GATE Examination

YEAR 2016

  • Q.1

    The model y = mx\(^2\) is to be fit to the data given below.

    x 1 \(\sqrt 2\) \(\sqrt3\)
    y 2 5 8

    Using linear regression, the value (rounded off to the second decimal place) of m is _______.

    Marks:2 Explanation
    • Answer:

      .Here, let us remake the table with the m values:

      x 1 \(\sqrt 2\) \(\sqrt 3\)
      y 2 5 8
      m 2 2.5 2.67

      In linear regression we try to transform a really complex data set into a more understandable curve. Here, we trying to transfrom the data to a curve of \(mx^2\).

      Looking at the table we can see that the general value of m should be between 2 and 2.67.

      The average of the m values = 2.39.

      Hence, the of m = 2.39.

       

  • Q.2

    The Lagrange mean-value theorem is satisfied for f (x) = x\(^3\) + 5 , in the interval (1, 4) at a value
    (rounded off to the second decimal place) of x equal to _______

    Marks:2 Explanation
    • Answer:

      .Lagrange's mean value theorem :

      If f(x) be a function such that f(x) is continuous in {a,b} abd f'(x) exists in (a,b), then there exists atleast one c, such that 

      \(f'(c) = \frac {f(b) \space - \space f(a)}{b \space - \space a}\)

      Here, \(f(x) = x^2 + 5\) and a= 1 and b= 4.

      \(f'(x) = 3x^2 \)

      Now, \(f'(c) = 3c^2\)

      \(=> 3c^2 = \frac {f(4) \space - \space f(1)}{4 \space - \space 1}\)

      \(=> 3c^2 = \frac{69 \space - \space 6}{3} = \frac{63}{3} = 21\)

      \(=> c^2 = \frac {21}{3} = 7\)

      \(=> c = \pm \sqrt 7 = \pm 2.646\)

      Since, only 2.646 lies within the range of [1,4], so correct answer is 2.65

  • Q.3

    Values of f (x) in the interval [0, 4] are given below.

    x 0 1 2 3 4
    f(x) 3 10 21 36 55

    Using Simpson’s 1/3 rule with a step size of 1, the numerical approximation (rounded off to the
    second decimal place) of \(\int _0^4 f(x)dx\) is _________.

    Marks:2 Explanation
    • Answer:

      General formula for Simpson's 1/3\(^{rd}\) rule is as below:

      \(\int ^{x_0+h}_{x_0} f(x)dx = \frac{h}{3}[(y_0 + y_n) + 4(y_1 + y_3 + y_5 + ....) + 2(y_2 + y_4 + y_6 + ....)]\)

      Here,  \(x_0 = 0 \space and \space h = 1\) .

      x 0 1 2 3 4
      f(x) 3 10 21 36 55
      y \(y_0\) \(y_1\) \(y_2\) \(y_3\) \(y_n\)

      Now, putting all the values in the formula:

      \(\int ^{4}_{0} f(x)dx = \frac{1}{3}[(3 + 55) + 4(10+36) + 2(21)]\)\(= \frac{1}{3} [58 + 4\times 46 + 42] = \frac {1}{3} [284] = 94.6667\)

      Hence, the answer should be 94.6 or 94.7.

  • Q.4

    A jacketed stirred tank with a provision for heat removal is used to mix sulphuric acid and water in
    a steady state flow process. H2SO4 (l) enters at a rate of 4 kg/h at 25\(^0\)C and H2O (l) enters at a rate
    of 6 kg/h at 10\(^0\)C. The following data are available:
    Specific heat capacity of water = 4.2 kJ kg\(^{-1}\)K\(^{-1}\).

     

    Specific heat capacity of aqueous solution of 40 mass% H\(_2\)SO\(_4\) = 2.8 kJ (kg solution)‒\(^{-1}\) K\(^{-1}\).
    Assume the specific heat capacities to be independent of temperature.
    Based on reference states of H\(_2\)SO\(_4\) (l) and H\(_2\)O (l) at 25\(^0\)C, the heat of mixing for aqueous solution
    of 40 mass% H\(_2\)SO\(_4\) = ‒ 650 kJ (kg H\(_2\)SO\(_4\))\(^{-1}\).

     

    If the mixed stream leaves at 40\(^0\)C, what is the rate of heat removal (in kJ/h)?

