac{y_Mx_P}{y_Px_M}\space=\space 10 \space \implies\space \frac{0.12\times x_P}{0.60\times x_M}\space=\space10\space\implies\frac{x_P}{x_M}\space=\space50 \space \implies\space x_M\space=\space\frac{x_P}{50}$$.   $$\to\space(1)$$

$$\alpha_{EP}\space=\space \frac{(\frac{y_E}{x_E})}{(\frac{y_P}{x_P})}\space=\space \frac{y_Ex_P}{y_Px_E}\space=\space 4 \space \implies\space \frac{0.28\times x_P}{0.60\times x_E}\space=\space4\space\implies\frac{x_P}{x_E}\space=\space\frac{60}{7} \space\implies\space x_E\space = \space (\frac{7}{60}) x_P$$.  $$\to\space (2)$$

Now ,  $$x_M\space +\space x_E\space+\space x_P\space =\space 1\space \implies (\frac{x_P}{50}\space +\space \frac{7x_P}{60}\space+\space x_P)\space=\space 1\space$$, using (1) & (2)

$$\implies\space (\frac{6x_P\space+\space35x_P\space + \space +\space 300x_P}{300})\space=\space 1\space\implies\space x_P\space=\space\frac{300}{341}\space=\space0.879\space\approx\space 0.88$$

• Q.4

In nucleate boiling, the pressure inside a bubble is higher than the pressure surrounding liquid. Assuming that both the liquid and vapour pressure are saturated, the temperature of the liquid will ALWAYS be

(A)

at $$100^0 C$$

(B)

lower than the temperature of the vapour

(C)

equal to the temperature of the vapour

(D)

higher than the temperature of the vapour

Marks:2 Explanation

The first bubble starts forming at point A of the boiling curve at preferential sites on the heating surface . The bubble form at an increasing rate of an increasing number of nucleation sites as we move along the boiling curve towards point C.

In a gravitational field, the bubble being lighter than the surrounding liquid will buoy away, however here we are considering a static equilibrium such that the bubble is relatively small in size and is held in place. Now if the bubble is in thermal equilibrium with the liquid, the temperature of the bubble and the liquid should be the same, i.e. $$T_b =T_l$$ .

Since from thermodynamic equilibrium consideration, the vapor must be at the saturation temperature corresponding to pressure in the bubble and we know from the equation of state that saturation temperature increases with pressure, the bubble or the liquid temperature should be higher than the saturation temperature corresponding to the pressure in the liquid. In other words, the liquid must be superheated.

• Q.5

Let $$I_{b\lambda}$$ be the spectral blackbody radiation intensity per unit wavelength about the wavelength $$\lambda$$. The blackbody radiation intensity emitted by a blackbody over all wavelengths is

(A)

$$dI_{b\lambda} \over d\lambda$$

(B)

$$d^2I_{b\lambda} \over d\lambda^2$$

(C)

$$\int_0^\infty I_{b\lambda}d\lambda$$

(D)

$$\int_0^\infty \lambda I_{b\lambda}d\lambda$$

Marks:2 Explanation

Spectral blackbody emissive power : It is the amount of radiation energy emitted by a blackbody at an absolute temperature T per unit time, per unit surface area, and per unit wavelength about the wavelength $$\lambda$$. It is given by Planck's law & expressed as

.$$E_{b\lambda}(\lambda,T)\space= \space \frac{C_1}{\lambda^5[exp(c_2/\lambda T)\space -\space 1]}$$

where $$C_1\space =\space 2\pi hc_0^2\space=\space 3.742\times10^8\space W.\mu m^4/m^2$$  ,  $$C_2\space=\space hc_0/k\space=\space 1.439\times 10^4\mu m.K$$

Integration of the spectral blackbody emissive power $$E_{b\lambda}$$ over the entire wavelength spectrum gives the total black-body emissive power $$E_b(T)\space= \space \int_0^\infty E_{b\lambda}(\lambda,T)d\lambda\space= \space\sigma T^4 \space\space\space\space (W/m^2)$$

• Q.6

A fluid flows over a heated horizontal plate maintained at temperature $$T_w$$. The bulk temperature of the fluid is $$T_{\infty}$$. The temperature profile in the thermal boundary layer is given by:

$$T = T_w + (T_w + T_\infty)[\frac{1}{2} (\frac{y}{\delta_t})^3 - \frac{3}{2}(\frac{y}{\delta_t})], \space \space\space\space\space\space\space\space$$  $$0\leq y \leq \delta_t$$

Here, y is the vertical distance from the plate, $$\delta_t$$ is the thickness of the thermal boundary layer  and k is the theram l conductivity of the fluid.

The local heat transfer coefficient is given by

(A)

$$k \over 2\delta_t$$

(B)

$$k \over \delta_t$$

(C)

$${3\over2} {k \over 2\delta_t}$$

(D)

$${2} {k \over 2\delta_t}$$

Marks:2 Explanation

.

• Q.7

Two machine M1 and M2 are able to execute any of four jobs P, Q, R, and S. The machine can perform one job on one object at a time. Jobs P, Q, R and S take 30 minutes, 20 minutes, 60 minutes and 15 minutes each respectively. There are 10 objects each requiring exactly 1 job. Job P is to be performed on 2 objects. Job Q on 3 objects, job R on 1 object and Job S on 4 objects. What is the minimum time needed to complete all the jobs?

(A)

2 hours

(B)

2.5 hours

(C)

3 hours

(D)

3.5 hours

Marks:2 Explanation

 Jobs Time Taken(min) Number of jobs Total time(mins) P 30 2 60 Q 20 3 60 R 60 1 60 S 15 4 60

For minimum time for completion of the jobs both machines need to works without break till jobs are complete.

First M1 and M2 will do the 1 jobs of P each simultaneously, so to complete the 2 jobs of P time taken is 30 minutes.

Next M1 and M2 will take 1 job each of Q which will take 20 minutes to complete 2 jobs of Q and upon completion M1 and M2 both will perform 1 job of Q together which will take 10 minutes.  Therefore all three jobs of Q will take 20 + 10 = 30 minutes.

Then M1 will be assigned the one job of R and at the same time M2 will be assigned all the 4 jobs of S consecutively. M1 will  take 60 minutes while M2 will also take 60 minutes to do all the 4 S jobs, thus they both will complete job R and S in 60 minutes.

Hence, total time taken to complete all the jobs  = 30 + 30 + 60 = 120 minutes i.e. 2 hours. (Option (A)

• Q.8

In a countercurrent stripping operation using pure steam, the mole ratio of a solute in the liquid stream is reduced from 0.25 to 0.05. The liquid feed flowrate, on a solute-free basis, is 3 mol/s. The equilibrium line for the system is given in the figure below.

The MINIMUM flowrate of pure steam for this process, rounded to 1 decimal place, is _________ mol/s.

Marks:2 Explanation

slope m = $$\frac{y_2-y}{x_0-x}\space =\space \frac{0.15\space -\space 0}{0.25\space - \space 0}\space =\space \frac{3}{5}$$ , $$\implies\space y_2\space =\space mx_0\space = \space \frac{3}{5}\times 0.25\space = \space0.15$$

Applying material balance for solute;

$$Gy_1\space+\space Lx_0\space=\space Gy_2\space +\space Lx_1$$

$$\implies G(y_1\space- \space y_2)\space=\space L(x_1\space-\space x_0)\space\implies \space G\space=\space \frac{L(x_1\space-\space x_0)}{y_1\space-\space y_2}$$

$$\implies\space G\space=\space \frac{3(0.05-0.25)}{0-0.15} \implies\space G\space=\space\implies\space G\space=\space4mol/s$$.

• Q.9

The real part of $$6e^{i\pi/3 }$$ is                  .

Marks:1 Explanation

The given complex number is in euler form $$re^{i\theta }$$

which can be written as $$rcos\theta + isin\theta$$ where real part is $$rcos\theta$$

Thus real part is $$6cos\frac{\pi}{3}$$ = 6*$$\frac{1}{2}$$ = 3

• Q.10

The number of positive roots of the function f(x) shown below in the range $$0 \space < \space x \space < \space 6$$ is                .            .

Marks:1 Explanation

Roots of a function f(x) is an input value that produces output zero i.e.

the value of x for which f(x) is 0.