    (A) 

    1802

    (B) 

    2558

    (C) 

    5702

    (D) 

    6458

    Marks:2 Explanation
    • Answer:

      .

  • Q.5

    An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed
    system. The work required for the irreversible compression is 1.5 times the work that is required for
    reversible compression from the same initial temperature and pressure to the same final pressure.
    The molar heat capacity of the gas at constant volume is 30 J mol\(^{-1}\) K\(^{-1}\) (assumed to be independent
    of temperature); universal gas constant, R is 8.314 J mol\(^{-1}\) K\(^{-1}\); ratio of molar heat capacities is
    1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in
    the irreversible compression case is _______.

    Marks:2 Explanation
    • Answer:

      .GIVEN: Initial Pressure \(P_1=3\space bar\)  ; Initial Temperature  \(T_1=\space 300K\) . Final Pressure \(P_2=6\space bar\).

                                        \(W_{Ir-reversible}\space =\space 1.5(W_{Reversible})\)      

           Work done in adiabatic compression = \(C_V\bigtriangleup T\) = \(C_V(T_2-T_1)\).

      Let \(T_2'\) be the temperature of gas at final state in the irreversible compression.

         \(\implies\space C_V(T_2'-T_1)\space =\space 1.5C_V(T_2-T_1)\) = 1.5\(C_V[T_1(\frac{P_2}{P_1})^{\frac{\gamma -1}{\gamma}}\space-\space T_1]\)

         \(\implies \space (T_2'-T_1)\space = \space 1.5\times T_1[(\frac{P_2}{P_1})^{\frac{\gamma -1}{\gamma }}-1]\)   \(=1.5\times30\times300[(\frac{6}{3})^{\frac{1.277-1}{1.277}}-1]=73.01\)

         \(\implies \space T_2'\space= \space 73.01+T_1=73.01+300=373.01K\).

  • Q.6

    A gas obeying the Clausius equation of state is isothermally compressed from 5 MPa to 15 MPa in
    a closed system at 400 K. The Clausius equation of state is \(P = {RT \over {v \space -\space b(T) }}\) where P is the pressure, T is
    the temperature, v is the molar volume and R is the universal gas constant. The parameter b in the
    above equation varies with temperature as below:\(b(T) = b_0 \space + \space b_1T \space with \space b_0 = 4 *10^{-5} m^3mol^{-1 } \space and \space b_1 = 1.35*10^{-7} \space m^3mol^{-1}K^{-1}. \) 

     The effect of pressure on the molar enthalpy (h) at a constant
    temperature is given by \(({\partial h \over \partial P})_T = v \space - \space T({\partial v \over \partial T})_P\)

    Let hi and hf denote the initial and final molar
    enthalpies, respectively. The change in the molar enthalpy \(h_f \space - \space h_i\) (in J mol\(^{-1}\), rounded off to the
    first decimal place) for this process is _______

    Marks:2 Explanation
    • Answer:

       Given: Clausius equation

       \(P=\frac{RT}{v-b(T)}\space \implies\space v\space = \space \frac{RT}{P}+b(T)=\frac{RT}{P}+b_o+b_1T\)\(\implies\space( \frac{\partial v}{\partial T})_P \space = \space \frac{R}{P}+b_1\).

      \((\frac{\partial h}{\partial P})_T= v-T(\frac{\partial v}{\partial T})_P=\frac{RT}{P}+b_o+b_1T-T(\frac{R}{P}+b_1)\)\(=b_0\).

      Thus , \(\bigtriangleup H=\bigtriangleup P\times b_o=(15-5)\times10^6\times(4*10^{-5})=400J/mol\).

                                                  

  • Q.7

    A binary system at a constant pressure with species ‘1’ and ‘2’ is described by the two-suffix
    Margules equation, \({{g^E}\over RT} = 3x_1x_2\)  , where \(g^E\) is the molar excess Gibbs free energy, R is the universal
    gas constant, T is the temperature and \(x_1,x_2\) are the mole fractions of species 1 and 2, respectively.
    At a temperature T, \({{g_1}\over RT} = 1 \space and \space {{g_2}\over RT} = 2 \) , where \(g_1 \space and \space g_2\) are the molar Gibbs free energies of pure
    species 1 and 2, respectively. At the same temperature, g represents the molar Gibbs free energy of
    the mixture. For a binary mixture with 40 mole % of species 1, the value (rounded off to the
    second decimal place) of \(g \over RT\) is _______