Here in the graph for given range of x ie 0$\bg_black <$x$\bg_black <$6, the function f(x) is

equal to zero 3 times . so number of roots are 3.

• Q.11

The marks obtained by a set of students are 38,84,45,70,75,60,48.

The mean & median marks, respectively , are

(A) 45 and 75 (B) 55 and 48 (C) 60 and 60 (D) 60 and 70
Marks:1 Explanation

• MEAN : The mean of a data set is the average of all the data values.

$$\bar{x}$$ = $$\frac{\sum x}{n}$$ = ($$\frac{38+84+45+70+75+60+48}{7}$$) = $$\frac{420}{7}$$ = 60.

Hence mean of above data is 60.

MEDIAN: The median of a data set is the value in the middle

when the data items are arranged in ascending order.

· For odd number of observations:

· the median is the middle value =  ($$\frac{n+1}{2}$$)th term.

· For even number of observations:

· the median is the average of the middle two values.

i.e.  median =     ($$\frac{\frac{n}{2}th term+\frac{n+2}{2}th term}{2}$$)

arranging the data in ascending order 38,45,48,60,70,75,84.

Total number of terms is 7, i.e. odd ,

so median =  ($$\frac{n+1}{2}$$)th term = ($$\frac{7+1}{2}$$)th term = 4 th term, which is 60.

Hence median of above data is 60.

• Q.12

Let $$\hat{i{}}$$ and $$\hat{j{}}$$ be the unit vectors in x and y directions, respectively. For the function

F(x,y) = $$x^3 + y^2$$

the gradient of function i.e. $$\bigtriangledown{F}$$ is given by

(A)

​​$$3x^2 \hat{i} - 2y\hat{j}$$

(B)

$$6x^2y$$

(C)

$$3x^2 \hat{i} + 2y\hat{j}$$

(D)

$$2y\hat{i} - 3x^2\hat{j}$$

Marks:1 Explanation

In rectangular coordinates, the gradient is given by

$$​​​​\bigtriangledown\phi(x,y,z)$$ =  $$\frac{\partial \phi}{\partial x}\hat{i}$$ + $$\frac{\partial\phi}{\partial y}\hat{j}$$ $$\frac{\partial\phi}{\partial z}{\hat k}$$

Here    $$\phi(x,y,z)$$ = $$x^3 + y^2$$

$$\frac{\partial\phi}{\partial x} = 3x^2$$  , $$\frac{\partial\phi}{\partial y} = 2y$$  , $$\frac{\partial\phi}{\partial z} = 0$$ .

Hence gradient of $$\phi(x,y,z)$$$$x^3 + y^2$$ is $$3x^2\hat{i} + 2y\hat{j}$$

• Q.13

The volumetric properties of two gases M and N are described by the generalized compressibility      chart which expresses the compressibility factor (Z) as a function of reduced pressure and reduced  temperature only. The operating pressure(P) and temperature(T) of two gases M and N along with      their  critical properties $$(P_C, T_C)$$ are given in the table below.                                                                $$Z_M$$ and $$Z_N$$ are the compressibility factor of the gases M and N under the given operating    conditions, respectively.The relation between $$Z_M$$ and $$Z_N$$ is

(A)

$$Z_m = 8Z_N$$

(B)

$$Z_M = 3Z_N$$

(C)

$$Z_M = Z_N$$

(D)

$$Z_M = 0.333Z_N$$

Marks:1 Explanation

First, we calculate reduced pressure and reduced temperature for gas M and N respectively.

For gas M : P = 25 bar , $$P_C$$ = 75 bar , T = 300K , $$T_C$$ = 150

So, reduced pressure Pr = $$\frac{P}{P_C}$$ = $$\frac{25}{75}$$$$\frac{1}{3}$$

Reduced temperature Tr = $$\frac{T}{T_C}$$  = $$\frac{300}{150}$$= 2

For gas N: P = 75bar , $$P_C$$ = 225 bar , T = 1000K , $$T_C$$ = 500K

Reduced pressure  Pr  = $$\frac{P}{P_C}$$ = $$\frac{75}{225}$$ = $$\frac{1}{3}$$

Reduced temperature Tr = $$\frac{T}{T_C} = \frac{1000}{500} = 2$$
We know that  Z = $$\frac{PV}{RT} = 1 + \frac{BP}{RT} = 1 + \frac{BP_c}{RT_C}(\frac{P_r}{T_r})$$

Thus  $$Z = \alpha (\frac{P_r}{T_r})$$   $$Z_M = \alpha(\frac{P_r}{T_r}) =$$  $$\frac{\frac{1}{3}}{2}$$ = $$\frac{1}{6}$$

Similarly                $$Z_N = \alpha(\frac{P_r}{T_r}) = \alpha(\frac{\frac{1}{3}}{2}) =\frac{1}{6}$$

Thus      $$Z_M = Z_N$$

• Q.14

Water is heated at atmospheric pressure from   $$40^0C$$   to   $$80^0C$$ using two different processes . In process I , the heating is done by a source at   $$80^0C$$   . In process II , the water is first heated from $$40^0C$$ to $$60^0C$$ by a source at $$60^0C$$  , and then from $$60^0C$$ to $$80^0C$$   by another source at $$80^0C$$   .

Identify the correct statement.

(A)

Enthalpy change of water in process I is greater than enthalpy change in process II

(B)

Enthalpy change of water in process II is greater than enthalpy change in process I

(C)

Process I is closer to reversibility

(D)

Process II is closer to reversibility

Marks:1 Explanation

The process II is closer to reversibility

• Q.15

In a venturi meter  $$\bigtriangleup P_1$$  and  $$\bigtriangleup P_2$$  are the pressure drops corresponding to volumetric  flowrates     $$Q_1$$ and  $$Q_2$$  .

If     $$\frac{Q_2}{Q_1} = 2$$ ;  then  $$\frac{\bigtriangleup P_2}{\bigtriangleup P_1}$$  equals

(A)

2

(B)

4

(C)

0.5

(D)

0.25

Marks:1 Explanation

Thus  $$Q = \alpha \sqrt{\bigtriangleup P}$$    ;  $$\frac{Q_2}{Q_1} = \sqrt{\frac{\bigtriangleup P_2}{\bigtriangleup P_1}}$$ $$\implies \frac{\bigtriangleup P_2}{\bigtriangleup P_1} = (\frac{Q_2}{Q_1})^2 = (2)^2 =4$$

• Q.16

The thickness of laminar boundary layer over a flat plate varies along the distance from the leading edge of the plate. As the distance increases , the boundary layer thickness

(A)

Increases

(B)

Decreases

(C)

Initially increases and then decreases

(D)

Initially decreases and then increases

Marks:1 Explanation

Its quite evident from the figure that boundary layer thickness increases as the distance increases over a flat plate.

• Q.17

A gas bubble ( gas density $$\rho_g = 2 kg/m^3$$, bubble diameter D = $$10^{-4}$$ m) is rising vertically through water  ( density $$\rho = 1000 kg/m^3$$; viscosity $$\mu = 0.001 Pa.s$$ ). Force balance on the bubble leads to the following equation,

$$\frac{dv}{dt} = -g\frac{\rho_g -\rho}{\rho_g} - \frac{18\mu}{\rho_gD^2}v$$

Where v is the velocity of the bubble at any given time t. Assume that the volume of the rising bubble does not change. The value of $$g = 9.81 m/s^2$$ .

The terminal rising velocity of the bubble (in cm/s) , rounded to 2 decimal places, is _______ cm/s .

Marks:1 Explanation

At some speed the drag or force of resistance will be equal to the gravitational pull on the object;

At this point the bubble ceases to accelerate i.e. $$\frac{dv}{dt} = 0$$  and continue to rise with constant velocity.

$$\frac{dv}{dt} = 0 = -g\frac{\rho_g-\rho}{\rho_g} - \frac{18\mu}{\rho_gD^2} v$$   $$\implies v = \frac{gD^2(\rho - \rho_g)}{18\mu}$$ $$= \frac{9.81\times(10^{-4})^2\times(1000-2)}{18\times0.001} = 5.439\times10^{-3} m/s = 0.54 m/s$$

• Q.18

The one dimensional unsteady heat conduction equation is

$$\rho C_p\frac{\partial T}{\partial t} = \frac{1}{r^n} \frac{\partial }{\partial r}( r^nk\frac{\partial T}{\partial r} )$$

Where T – temperature , t – time , r – radial position , k – thermal conductivity ,   $$\rho$$  - density and

$$C_p$$- specific heat.  For the cylindrical coordinate system , the value of n in the above equation is

(A)

0

(B)

1

(C)

2

(D)

3

Marks:1 Explanation

Thus for cylindrical co-ordinates n = 1

• Q.19

In a heat exchanger , the inner diameter of a tube is 25mm and its outer diameter is 30mm . The overall heat transfer coefficient based on the inner area is 360 $$W/m^2.^{\circ} C$$ . Then the overall heat transfer coefficient based on the outer area, rounded to the nearest integer, is ______$$W/m^2.^{\circ}C$$ .