    Marks:2 Explanation
    • Answer:

      GIVEN:   \(\frac{g_1}{RT}=1\space ,\space \frac{g_2}{RT}=2\space ,\space\space\space\space\space\space x_1=0.4\space,\space x_2=0.6.\)

        As  \(\bigtriangleup G_{mix}=RT\sum x_ilnx_i.\)

      \(\implies \space G^{ig}-\sum x_iG_i^{ig}=RT\sum x_ilnx_i.\)

      \(\implies \space G^{ig}\space = \space\sum x_iG_i^{ig}+RT\sum x_ilnx_i\).

      \(\space\implies \frac{G^{ig}}{RT}\space =\space \frac{\sum x_iG_i^{ig}}{RT}+\sum x_ilnx_i\).

      Also \(\frac{G^E}{RT}\space\space\space=\space\space\space\frac{G}{RT}\space-\space\frac{G^{ig}}{RT}\)    \(\therefore\space \frac{G}{RT}\space=\space \frac{G^E}{RT}+\frac{G^{ig}}{RT}\)

      \(\implies \space \frac{G}{RT}\space=\space \frac{G^E}{RT}\space+\space\frac{\sum x_iG_i^{ig}}{RT}\space+\space x_ilnx_i\space=\space 3x_1x_2\space+\frac{x_1g_1}{RT}\space+\frac{x_2g_2}{RT}\space+x_1lnx_1\space+x_2lnx_2\).

      \(\implies \space \frac{G}{RT} \space =\space3\times0.4\times0.6\space+\space0.4\times1\space+0.6\times2\space+0.4\times ln0.4\space+0.6\times ln0.6\)

      \(\implies \space \frac{G}{RT}\space=\space1.65\).  

  • Q.8

    Water (density=1000 kg m\(^{-3}\)) is pumped at a rate of 36 m\(^3\)/h, from a tank 2 m below the pump, to an
    overhead pressurized vessel 10 m above the pump. The pressure values at the point of suction from
    the bottom tank and at the discharge point to the overhead vessel are 120 kPa and 240 kPa,
    respectively. All pipes in the system have the same diameter. Take acceleration due to gravity,
    g = 10 m s\(^{-2}\). Neglecting frictional losses, what is the power (in kW) required to deliver the fluid?

    (A) 

    1.2

    (B) 

    2.4

    (C) 

    3.6

    (D) 

    4.8

    Marks:2 Explanation
    • Answer:

                                    

      Applying Bernoulli's Theorem at (1) & (2) , 

           \(\frac{P_1}{\rho }+\frac{V_1^2}{2}+gZ_1+\eta W_P\space=\space\frac{P_2}{\rho }+\frac{V^2}{2}+gZ_2+h_f\)

      \(\implies \space\eta W_P\space=\space \frac{(P_2-P_1)}{\rho }+g(Z_2-Z_1)\space= \space \frac{(240-120)\times 10^3}{1000}+10\times[(10-(-2))]\space=\space 240 \space(\frac{m^2}{s^2})\)

      Power delivered to fluid = \(\dot{m}\times \eta W_P\space=\space\rho\times Q\times \eta W_P\space =\space 1000\times0.01\times240\space= \space 2400W\space=\space2.4KW\)

  • Q.9

    An agitated cylindrical vessel is fitted with baffles and flat blade impellers. The power number for
    this system is given by \(N_p = {{P \over \rho\space n^3 D^5}}\) where P is the power consumed for the mixing, ρ is the
    density of the fluid, n is the speed of the impeller and D is the diameter of the impeller. The
    diameter of the impeller is 1/3\(^{rd}\) the diameter of the tank and the height of liquid level is equal to the
    tank diameter. The impeller speed to achieve the desired degree of mixing is 4 rpm. In a scaled up
    design, the linear dimensions of the equipment are to be doubled, holding the power input per unit
    volume constant. Assuming the liquid to be Newtonian and \(N_p\) to be independent of Reynolds
    number, what is the impeller speed (in rpm) to achieve the same degree of mixing in the scaled up
    vessel?