Marks:1 Explanation

Rate of heat transfer between two fluids can be expressed as :

$$\dot{Q} = UA(T_h - T_c) = U_OA_O(T_h - T_c) = U_iT_i(T_h - T_c)$$ , where $$(T_h - T_c)$$ is the driving force.

U - Overall heat transfer coefficient.

$$U_O$$ - Heat transfer coefficient based on outside area .

$$U_i$$ - Heat transfer coefficient based on inside area .

A - Surface area through which heat transfer takes place.

Thus ;   $$U_OA_O = U_iA_i \implies \frac{U_O}{U_i} = \frac{A_i}{A_O} = \frac{\pi D_iL}{\pi D_O L} = \frac{D_i}{D_O}$$

$$\implies U_O = (\frac{D_i}{D_O})\times U_i = (\frac{25}{30})\times360 = 300 W/m^2.^{\circ}C$$

• Q.20

Which of the following conditions are valid at the plait point?

P) Density difference between the extract and raffinate phases is zero.

Q) Interfacial tension between the extract and raffinate phases is zero.

R) Composition difference between the extract and raffinate phases is zero.

(A)

P and Q only

(B)

Q and R only

(C)

P and R only

(D)

P, Q and R

Marks:1 Explanation

The Plait Point P, is the intersection of the raffinate-phase and extract-phase boundary curves. At this point, the equilibrium phases become coincident and no separation can be made at that point.

So , all options are correct.

• Q.21

The composition of vapor entering  a tray in a distillation column is 0.47. The average composition of the vapor leaving the tray is 0.53 . the equilibrium composition of the vapor corresponding to the leaving tray is 0.52 . The equilibrium composition of the vapor corresponding are expressed in mole fraction of the more volatile component.

The Murphree efficiency based on the vapour phase , rounded to the nearest integer , is_________%.

Marks:1 Explanation

It is defined for each tray according to the separation achieved on each tray. This can be based on either the liquid phase or the vapor phase.

For a given component, it is equal to the change in actual concentration in the phase, divided by the change predicted by equilibrium condition.

Referring to the figure below, we have

$$E_{MV} = \frac{Y_n - Y_{n+1}}{ Y^{\ast}_n - Y_n+1} = \frac{0.53 - 0.47}{0.52 - 0.47} = \frac{0.06}{0.05} = 1.20$$ ;  %$$E_{MV} = 1.20\times100 = 120$$%

• Q.22

Consider steady state mass transfer of a solute A from a gas phase to a liquid phase . The gas phase bulk and interface mole fractions are $$Y_{A,G}$$  and $$Y_{Ai}$$  respectively . The liquid phase bulk and interface mole fractions are $$X_{A,L}$$  and $$X_{A,i}$$  , respectively. The ratio  $$\frac{X_{A,i} - X_{AL}}{Y_{A,G} - Y_{A,i}}$$  is very close to zero.                           This implies that mass transfer resistance  is

(A)

Negligible in the gas phase only

(B)

Negligible in the liquid phase only

(C)

Negligible in both phases

(D)

Considerable in both the phases

Marks:1 Explanation

As evident from the graph , the slope of the curve m = $$-\frac{k_x}{k_y} = \frac{y_{Ab} - y_{Ai}}{x_{Ab} - x_{Ai}}$$

From the given condition i.e. $$\frac{X_{A,i} - X_{AL}}{Y_{A,G} - Y_{A,i}} \approx 0$$  , thus slope $$m\to\infty$$

This shows that mass transfer resistance is completely controlled by gas phase only & there is negligible resistance in liquid phase , making the gas highly soluble.

• Q.23

The following reaction rate curve is shown for a reaction  $$A\to P$$. Here ($$-r_A$$) and $$X_A$$ represent reaction rate and conversion, respectively . The feed is pure A and 90% conversion is desired.

Which among the following reactor configurations gives the lowest total volume of the reactor(s)?

(A)

CSTR followed by PFR

(B)

Two CSTR in series

(C)

PFR followed by CSTR

(D)

A single PFR

Marks:1 Explanation

The critical point in the graph is $$X_A = 0.5$$ , from the graph it is quite clear that conversion from          $$X_A = 0$$  to $$X_A = 0.5$$ ;     Area occupied by CSTR reactor < Area occupied by PFR

$$X_A = 0.5$$  to $$X_A = 1$$ ;      Area occupied by PFR reactor < Area occupied by CSTR.

Thus best arrangement of reactor would be CSTR followed by PFR.

• Q.24

Consider a first order catalytic reaction in a porous catalyst pellet.                                                          Given R – characteristic length of the pellet ; $$D_e$$ =  effective diffusivity ;  $$K_c$$ -  mass transfer coefficient ;   $$K_1$$- rate constant based on volume of the catalyst pellet ; $$C_s$$ -  concentration of reactant on the pellet surface.The expression of Thiele modulus is

(A)

$$\frac{K_cR}{D_e}$$

(B)

$$R\sqrt\frac{K_1}{D_e}$$

(C)

$$R\sqrt\frac{K_1C_s}{D_e}$$

(D)

$$R\sqrt\frac{D_e}{K_1}$$

Marks:1 Explanation

Thiele Modulus : $$\Phi = \sqrt\frac{K_1R^2}{D_e} = R\sqrt\frac{K_1}{D_e}$$ $$= \frac{reaction.rate}{diffusion. rate}$$

Note for small values of Thiele modulus, the reaction rate is small compared to the diffusion rate, and the pellet concentration becomes nearly uniform.

For large values of Thiele modulus, the reaction rate is large compared to the diffusion rate, and the reactant is converted to product before it can penetrate very far into the pellet.

• Q.25

For a solid – catalyzed gas phase reversible reaction , which of the following statements is ALWAYS TRUE ?

(A)

(B)

Desorption is rate limiting

(C)

Solid catalyst doesnot effect equilibrium conversion

(D)

Temperature doesnot effect equilibrium conversion

Marks:1 Explanation

Solid catalyst doesnot effect equilibrium conversion

• Q.26

Match the variables in Group – I with the instruments in Group – 2.

 Group 1 Group 2 P) Temperature I) Capacitance Probe Q) Liquid level II) McLeod gauge R) Vacuum III) Chromatograph S) Concentration IV) Thermistor

(A)

P-IV,Q-III,R-II,S-I

(B)

P-I,Q-II,R-IV,S-III

(C)

P-IV,Q-I,R-II,S-III

(D)

P-III,Q-II,R-II,S-IV

Marks:1 Explanation

A capacitance probe is a point level indicator, commonly used for high or low level detection in a bin, tank or silo.

A McLeod gauge is a scientific instrument used to measure very low pressures, down to 10−6 Torr.

Chromatography is a laboratory technique for the separation of a mixture.

A thermistor is a temperature-sensing element composed of sintered semiconductor material which exhibits a large change in resistance proportional to a small change in temperature. Thermistors usually have negative temperature coefficients which means the resistance of the thermistor decreases as the temperature increases.

So the correct option is P-IV,Q-I,R-II,S-III .

• Q.27

The cost of new pump ( including installation ) is 24,000 Rupees. The pump has a useful life of 10 years . Its salvage value is 4000 Rupees . Assuming straight line depreciation , the book value of the pump at the end of  $$4^{th}$$ year , rounded to the nearest integer, is  _______ Rupees.

Marks:1 Explanation

Given:

Cost of pump, V = Rs 24,000 ;

Salvage value of pump, V$$_S$$ = Rs 4000 ;

Useful life of pump, n = 10 yrs

Average depreciation per year, d = $$\frac{V-V_S}{n} = \frac{24000-4000}{10}=Rs2000$$

Book value of the pump at the end of $$4^{th}$$ yr = $$V - a\times d = 24000 - 4\times 2000 = Rs16000$$

• Q.28

Match the polymerization process in Group-1 with the polymers in Group-2.

 Group-1 Group-2 P) Free radical polymerization I) Nylon 6,6 Q) Ziegler Natta polymerizaion II) Polypropylene R) Condensation polymerization III) Poly vinyl chloride

Choose the correct set of combinations.

(A)

P-I, Q-II, R-III

(B)

P-III, Q-II, R-I

(C)

P-I, Q-III, R-II

(D)

P-II, Q-I, R-III

Marks:1 Explanation

Nylon 6,6: nylon is made by a reaction which is a step-growth polymerization, and a condensation polymerization.

Polypropylene: Polypropylene can be made from the monomer propylene by Ziegler-Natta polymerization.

Polyvinyl chloride:PVC is a vinyl polymer. It's similar to polyethylene, but on every other carbon in the backbone chain, one of the hydrogen atoms is replaced with a chlorine atom. It's produced by the free radical polymerization of vinyl chloride.

• Q.29

The purpose of methanation reaction used in ammonia plants is to

(A)

remove CO as it is a catalyst poison

(B)

increase the amount of hydrogen

(C)

remove sulphur as it is a catalyst poison

(D)

utilize methane as a catalyst for ammonia synthesis

Marks:1 Explanation

carbon oxides are poison for ammonia synthesis catalyst. Methanation is the final stage of purification of synthesis gas after $$CO_2$$ removal to reduce carbon oxides to trace levels.                                                  Methanation reactions:

$$CO + 3H_2 \iff CH_4 +H_2O$$ ;       $$\bigtriangleup H = -206 KJ/mol$$

$$CO_2 + 4H_2 \iff CH_4 +2H_2O$$ ;     $$\bigtriangleup H = -165KJ/mol$$

• Q.30

For initial value problem

$$dx\over dt$$ = sin(t),   x(0) = 0

the value of x at t  = $$\pi /3$$ is _____________.

Marks:2 Explanation

We have , .    $$dx\over dt$$ = sin(t),   x(0) = 0

Integrating , we get $$\int dx = \int sin(t)dt \implies x = -cos(t) + c$$

Applying initial value conditions, we get $$x(0) = -cos(0) + c \implies 0 = -1+c \implies c=1$$

Thus at t = $$\frac{\pi}{3}$$ $$x(\frac{\pi}{3}) = -cos(\frac{\pi}{3}) + 1 = -\frac{1}{2} +1 = \frac{1}{2} = 0.5$$

• Q.31

The Laplace transform of a function is  $$s+1 \over s(s+2)$$

The initial and final values, respectively, of the function are

(A)

0 and 1

(B)

1 and $$1 \over 2$$

(C)

$$1 \over 2$$ and 1

(D)

$$1 \over 2$$ and 0

Marks:2 Explanation

.Given ; Laplace transform = $$\frac{s+1}{s(s+2)}$$

The initial value of the function is   $$\lim_{s\to\infty}s\times f(s) = \lim_{s\to\infty}\frac{s(s+1)}{s(s+2)} = \lim_{s\to\infty}\frac{s(1+\frac{1}{s})}{s(1+\frac{2}{s})} = 1$$

The final value of the function is      $$\lim_{s\to0}s\times f(s) = \lim_{s\to0}\frac{s(s+1)}{s(s+2)} = \lim_{s\to0}\frac{s+1}{s+2} = \frac{1}{2}$$

• Q.32

Match the problem type in Group-1 with the numerical method in Group-2.

 Group-1 Group-2 P) System of linear algebraic equations I) Newton-Raphson Q) Non-linear algebraic equations II) Gauss-Siedel R) Ordinary differential equations III) Simpson's rule S) Numerical integration IV) Runge-Kutta

Choose the correct set of combinations.

(A)

P-II, Q-I, R-III, S-IV

(B)

P-I, Q-II, R-IV, S-III

(C)

P-IV, Q-III, R-II, S-I

(D)

P-II, Q-I, R-IV, S-III

Marks:2 Explanation

P) System of linear algebraic equations     $$\implies$$  II) Gauss-Siedel

Q) Non-linear algebraic equations               $$\implies$$ I) Newton-Raphson

R) Ordinary differential equations               $$\implies$$  IV) Runge-Kutta

S) Numerical integration                               $$\implies$$ III) Simpson's rule

• Q.33

A box has 6 red balls and 4 white balls . A ball is picked at random and replaced in the box, after which a second ball is picked.

The probability of both balls being red , rounded to 2 decimal places is __________.

Marks:2 Explanation

.The probability of both ball being red = $$\frac{^6C_1}{^10C_1}\times\frac{^6C_1}{^10C_1} = \frac{6}{10}\times\frac{6}{10} = \frac{36}{100} = 0.36$$

• Q.34

An aqueous salt – solution enters a crystallizer operating at steady state at $$25^0C$$. The feed temperature is  $$90^0C$$ and the salt concentration in the feed is 40 weight %. The salt crystallizes as a pentahydrate . The crystals and the mother liquor leave the crystallizer. The molecular weight of the anhydrous salt is 135 . The solubility of the salt at   is 20 weight %.

The feed flowrate required for a production rate of 100 Kg/s of the hydrated salt , rounded to the nearest integer, is  __________ kg/s .

Marks:2 Explanation

Weight fraction of salt crystals ,$$x_C$$ =                                                                                                                    $$\frac{molecular weight of anhydrous salt}{molecular weight of anhydrous salt+weight of hydrated water} = \frac{135}{135+18\times 5} = \frac{135}{225} =\frac{3}{5} = 0.6$$

Applying material balance over crystallizer;

Feed = Crystals + Saturated solution $$\implies$$F = 100 +  M                     (1)

Salt balance over the crystallizer;

$$F\times x_F = C\times x_c + M\times x_M \implies$$$$F\times 0.4 = 100\times 0.6 + M\times 0.2$$         (2)

$$\implies F\times 0.4 = 100\times 0.6 + (F-100)\times 0.2$$  $$\to$$   using (1)

$$\implies F\times 0.4 = 100\times 0.6 - 100\times 0.2 + F\times 0.2$$

$$\implies F\times (0.4-0.2) = 100\times (0.6-0.2)$$

$$\implies F = 100\times \frac{0.4} { 0.2} = 200$$ Kg/s

• Q.35

Reaction $$A\to B$$   is carried out in a reactor operating at steady state and 1 mol/s of pure A at 425$$^0C$$  enters the reactor. The outlet stream leaves the reactor at 325 $$^0C$$  .The heat input to the reactor is 17 KW. The heat of reaction at the reference temperature of 25$$^0C$$  is 30 KJ/mol . The specific heat capacities ( in KJ/mol.K) of A and B are 0.1 and 0.15 respectively .

The molar flowrate of B leaving the reactor, rounded to 2 decimal places , is  ________ mol/s.

Marks:2 Explanation

$$\bigtriangleup H = 30 KJ/mol$$ ( +ve , Endothermic reaction )

$$A\to B$$

t = 0            1        0                                                     $$C_P$$(KJ/mol.K)               A =  0.1  ;  B = 0.15

t = t            1-x       x      ;

By Heat balance ;

Heat in - Heat output  + $$\bigtriangleup H_{rxn}$$  + Q   = 0  ( Taking reference temperature of 25$$^0C$$ )

$$\implies$$  $$F_{A^0}C_{PA}(T_{in} - 25 ) - \{ F_A C_{PA} (T_{out} - 25) + F_B C_{PB}(T_{out}-25)\} -30\times x+17 = 0$$

$$\implies$$ $$1\times 0.1\times (425 - 25 ) - \{(1-x)\times0.1\times (325 - 25) + x\times 0.15(T_{out}-25)\} -30\times x+17 = 0$$

$$\implies$$ $$40 - \{ 30-30\times x + 45\times x\} -30\times x+17 = 0$$

$$\implies$$  $$40 - \{ 30+15\times x\} -30\times x+17 = 0$$

$$\implies 40 - 30 -15\times x -30\times x+17 = 0$$

$$\implies 27 - 45\times x = 0$$

$$\implies x = \frac{27}{45} = \frac{3}{5} = 0.6$$

Thus ; $$F_B = x = 0.6 mol/s , F_A = (1 - x) = ( 1-0.6) = 0.4 mol/s$$

$$\star$$ Note : In endothermic reaction , $$\bigtriangleup H_{rxn}(+_{ve})$$ i.e. heat consumed by system .