    (A) 

    0.13

    (B) 

    1.26

    (C) 

    2.52

    (D) 

    3.82

    Marks:2 Explanation
    • Answer:

      Given : Power No \(N_p = {{P \over \rho\space n^3 D^5}}\), Thus  \(P\space=\space\alpha N^3D^5\).

      Volume of the tank , \(V\space=\space\frac{\pi}{4}D_v^2h\space=\space\frac{\pi}{4}D_v^3\space\space\space\space\space[\because \space D_v=h]\)  Thus  \(V\space\space \alpha \space\space D_v^3\).

      Thus \(\frac{P}{V}\space\space=\space\space\alpha\space[\frac{N^3D_i^5}{D_v^3}]\space\space\space D_i-Impeller\space Diameter\space\space D_v-Tank\space Diameter\)

      Since \(\frac{D_{i2}}{D_{i1}}\space=\frac{D_{v2}}{D_{v1}}\space=2\)  ( Linear dimensions of the equipment are doubled ).

      \([\frac{N^3D_i^5}{D_v^3}]_2\space\space\space\space=\space\space\space\space[\frac{N^3D_i^5}{D_v^3}]_1\)

      \(\implies N_2^3\space=(N_1^3)(\frac{D_{i1}}{D_{i2}})^5(\frac{D_{v2}}{D_{v1}})^3=(4^3)(\frac{1}{2})^5(2)^3=16\).

      \(\implies \space N_2\space=\space(16)^{\frac{1}{3}}=2.52\space rpm\).

  • Q.10

    Consider a rigid solid sphere falling with a constant velocity in a fluid. The following data are
    known at the conditions of interest: viscosity of the fluid = 0.1 Pa s, acceleration due to gravity =
    10 m s\(^{-2}\), density of the particle = 1180 kg m\(^{-3}\) and density of the fluid = 1000 kg m\(^{-3}\). The diameter
    (in mm, rounded off to the second decimal place) of the largest sphere that settles in the Stokes’ law
    regime (Reynolds number \(\leq\) 0.1), is _______

    Marks:2 Explanation
    • Answer:

      Since \(Re_N\leq0.1\) , it is Stoke's regime. \(\rho_S=1180Kg/m^3\space,\space\rho=1000Kg/m^3,\space\mu\space=0.1Pa.s,\space g=10m/s^2\)

      Terminal velocity , \(U_t=\frac{D_P^2(\rho_S-\rho)g}{18\mu}=\frac{D_P^2(1180-1000)\times 10}{18\times 0.1}=1000D_P^2\)  (1)

      \(Re_N=0.1\space\implies\space\frac{\rho U_tD_P}{\mu}=0.1\space\implies \space\frac{1000\times1000D_P^2\times D_P}{0.1}=0.1\space\implies\space D_P^3\space=\space 1\times 10^{-8}\)

      \(\implies\space D_P^3\space=\space1\times 10^{-8}\space\implies\space D_P\space= \space(10^{-8})^{\frac{1}{3}}\space=\space2.15mm\)

  • Q.11

    The characteristics curve (Head – Capacity relationship) of a centrifugal pump is represented by the
    equation \(\bigtriangleup H_{pump} = 43.8 \space - \space 0.19Q\) , where \(\bigtriangleup H_{pump}\) is the head developed by the pump (in m) and
    Q is the flowrate (in m\(^3\)/h) through the pump. This pump is to be used for pumping water through a
    horizontal pipeline. The frictional head loss \(\bigtriangleup H_{piping}\) piping (in m) is related to the water flowrate \(Q_L\) (in
    m\(^3\)/h) by the equation \(\bigtriangleup H_{piping} = 0.0135Q_L^2 \space + \space 0.045Q_L\) . The flowrate (in m\(^3\)/h, rounded off to the
    first decimal place) of water pumped through the above pipeline, is _______

    Marks:2 Explanation
    • Answer:

      Applying Bernoulli's equation :

       Head developed by the pump = Head utilize by the Pipe;

      Thus  \(\bigtriangleup H_{pump}\space\space\space\space=\space\space\space\space \bigtriangleup H_{piping}\)

      \(\implies \space 43.8-0.19Q\space=\space0.0135Q^2+0.045Q\space\space\space \implies\space0.0135Q^2+0.235Q-43.8=0\)

      \(\implies \space\space Q=48.9\space m^3/h\)