Hence $$\bigtriangleup H_{rxn}$$ term will be negative .

• Q.36

The pressure of a liquid is increased isothermally. The molar volume of the liquid decreases from            $$50.45\times10^{-6} m^3/mol$$ to $$48\times10^{-6} m^3/mol$$ during this process. The isothermal compressibility of the        liquid is  $$10^{-9} Pa^{-1}$$  , which can be assumed to be independent of the pressure.

The change in the molar Gibbs free energy of the liquid , rounded to nearest integer, is   _____ J/mol.

Marks:2 Explanation

.Isothermal compressibility  $$K = -\frac{1}{V}(\frac{\partial V}{\partial P})_T$$   $$\to (1)$$

Change in Gibbs free energy $$dG = VdP - SdT$$

At constant temperature ,       $$dG = VdP = - \frac{1}{K}(\frac{\partial V}{\partial P})dP$$

$$\therefore$$   $$dG = - \frac{1}{K}(\partial V)$$

$$dV = 10^{-6}\times ( 48 - 50.45) = -2.45\times 10^{-6} m^3/mol$$

Thus $$\bigtriangleup G = - (\frac{-2.45\times 10^{-6}}{10^{-9}}) = 2450 J/mol$$

• Q.37

A sparingly soluble gas (solute ) is in equilibrium with a solvent at 10 bar . The mole fraction of the solvent in the gas phase is 0.01 . At the operating temperature and pressure, the fugacity coefficient of the solute in the gas phase and the Henry’s law constant  are 0.92 and 1000 bar , respectively .

Assume that the liquid phase obeys Henry’s law.

The MOLE PERCENTAGE of the solute in the liquid phase , rounded to 2 decimal places, is ________.

Marks:2 Explanation

.Given : P = 10 bar ; $$\hat{\phi}_i$$ = 0.92 , $$K_H$$ = 1000 bar

Mole fraction of solvent, in the gas phase = 0.01

Mole fraction of solute , in the gas phase = (1 - 0.01) = 0.99

For sparingly soluble gas, Rault's law can be modified as

$$f_i\times y \space = \space K_h\times x$$

$$\implies \space \hat{\phi}_iPy \space=\space K_hx$$

$$\implies \space x\space=\space \frac{\hat{\phi}_iPy}{K_h}$$      $$\implies \space x\%\space=\space \frac{\hat{\phi}_iPy}{K_h}\times 100$$

$$\implies \space x\%\space=\space \frac{0.92\times10\times0.99}{1000}\times 100 =\space0.9108 \space\approx\space0.91\%$$

• Q.38

The vapour pressure of a pure substance at a temperature T is 30 bar. The actual and ideal gas values of  g/RT for the saturated vapour at this temperature T and 30 bar are 7.0 and 7.7 respectively . Here ,   g is the molar Gibbs free energy and R is the universal constant.

The fugacity of the saturated liquid at these conditions , rounded to  1 decimal places , is  ______ Pa.

Marks:2 Explanation

.Given : P = 30 bar $$\frac{G}{RT}\space=\space7.0\space\frac{G^{ig}}{RT}\space=7.7$$

We kow that      $$G^R\space=\space RTln(\frac{f_i}{P})\space\implies\space(G\space-\space G^{ig})\space = RTln(\frac{f_i}{P})$$

$$\implies\space(\frac{G}{RT} \space-\space \frac{G^R}{RT})\space = ln(\frac{f_i}{P})\space\implies\space (7.0\space- 7.7) =\space ln(\frac{f_i}{P})$$

$$\implies\space ln(\frac{f_i}{P})\space=\space -0.7\space \implies\space \frac{f_i}{P}\space = \space exp^{-0.7}$$     $$\implies\space f_i\space = \space P\times exp^{-0.7} \space =\space 30\times exp{-0.7}\space =\space 14.897\space \approx\space 14.9Pa$$

• Q.39

Oil is being delivered at a steady flowrate through a circular pipe of radius $$1.25\times10^{-2}m$$ and length 10m. the pressure drop across the pipe is 500 Pa .

The shear stress at the pipe wall, rounded to 2 decimal places, is  _______ Pa.

Marks:2 Explanation

Net force balance, gives

$$\Sigma F \space =\space \pi r^2P\space -\space\pi r^2(P+dP)\space -\space (2\pi rdl)\tau\space =\space 0$$

Simplifying this equation and dividing by  $$\pi r^2dL$$  gives ,

$$\frac{dP}{dL}\space +\space (\frac{2\tau}{r})\space=\space 0$$

In steady flow, either laminar or turbulent, the pressure at any given cross - section of a stream tube is constant, so that $$\frac{dP}{dL}$$ is independent of r.  For entire cross section of tube the equation could be written by taking $$\tau\space =\space \tau_w$$ and $$r\space=\space r_w$$ ;    $$\frac{dP}{dL}\space +\space (\frac{2\tau_w}{r_w})\space=\space 0$$

Thus $$\tau_w\space =\space\frac{r_w}{2}(\frac{dP}{dL})$$ $$\implies\space\tau_w\space=\space\frac{1.25\times 10^{-2}}{2}(\frac{500}{10})\space=\space 0.3125 \space Pa$$

• Q.40

The following table provides four sets of Fanning friction factor data, for different values of Reynolds number (Re) and roughness factor ($$k \over D$$).

 Re $$10^2$$ $$10^3$$ $$10^5$$ $$10^5$$ $$k \over D$$ 0 0.001 0 0.001 Set I f 0.16 0.016 $$16 \times 10^{-5}$$ $$16\times 10^{-5}$$ Set II f 0.016 0.16 0.0055 0.0045 Set III f 0.16 0.016 0.0045 0.0055 Set IV f 0.0045 0.0055 0.016 0.16

Which of the above sets of friction factor data is correct?

(A)

Set I

(B)

Set II

(C)

Set III

(D)

Set IV

Marks:2 Explanation

For laminar flow , friction factor  f  = $$\frac{16}{N_{Re}}$$   upto  $$N_{Re}\space=\space 2100$$

For turbulent flow, friction factor f = $$\frac{0.046}{N_{Re}^{0.2}}$$  Upto $$N_{Re}\space =\space 1\times10^6$$

From the given data set, we have $$N_{Re}\space=\space100\space\&\space N_{Re}\space=\space1000\space$$ which lies in laminar zone & $$N_{Re}\space=\space 10^5$$lies in turbulent zone.

for $$N_{Re}\space=\space100,\space f\space =\space 0.16$$   &  for $$N_{Re}\space=\space1000,\space f\space =\space 0.016$$

for $$N_Re\space=\space10^5\space, f\space=\space \frac{0.046}{(10^5)^{0.2}}\space =\space 0.0046$$

Thus , set III satisfy these data.

• Q.41

A propeller  (diameter D = 15 m) rotates at N = 1 revolutions per seconds (rps). To understand the flow around the propeller, lab scale model is made. Important parameters to study the flow are velocity of the propeller tip  ($$V\space =\space \pi ND$$), diameter D and acceleration due to gravity (g). The lab-scale model is $$1/100^{th}$$ of the actual size of the propeller.

The rotation speed of the lab scale model,  to the nearest integer shoulf be _____________ rps.

Marks:2 Explanation

.$\bullet$ The controversial question of GATE 2017.

$\bullet$  Scale up procedure

(1) Calculate the scale up ratio R. Assuming that the original vessel is a standard cylinder with $$D_{T1}\space=\space H_1$$, the volume $$V_1$$ is

$$V_1\space=\space (\frac{\pi D_{T1}^2}{4})(H_1)\space=\space(\frac{\pi D_{T1}^3}{4})$$

Then the ratio of the volumes is

$$\frac{V_2}{V_1}\space =\space \frac{\pi D_{T2}^3/4}{\pi D_{T1}^3 /4}\space=\space\frac{D_{T2}^3}{D_{T1}^3}$$

The scale-up ratio is then

$$R\space= \space (\frac{V_2}{V_1})^{1/3}\space=\space \frac{D_{T2}}{D_{T1}}$$

(2) Using this value of R, apply to all of the dimensions in Table as shown in fig above, to calculate the new dimensions. For example, $$D_{a2}\space=\space RD_{a1}, \space \space \space J_2\space =\space RJ_{1} ......$$

(3) Scale up rule to determine the agitator speed

$$N_2\space=\space N_1(\frac{1}{R})^n\space = \space N_1(\frac{D_{T1}}{D_{T2}})^n$$

where n=1 for equal liquid motion, n = $$\frac{3}{4}$$ for equal suspension of solids,

and n = $$\frac{2}{3}$$ for equal rates of mass transfer ( which is equivalent to equal power per unit volume ).

• Q.42

Size analysis was carried out on a sample of gravel. The data for the mass fraction ($$x_i$$) and average particle diameter ($$D_{pi}$$) of the fraction is given in the table below:

 $$x_i$$ $$D_{pi}$$(mm) 0.2 5 0.4 10 0.4 20

The mass mean diameter of the sample, to the nearest integer, is _______mm.

Marks:2 Explanation
• Answer:  $$x_i$$ $$D_{pi}(mm)$$ $$x_i\times​​ D_{pi}$$ 0.2 5 1 0.4 10 4 0.4 20 8

.$\bullet$ Mass mean diameter = $$\frac{\sum(x_i\times D_{pi})}{\sum x_i}\space =\space\frac{13}{1.0}\space=\space 13mm$$

• Q.43

The following liquid phase second-order reaction is carried out in an isothermal CSTR at steady state               $$A\space \to \space R\space\space\space\space(-r_A)\space=\space0.005C_A^2mol/m^3.hr$$

where $$C_A$$is the concentration of the reactant in the CSTR. The reactor volume is $$2m^3$$ , the inlet flowrate is $$0.5mol/m^3.hr$$ and the inlet concentration of the reactant is $$1000 mol/m^3$$.

The fractional conversion , rounded to 2 decimal places, is ______________ .

Marks:2 Explanation

Given:       $$V_{Reactor} \space=\space2m^3\space\space\space v_0\space=\space 0.5m^3/hr$$  $$C_{A0}\space\space=\space1000mol/m^3$$

Design equation of a CSTR :

$$\frac{V}{v_0}\space=\space\frac{C_{A0}X_A}{-r_A}$$   $$\implies\space \frac{2}{0.5}\space=\space \frac{C_{A0}X_A}{0.005C_A^2}\space =\space \frac{C_{A0}X_A}{0.005C_{A0}^2(1-X_A)^2}\space=\space\frac{X_A}{0.005C_{A0}(1-X_A)^2}$$ ;

$$\implies \space\frac{X_A}{(1-X_A)^2}\space =\space \frac{2\times0.005C_{A0}}{0.5}\space=\space 0.02C_{A0}\space=\space0.02\times1000\space= \space 20$$;

$$\implies\space X_A\space=\space 20(1-X_A)^2\space=\space20(1+X_A^2-2X_A)$$

$$\implies\space 20X_A^2\space -41X_A\space +20\space=\space0$$

$$\implies\space X_A = {-(-41) \pm \sqrt{(-41)^2-4\times 20\times 20} \over 2\times 20}\space=\space\frac{41\pm \sqrt{81}}{40}\space=\frac{41\pm 9}{40}$$;

Since $$X_A\space\neq\space>1$$, so neglecting positive sign; $$X_A\space=\space\frac{41-9}{40}\space=\space\frac{32}{40}\space=\space0.80$$

• Q.44

The reversible reaction of t-butyl alcohol(TBA) and ethanol(EtOH) to ethyl t-butyl ethr (ETBE) is

TBA  +  EtOH  $$\rightleftharpoons$$    ETBE  +  Water

The equilibrium constant for this reaction is $$K_c\space=\space1$$ . Initially, 74g of TBA is mixed with 100g of aqueous solution containing 46 weight % ethanol. The molecular weights are: 74g/mol for TBA,46g/mol for EtOH,102g/mol for ETBE, and 18g/mol for water.

The mass of ETBE at equilibrium, rounded to 1 decimal place, is _______ g.

Marks:2 Explanation

Given: Initially , TBA = 74g , EtOH = 46%0f 100 = 46 g , $$K_c\space=\space1$$ , Water = (100-46) = 54g.

TBA     +        EtOH           $$\rightleftharpoons$$        ETBE       +          Water

t = o         $$(\frac{74}{74})=1mol$$     $$(\frac{46}{46})\space=\space1mol$$                0                     $$(\frac{54}{18})\space=3mol$$

At equilibrium ;    t = t           (1-x)mol           (1-x)mol                       x mol                 (3+x)mol

$$K_c\space=\space\frac{[ETBE][water]}{[TBA][EtOH]}\space=\space\frac{(x)(3+x)}{(1-x)(1-x)}\space=\space1$$

$$\implies\space(3x+x^2)\space=\space(1-x)^2\space=\space(1+x^2-2x)$$

$$\implies\space(5x)\space=\space1\space\implies\space x \space=\space \frac15\space=0.2$$

so, moles of ETBE formed at equilibrium = x mol = 0.2 mol.

weight of ETBE formed = $$mole\times molecular\space weight\space=\space 0.2\times 102\space=\space20.4 g/mol$$

• Q.45

The tranfer function of a system is

$$\frac{1}{4s^2+1.2s+1}$$

For a unit step increase in the input, the fractional overshoot , rounded to 2 decimal places, is ____.

Marks:2 Explanation

General form of 2nd order function is  $$\frac{K_p}{\tau^2s^2+2\xi \tau s +1}$$

comparing from the given transfer function, we get  $$K_p\space=\space 1,\space\tau^2\space=\space4,\space2\xi \tau\space=\space1.2$$

Thus $$\tau\space=\space 2,\space \xi \space=\space0.3$$

Overshoot = $$exp(\frac{-\pi \xi}{\sqrt{1-\xi^2}})\space=\space exp(\frac{-0.3\pi}{\sqrt{1-0.3^2}}) \space=\space exp(-0.988)\space =0.37$$

• Q.46

$$\implies\space\frac{K_c}{\omega}\space=\space1,\space\implies \space K_c \space=\space \omega \space=\frac{\pi}{4}\space=\space0.79$$The open loop transfer function of a process with proportional controller (gain $$K_c$$) is

$$G_{OL}\space=\space K_c\frac{e^{-2s}}{s}$$

Based on the Bode criterion for closed-loop stability, the ultimate gain of the controller, rounded to 2 decimal places, is_____________.

Marks:2 Explanation

The given transfer function is a pure dead time capacitive process.

phase shift = $$\phi \space=\space (\frac{-\pi}{2}) + \space(-2\omega)$$

Let us find crossover frequency $$\omega _{co}$$ i.e. where $$\phi\space=\space-\pi$$

$$\implies\space-\pi \space=\space(-\frac{\pi}{2})\space+(-2\omega)\space \implies\space\omega_{co}\space=\frac{\pi}{4}$$

Amplitude ratio of the given transfer function , AR = $$\frac{K_c}{\omega}$$

For ultimate gain AR = 1, $$\implies \space \frac{K_c}{\omega}\space =\space 1\implies\space K_c\space=\space \omega \space =\space \frac{\pi}{4}\space= \space0.79$$

.

• Q.47

The characteristic equation of a closed loop system is

$$6s^3+11s^2+6s+(1+k)\space=\space0$$,  where K > 0

The value of K beyond which the system becomes unstable, rounded to the nearest integer, is ___.

Marks:2 Explanation

By Routh-Hurwitz criterion for stability , the corresponding routh array can be formed

 Row 1 6 6 Row 2 11 (1+K) Row 3 $$\frac{11\times 6-6\times (1+K)}{11}$$ 0 Row 4 (1+K)

.The elements of the first column are $$[ 6,11,\frac{11\times 6-6\times (1+K)}{11},(1+k)]$$

The system could be unstable if third element becomes less than 0.

i.e. $$\frac{11\times 6-6\times (1+k)}{11}\space=\space0 \space\implies K\space = \space \frac{60}{6}\space=\space 10$$

Thus the system could be unstable if value of K lies beyond 10.

• Q.48

A bond has a maturity value of 20,000 Rupees at the end of 4 years. The interest is compounded at the rate of 5% per year.