  • Q.12

    Water flows through a smooth circular pipe under turbulent conditions. In the viscous sub-layer, the
    velocity varies linearly with the distance from the wall. The Fanning friction factor is defined as,
    \(f = {\tau_w\over\rho\bar{u}^2/2} , \space where \space \tau_w\) is the shear stress at the wall of the pipe, \(\rho\) is the density of the fluid and \(\bar{u}\) is
    the average velocity in the pipe. Water (density = 1000 kg m\(^{-3}\), viscosity = \(1 \times 10^{-3} kg\space m^{-1}\space s^{-1} \) flows
    at an average velocity of 1 m /s through the pipe. For this flow condition, the friction factor f is
    0.005. At a distance of 0.05 mm from the wall of the pipe (in the viscous sub-layer), the velocity (in
    m /s, rounded off to the third decimal place), is _______

    Marks:2 Explanation
    • Answer:

      For thin viscous sub layer , \(\tau_o= -\mu\frac{dv}{dy},\space\space\space\space\space\space\tau_o\) Assumed to be constant .

      On integrating , \(\tau_oy\space=\space\mu v\) _____ (1)

      \(v^\ast\space=\space\sqrt{\frac{\tau_o}{\rho}}\space\implies\space\tau_o\space=\space\rho (v^{\ast} )^2\)              \(v^{\ast}\space=\space friction\space velocity\) .

      Substituting in (1) we get  ,    

      \(\rho(v^{\ast})^2y\space=\space\mu v\)    or  \(\frac{v}{v^{\ast}} ​​​​\space= ​​​​\space\frac{yv^{\ast}}{(\frac{\mu}{\rho})}\)   ________ (2)

      Left hand side , dimensionless ratio can be written as  \(v^{+}​​​​\space=​​​​\space v \sqrt{\frac{\rho}{\tau_o}}\).

      Right hand side , dimensionless ratio can be written as \(y^+​​​​\space=​​​​\space\frac{\sqrt{\tau_o \rho }}{\mu}y\)

      For viscous-sublayer    

         \(v^+​​​​\space=​​​​\space y^+\)

       \(\implies\space v\sqrt{\frac{\rho}{\tau_o}}\space=\space\frac{\sqrt{\tau_o \rho}}{\mu}y \space\implies\space v \space =\space \frac{\tau_o y}{\mu}\)    . Also    \(f\space=\space\frac{\tau}{(\frac{\rho \bar{v}^2}{2})}\space\implies\space\tau\space =\space\frac{f\rho \bar{v}^2}{2}\)

      Thus      \(v\space=\space\frac{f \rho v^2y}{2\mu}\)    ____________ (3)

      \(f=0.005,\space\bar{v}=1m/s,\space y=0.05mm=0.05\times 10^{-3}m,\space\mu =1\times 10^{-3}Kg/m.s,\space \rho = 1000Kg/m^3\)

      Putting above data in (3) we get

      \(v\space=\space \frac{0.005\times 1000\times 1^2\times0.05\times 10^{-3}}{2\times 10^{-3}}\space =0.125m/s\) 

  • Q.13

    In a 1-1 pass shell and tube exchanger, steam is condensing in the shell side at a temperature (T\(_s\)) of
    135\(^0\)C and the cold fluid is heated from a temperature (T\(_1\)) of 20\(^0\)C to a temperature (T\(_2\)) of 90\(^0\)C.
    The energy balance equation for this heat exchanger is

    \(\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space ln{{T_s\space - \space T_1}\over {T_s \space - \space T_2}} = {UA \over {\dot{m}c_p}}\)

    where U is the overall heat transfer coefficient, A is the heat transfer area,\(\dot{m}\) is the mass flow rate
    of the cold fluid and \(c_p\) is its specific heat. Tube side fluid is in a turbulent flow and the heat
    transfer coefficient can be estimated from the following equation:

    \(Nu = 0.023\space (Re)^{0.8} \space (Pr)^{1/3}\)

    where Nu is the Nusselt number, Re is the Reynolds number and Pr is the Prandtl number. The
    condensing heat transfer coefficient in the shell side is significantly higher than the tube side heat
    transfer coefficient. The resistance of the wall to heat transfer is negligible. If only the mass flow
    rate of the cold fluid is doubled, what is the outlet temperature (in \(^0\)C) of the cold fluid at steady
    state?

     

    (A) 

    80.2

    (B) 

    84.2

    (C) 

    87.4

    (D) 

    88.6

    Marks:2 Explanation
    • Answer:

      Overall heat transfer coefficient is \(\frac{1}{U}\space=\space\frac{1}{h_i}+\frac{1}{h_o}\space\space\space\space\space\space\space\space\space\space \because h_o>>h_i\)  \(\frac{1}{U}\space=\space\frac{1}{h_i}\space\implies \space U\space=\space h_i\)

      \(Nu\space​​=​​​​\space 0.023(Re)^{0.8}(Pr)^{1/3}\space\implies\space\frac{h_i D_i}{K}=\space 0.023(\frac{\rho v D}{\mu})^{0.8}(\frac{\mu C_P}{K})^{1/3}\)

      \(Re\space=\space \frac{\rho vD}{\mu}\space=\space\frac{\dot{m}D}{A \mu}\space=\frac{\dot{m}D}{\frac{\pi}{4}D^2 \mu}\space=\frac{4 \dot{m}}{\pi D \mu }\)     

      \(\frac{h_i D_i}{K}=\space 0.023(\frac{4\dot{m}}{\pi D\mu})^{0.8}(\frac{\mu C_P}{K})^{1/3}\)    \(\implies\space h_i\space=\space K(\dot{m})^{0.8}\)   [ Since all other parameters are constant.]

      \(ln\frac{(T_S-T_1)}{(T_S-T_2)}\space= \frac{UA}{\dot{m}C_P}\space= \frac{h_iA}{\dot{m}C_P}\space=\space\frac{K\dot{m}^{0.8}A}{\dot{m}C_P}\space =\space \frac{k'}{\dot{m}^{0.2}}\)   _________  ( modified energy balance equation ).

      CASE I :

      \(T_S=135^0C,\space T_1=20^0C,\space T_2=90^0C \space, Mass\space flowrate\space is \space \dot{m}\)

      CASE II:

        \(T_S=135^0C,\space T_1=20^0C,\space T_2=? \space, Mass\space flowrate\space is \space 2\dot{m}\)

      By CASE I  :       \(ln\frac{(135-20)}{(135-90)}\space= \space \frac{K'}{\dot{m}^{0.2}}\)        _________  (1)

      By CASE II:        \(ln\frac{(135-20)}{(135-T_2)}\space= \space \frac{K'}{(2\dot{m})^{0.2}}\)     _________   (2)

      Dividing (1) by (2) we get 

      \(\frac{ln\frac{(135-20)}{(135-90)}}{ln\frac{(135-20)}{(135-T_2)}}\space=\space \frac{K'/\dot{m}^{0.2}}{K'/(2\dot{m})^{0.8}}\space=\space 2^{0.2}\)

      \(\implies \space \frac{ln(\frac{115}{45})}{ln(\frac{115}{135-T_2 })}\space=\space 2^{0.2}\)   ,  Solving we get  \(T_2=84.2^0C\)

  • Q.14

    In an experimental setup, mineral oil is filled in between the narrow gap of two horizontal smooth
    plates. The setup has arrangements to maintain the plates at desired uniform temperatures. At these
    temperatures, ONLY the radiative heat flux is negligible. The thermal conductivity of the oil does
    not vary perceptibly in this temperature range. Consider four experiments at steady state under
    different experimental conditions, as shown in the figure below. The figure shows plate
    temperatures and the heat fluxes in the vertical direction.

    What is the steady state heat flux (in W m\(^{-2}\)) with the top plate at 70\(^0\)C and the bottom plate at
    40\(^0\)C?

    (A) 

    26

    (B) 

    39

    (C) 

    42

    (D) 

    63

    Marks:2 Explanation
    • Answer:

      .