The investment to be made, rounded to the nearest integer, is _______ Rupees.

Marks:2 Explanation

Given maturity value S = Rs 20,000 , n = 4 yrs , i = 5%

Let investment made be P .  Maturity value $$S \space =\space P(1+i)^n$$ $$\implies\space P \space=\space\frac{S}{(1+i)^n}\space=\space \frac{20000}{(1+0.05)^4}\space=\space Rs16454$$

• Q.49

The total cost ($$C_T$$) of an equipment in terms of the operating variables x and y is

$$C_T\space=\space 2x+\frac{12000}{xy}+y+5$$

The optimal value of $$C_T$$, rounded to 1 decimal place, is ________ .

Marks:2 Explanation

.Given cost of equipment  $$C_T\space=\space 2x+\frac{12000}{xy}+y+5$$

Differentiating w.r.t x , We get  $$\frac{\partial C_T}{\partial X}\space=\space 2-\frac{12000}{x^2y}$$

Differentiating w.r.t y , We get  $$\frac{\partial C_T}{\partial y}\space=\space1-\frac{12000}{xy^2}$$

For optimum value $$\frac{\partial C_T}{\partial X}\space=\space0\space,\space\frac{\partial C_T}{\partial y}\space=\space0$$

$$\implies \space\frac{\partial C_T}{\partial X}\space=0\space\implies 2\space-\space\frac{12000}{x^2y}\space=\space0\implies\space x^2y\space=6000$$      (1)

$$\implies \space\frac{\partial C_T}{\partial y}\space=0\space\implies 1\space-\space\frac{12000}{xy^2}\space=\space0\implies\space xy^2\space=12000$$     (2)

Divide (1) by (2) , we get $$\frac{x^2y}{xy^2}\space=\space \frac{6000}{12000}\space=\space\frac{1}{2}\space\implies\space\frac{x}{y}\space =\space \frac{1}{2}\implies\space y\space =\space 2x$$

$$x^2y\space=\space6000\space\implies x^2(2x)\space= \space6000\implies x\space=\space14.42$$

$$xy^2\space=\space12000\space\implies (y/2)(y)^2\space= \space6000\implies y\space=\space28.8$$

Thus optimum value of $$C_T\space=\space 2x+\frac{12000}{xy}+y+5\space= 2\times 14.42+\frac{12000}{14.42\times28.8}+22.89+5\space=\space91.5$$

• Q.50

Match the equipment in Group-1 with the process in Group-2

 Group-1 Group-2 P) Fluidized bed I) Paper-making Q) Multistage adiabatic reactor with inter-stage cooling II) Sodium hydroxide manufacture R) Fourdrinier machine III) $$SO_2$$ oxidation S) Diaphragm cell IV) Catalytic cracking

(A)

P-IV.Q-III,R-I,S-II

(B)

P-IV.Q-III,R-II,S-I

(C)

P-III.Q-IV,R-I,S-II

(D)

P-III.Q-IV,R-II,S-I

Marks:2 Explanation

 Group-1 Group-2 P) Fluidized bed I) Paper-making Q) Multistage adiabatic reactor with inter-stage cooling II) Sodium hydroxide manufacture R) Fourdrinier machine III) $$SO_2$$ oxidation S) Diaphragm cell IV) Catalytic cracking

Above table can be matched in the following manner

(P) Fluidized bed                 $$\to$$                                                   (IV) Catalytic Cracking

(Q) Multistage adiabatic reactor with inter-stage cooling $$\to$$ (III) $$SO_2$$ Oxidation

(R) Fourdrinier machine      $$\to$$                                                   (I) Paper-Making

(S) Diaphragm cell               $$\to$$                                                   (II) Sodium Hydroxide Manufacture

• Q.51

The following gas-phase reaction is carried out in a constant-volume isothermal reactor

$$A\space+\space B \space\to\space R\space+\space S$$

The reactants A and B as well as the product S are non-condensable gases. At the operating temperature, the saturation pressure of the product R is 40 Kpa.

Initially, the batch reactor contains equimolar amounts of A and B ( and no products) at a total pressure of 100 Kpa. The initial  concentrations of the reactants are $$C_{A,0}=C_{B,0}=12.5\space mol/m^3$$.The rate of reaction is given by $$(-r_A)=0.08C_AC_B\space mol/m^3.S.$$

The time at which R just start condensing, rounded to 1 decimal place, is ________ S.

Marks:2 Explanation

The given reaction is      $$A\space +\space B\to\space R\space+\space S$$ .

let N moles of A and B are present initially in the reactor.

$$A\space\space +\space\space B\to\space\space R\space+\space\space S$$

At t = 0         N        N        0       0

At t = t        (N-Nx)     (N-Nx)   Nx     Nx

Total moles after time t = Nx+Nx+N-Nx+N-Nx = 2N

Partial pressure of R at time t = $$x_R\times P_t\space= (\frac{Nx}{2N})\times P_t\space=\space (\frac{x}{2})P_t\space=\space P_R$$

At 40KPa(Saturation pressure of R) , R will start condensing.

$$P_t\space=\space100KPa$$ , $$P_R\space=\space (\frac{x}{2})P_t\space=\space 40\space \implies\space x\space =\space \frac{80}{P_t}\space=\space \frac{80}{100}\space=\space0.8$$

$$C_A\space=\space C_{A0}(1-X)\space=\space 12.5(1-0.8)\space=\space 2.5mol/m^3$$

For batch reactor, the general equation is      $$\frac{dC_A}{dt}\space=\space -r_A$$

$$\implies \int_0^t dt\space=\space\int_{C_A0}^{C_A} \frac{dC_A}{-r_A}\space \implies t\space=\space \int_{C_{A0}}^{C_A}\frac{dC_A}{0.08C_AC_B}\space=\space\int_{C_{A0}}^{C_A}\frac{dCA}{0.08C_A^2}$$ ( $$\because C_A\space=\space C_B$$ )

$$\implies \space t\space=\space\frac{1}{0.08}(\frac{1}{C_A}\space -\space \frac{1}{C_{A0}} )\space=\space\frac{1}{0.08}(\frac{1}{2.5}\space -\space \frac{1}{12.5} )\space \implies\space t\space =\space 4s$$

• Q.52

The bacteria in milk are destroyed when  _____________  heated to 80 degree Celsius.

(A)

would be

(B)

will be

(C)

is

(D)

was

Marks:1 Explanation

Here two sentences “The bacteria in milk are destroyed” and “  ________  when heated to 80 degree Celsius” are conjugated using “when” so both sentence need to be consistent i.e. having same tense and verb forms.

We can see here in “The bacteria in milk are destroyed” auxiliary verb “be” (present form) + verb 3 is used so the second part of the sentence will also have the same form. So, in the blank space it is clear a present form of “be” needs to be placed. From the given options, the correct form is option (C) “is”.

• Q.53

________________with someone else’s email account is now a very serious offence.

(A)

Involving

(B)

Assisting

(C)

Tampering

(D)

Incubating

Marks:1 Explanation

“Involving” – Involving does not clearly signify un-permitted participation in someone else’s account.

“Assisting” – Assisting does not signify anything illicit with someone else’s account.

“Tampering” – Tampering means unwanted or un-permitted change in someone else’s account which is illegal.

“Incubating” – Incubating does not provide any context with illegal involvement.

So, it can be concurred that “tampering” is correct option to be filled in the blank.

• Q.54

Consider the following sentences:

All benches are beds. No bed is a bulb. Some bulbs are lamps.

Which of the following can be inferred?

1. Some beds are lamps.
2. Some lamps are beds.
(A)

Only i                                   (B)

Only ii

(C)

Both i and ii

(D)

Neither i nor ii

Marks:1 Explanation

Consider the statement “All benches are beds”, Venn diagram for the given statement will be as follows:

Again, “No bed is a bulb”:

Third statement “Some bulbs are lamps” can be represented in two ways as it is not 100 % clear statement:

Statement i-  Some beds are lamps.

Here from first figure we can see that it is true but according to the second figure it is false, as we can figure out conclusively we will consider this statement (i) as not true.

Statement ii- Some lamps are bed.

Again, we can see that in first figure above statement is evaluated as true while according to second figure it comes out as false. So, it cannot be said conclusively that Statement ii is not true.

So, option (D) is correct.