  • Q.15

    The space between two hollow concentric spheres of radii 0.1 m and 0.2 m is under vacuum.
    Exchange of radiation (uniform in all directions) occurs only between the outer surface (S\(_1\)) of the
    smaller sphere and the inner surface (S\(_2\)) of the larger sphere. The fraction (rounded off to the
    second decimal place) of the radiation energy leaving S\(_2\), which reaches S\(_1\) is _______

    Marks:2 Explanation
    • Answer:

                                                       

      GIVEN :    \(r_1=0.1m ,\space r_2=0.2m\)     \(F_{11}=0,\space F_{11}+F_{12}=1,\space \implies F_{12}=1\)

      By Reciprocity theorem : \(A_1F_{12}=A_2F_{21}\implies\space F_{21}=\frac{A_1}{A_2}F_{12}=\frac{4\pi r_1^2}{4\pi r_2^2}=(\frac{r_1}{r_2})^2=(\frac{0.1}{0.2})^2=\frac{1}{4}=0.25\) 

        

  • Q.16

    A binary distillation column is to be designed using McCabe Thiele method. The distillate contains
    90 mol% of the more volatile component. The point of intersection of the q-line with the
    equilibrium curve is (0.5, 0.7). The minimum reflux ratio (rounded off to the first decimal place) for
    this operation is _______

    Marks:2 Explanation
    • Answer:

                                          

      Equation of rectification column : \(y=\frac{R}{R+1}x+\frac{1}{R+1}x_D \implies (R+1)y=Rx+x_D\)

      \(\implies \space Ry-Rx=x_D-y\space \implies R=\frac{x_D-y}{y-x}=\frac {0.9-0.7}{0.7-0.5}=1.\)

      Hence, minimum reflux ratio is 1.

  • Q.17

    Solute C is extracted in a batch process from its homogenous solution of A and C, using solvent B. The combined composition of the feed and the extracting solvent is shown in the figure below as point M, along with the tie line passing through it. The ends of the tie line are on the equilibrium curve.

    What is the selectivity for C? 

    (A) 

    3.5

    (B) 

    7

    (C) 

    10.5

    (D) 

    21

    Marks:2 Explanation
    • Answer:

                   

      Selectivity for C = \(\frac{(\frac{weight\space fraction \space of\space C}{weight\space fraction\space of\space A})_{Extract \space E}}{(\frac{weight\space fraction \space of\space C}{weight\space fraction\space of\space A})_{Raffinate \space R}}\)  =  \(\frac{(0.3/0.1)}{(0.2/0.7)}\space=\space \frac{21}{2}\space=10.5\)

  • Q.18

    At \(30^0C\)the amounts of acetone adsorbed at partial pressure of 10 and 100 mmHg are 0.1 and 0.4 kg acetone/kg activated carbon, respectively. Assume Langmuir isotherm describes the adsorption of acetone on activated carbon. What is the amount of acetone adsorbed (in kg per kg of activated carbon) at a partial pressure of 50 mmHg and \(30^0C\)?

    (A) 

    0.23

    (B) 

    0.25

    (C) 

    0.3

    (D) 

    0.35

    Marks:2 Explanation
    • Answer:

      Langmuir isotherm :  \(\frac{x}{m}\space=\space \frac{aP}{1+bP}\)  

      Given : At P = 10 mmHg , x/m = 0.1 & at P = 100 mmHg , x/m = 0.4 .

      Thus \(0.1 \space = \space \frac{a\times 10}{1+b\times 10}\space \implies \space 0.1+b\space =\space 10a\)  _______  (1).

               \(0.4\space= \space \frac{a\times 100}{1+b\times 100}\space \implies \space 0.4+40b\space = \space100a\) ______ (2).

      Solving (1) & (2) , we get  a = 0.012 , b = 0.02.

      Then at P = 50 mmHg , \(\frac{x}{m}\space=\space\frac{0.012\times 50}{1+0.02\times 50}\space = 0.30 Kg/Kg\space activated\space carbon\).

  • Q.19

    Consider the folowing two cases for a binary mixture of ideal gasese A and B under steady state conditions. In Case 1, the diffusion of A and B occurs through non-diffusiong B. In Case 2, equimolal counter diffusion of A and B occurs. In both the cases, the total pressure is 100 kPa and the partial pressure of A at two points separated by a distance of 10 mm are 10 kPa and 5 kPa.  Assume that the Fick's first law of diffusion is applicable. What is the ratio of molar flux of A in Case 1 to that in Case 2?

    (A) 

    0.58

    (B) 

    1.08

    (C) 

    1.58

    (D) 

    2.18

    Marks:2 Explanation
    • Answer:

      CASE 1 : Diffusion of A through non-diffusing B .

                         \(N_A\space=\space \frac{D_{AB}P}{RTz}ln\frac{P-P_{A2}}{P-P_{A1}}\)

      CASE 2: Equimolal counter diffusion of A&B occurs.