• Q.55

If the radius of a right circular cone is increased by 50% its volume increases  by

(A)

75%

(B)

100%

(C)

125%

(D)

237.5%

Marks:1 Explanation

Volume of a right circular cone is  $$V\space=\space\frac{\pi r^2h}{3}$$

Now, radius is increased by 50%, therefore new radius (rnew) = r + 0.5r = 1.5r

Since, nothing is said about the other dimensions of the cone so we will take them to be constant.

New volume of the cone is (Vnew) $$=\space\frac{\pi(1.5r)^2h}{3}\space=\space 2.25(\frac{\pi r^2h}{3})$$$$\space=\space 2.25V$$

Increase in volume  = (Vnew – V) / V  =  $$(\frac{2.25V-V}{V})\space=\space 1.25$$

Now, percentage increase in volume = $$1.25\times100\space=\space125\%$$

• Q.56

The following sequence of numbers is arranged in increasing order: 1, x, x, x, y, y, 9, 16, 18. Given that the mean and median are equal to twice of mode, the value of y is

(A)

5

(B)

6

(C)

7

(D)

8

Marks:1 Explanation

Given sequence is:           1, x, x, x, y, y, 9, 16, 18.

Mean of a sequence of numbers is found by dividing the sum of all the numbers in the sequence by number of observations.

$$mean = (∑x)/n$$

Where, n is the number of observations and $$\sum$$x  is the sum of the observations.

Mean  $$= \frac{(1 + x + x + x + y + y + 9 + 16 + 18))}{9}$$

Median of the sequence is found by finding the middle term of given sequence when it is arranged in ascending order.

Case 1:                                                                                                                                                                 The middle term of sequence having odd number of observations is $$\frac{(n+1)}{2}$$ th term

Case 2:                                                                                                                                                                 But, when the number of observations are even then the median is found by the following formula

$$median= \frac {(n/2 th term) +((n+2)/2 th term) }{2}$$

Here, number of observations is odd which is 9 so median is  $$\frac{9+1}{2}$$ i.e. 5th term. So, here median is y.

Mode is the maximum repeated observation, and if all the observations are repeated equal number of times then there are no modes.

Here, mode = x.

According to the question mean and median are equal and twice of the mode.                                                   So, mean = median = y  and   y = 2 * mode = 2x.

Now, mean $$= \frac{(1 + x + x + x + y + y + 9 + 16 + 18)}{9} = y$$

$$\implies \frac{(1 + x + x + x + 2x + 2x + 9 + 16 + 18)}{9} = 2x$$        $$\implies 44 +7x = 18x$$  $$\implies 44 = 11x$$

$$\implies x=4$$    Therefore, y = 2x = 8

• Q.57

The old concert hall was demolished because of fears of the foundation would be affected by the construction of the new metro line in the area. Modern technology for underground metro construction tried to mitigate the impact of pressurized air pockets created by the excavation of large amounts of soil. But even with these safeguards, it is feared that the soil below the concert hall would not be stable.

From this, one can infer that

(A)

the foundation of old buildings creates pressurized air pockets underground, which are difficult to handle during metro construction.

(B)

metro construction has to be done carefully considering its impact on the foundations of existing buildings.

(C)

old buildings in an area form an impossible hurdle to metro construction in that area.

(D)

pressurized air can be used to excavate large amounts of soil from undergrounds areas.

Marks:2 Explanation

In the paragraph given in the question it is said that foundation of the old concert hall would be affected by the construction of the new metro line in the area, in the first line itself and also that excavation of soil for the construction of the metro line creates pressurized pockets which are problematic for the foundation of the buildings nearby and also that even modern technologies used for the construction of metro lines are unable to mitigate the problem presented by the creation of pressurized air pockets during excavation.

After understanding the context of the paragraph given to us we examine the options presented to us one by one:

(A) the foundation of old buildings creates pressurized air pockets underground, which are difficult to handle during metro construction. – This option is incorrect as it says that foundation of the old buildings is the one creating pressurized air pockets instead of the new metro line.

(B) metro construction has to be done carefully considering its impact on the foundations of existing buildings. – This option is correct as it is in sync with the central idea of the given para.

(C) old buildings in an area form an impossible hurdle to metro construction in that area. – This option is incorrect as it is clearly stated in the para that it’s not that old buildings are the hurdles in the construction of the metro but it’s the construction of new metro line that are endangering the foundation of the old buildings.

(D) pressurized air can be used to excavate large amounts of soil from undergrounds areas. – This option is also incorrect is suggests using of pressurized air pockets for the excavation of the soil from underground areas, which is not given in the paragraph.

• Q.58

Students applying for hostel rooms are allotted rooms in order of seniority. Students already staying in a room will move if they get a room in their preferred list. Preference of lower ranked applicants are ignored during allocation.

Given the data below, which room will Ajit stay in:

 Names Student seniority Current room Rooms preference Amar 1 P R, S, Q Akbar 2 None R, S Anthony 3 Q P Ajit 4 S Q, P, R
(A)

P

(B)

Q

(C)

R

(D)

S

Marks:2 Explanation

As we can see in the given table that Amar is at priority number 1 so, his preference will be considered first and his first preference is R therefore he will be allotted room R.

In order of priority Akbar will be considered second and his first preference is R, but it is already allotted to Amar (who is priority 1) so, his second preference will be considered which is S and un allotted, therefore Akbar will be allotted room S.

Anthony will be considered in after Akbar (as his priority order is 3) and his only preference is P and is unallotted so he will be allotted room P.

Lastly Ajit will be allotted the remaining unallotted room which is Q.

Hence, the room allotment table will be as below:

 Names Student seniority Current room Rooms allotted Amar 1 P R Akbar 2 None S Anthony 3 Q P Ajit 4 S Q
• Q.59

The last digit of $$(2171)^7 + (2172)^9 + (2173)^{11} + (2174)^{13 }$$ is

(A)

2

(B)

4

(C)

6

(D)

8

Marks:2 Explanation

$$(2171)^7 + (2172)^9 + (2173)^{11} + (2174)^{13 }$$

Just replace all the numbers in the question by their unit place numbers in the calculator you will get the result i.e. $$1^7 + 2^9 + 3^{11} + 4^{13}$$ = 67286524.

Unit digit of 67286524 is 4 , so correct answer is 4.

• Q.60

The bar graph below shows the output of five carpenters over one month, each of whom made different items of furniture: chairs, tables, and beds:

Consider the following statements.

1. The number of beds made by carpenter C2 is exactly the same as the number of tables made by carpenter C3.
2. The total number of chairs made by all carpenters is less than the total number of tables.

Which one of the following is true?

(A)

Only i

(B)

Only ii

(C)

Both i and ii

(D)

Neither i nor ii

Marks:2 Explanation

Let’s change the graphical data to a more readable numerical data.

 C1 Bed 3 Table 7 Chair 2 C2 8 2 10 C3 2 8 5 C4 3 3 2 C5 1 10 4

As we can see that number of beds made by C2 is equal to the number of tables made by C3, so statement i is correct.

Now the total number of chairs made is = (2 + 10 + 5 + 2 + 4) = 23

Total number of tables made is = (7 + 2 + 8 + 3 + 10) = 30

Here, total number of chairs are less than total number of tables so, statement ii is also correct.

Both the statements are correct.

Hence, option (C) is correct.

• Q.61

In a batch adsorption process, 5g of fresh adsorbent is used to treat 1 litre of an aqueous phenol solution. The initial phenol concentration is 100 mg/liter. The equilibrium relation is given by

$$q^*\space=\space 1.3C$$

where $$q^*$$ is the amount of phenol per gram of adsorbent: and C is the concentration of phenol in mg/liter in the aqueous solution.

When equilibrium is attained between the adsorbent and the solution, the concentration of phenol in the solution, rounded to 1 decimal place, is ___________ mg/liter.

Marks:2 Explanation
By equilibrium relation :       $$q^*\space =\space 1.3C$$  $$\implies​​​​​​\space\frac{x}{5}\space = \space 1.3(\frac{100-x}{1})$$
$$\implies \space x\space = \space 6.5(100\space-\space x)\space \implies\space x\space =\space650\space-\space6.5x\space\implies\space 7.5x\space =\space 650\space\implies\space x\space=\space \frac{650}{7.5}\space =\space 86.7mg/liter$$