                       \(N_A\space=\space \frac{D_{AB}(P_{A2}-P_{A1})}{RTz}\)

      Thus  \(\frac{[N_A]_1}{[N_A]_2}\space =\space \frac{Pln\frac{(P-P_{A2})}{(P-P_{A1})}}{(P_{A2}-P_{A1}}\space=\space\frac{100ln\frac{(100-5)}{(100-10)}}{(10-5)}\space=\space 1.08\)

  • Q.20

    The liquid phase reversible reaction  \(A\rightleftharpoons B\)  is carried out in an isothermal CSTR operating under steady state conditions. The inlet stream does not contain B and the concentration of A in the inlet atream is 10 mol/lit. The concentrations of A at the reator exit, fir residence times 1 sec and 5 sec are 8 mol/lit and 5 mol/lit, respectively. Assume the forward and backward reactions are elementary following the forst order rate law. Also assume that the system has constant molar density. The rate constant of the forwad reaction (in sec\(^{-1}\), rounded of to the third decimal place) is __________.

    Marks:2 Explanation
    • Answer:

      GIVEN: \(A\rightleftharpoons B\space\space At\space\tau =1s,C_A=8mol/lit\space\& \space \tau = 5s,C_A=5mol/lit\)

      \((-r_A)=K_1C_A-K_2C_B=K_1C_A-K_2(C_{AO}-C_B)\space\space[\because C_{AO}+C_{BO}=C_A+C_B]\)

      IN CSTR \(\tau\space= \space \frac{C_{AO}X_A}{(-r_A)}\space =\space \frac{(C_{AO}-C_A)}{K_1C_A-K_2(C_{AO}-C_A)}\space\implies \space K_1C_A-K_2(C_{AO}-C_A)\space=\space \frac{C_{AO}-C_A)}{\tau}\) __ (1)

      For \(\tau\space=\space 1s,C_A=8mol/lit\) equation (1) reduces to \(8K_1-2K_2=2\)   ___ (2)

      For \(\tau=5s,C_A=5mol/lit\) equation (1) reduces to  \(5K_1-5K_2=1\)  _____ (3)

      Solving (2) & (3) we get rate of forward reaction \(K_1=0.267s^{-1}\).

  • Q.21

    A liquid phase irreversible reaction \(A \to B\) is carried out in an adiabatic CSTR operating under steady state conditions. The reaction is elementary and follows the first order rate law. For this reaction, the figure below shows the conversion (\(X_A\)) of A as the function of temperature (T) for different values of rate of reaction (\(-r_A \space in\space mol\space m^{-3}s^{-1}\)) denoted by the numbers to the left of each curve. The figure can be used to determine the rate of the reaction at a particular temperature, for a given conversion of A.

    The inlet stream does not contain B and the concentration of A in the inlet stream is 5 mol/m\(^3\). The  molar feed rate of A is 100 mol/s. A steady state energy balance for this CSTR results in the following relation: \(T = 350\space +\space 25X_A\), where T is the temperature (in K) of the exit stream and \(X_A\)is the conversion of A in CSTR. FOr an exit conversion of 80% of A, the volume (in m\(^3\), rounded off to the first decimal place) of CSTR required is _______.

    Marks:2 Explanation
    • Answer:

      GIVEN: \(C_{AO}=5 mol/m^3,\space F_{AO}=100mol/s,\space X_A=0.8,\space T=350+25X_A=350 +25\times 0.5=370K\)

      From the given graph , for \(X_A=0.8,\space T =370K ,\space (-r_A)=10mol/m^2s\)

      In CSTR \(\frac{V}{F_{AO}}\space =\space \frac{X_A}{-r_A}\space \implies\space V=\space \frac{F_{AO}X_A}{-r_A}\space= \frac{100\times 0.8}{10}\space =8m^3\)

  • Q.22

    A porous pellet with Pt dispersed in it is used to carry out a catalytic reaction. Following two
    scenarios are possible.
    Scenario 1: Pt present throughout the pores of the pellet is used for catalyzing the reaction.
    Scenario 2: Pt present only in the immediate vicinity of the external surface of the pellet is used for
    catalyzing the reaction.
    At a large value of Thiele modulus, which one of the following statements is TRUE?

    (A) 

    Since the reaction rate is much greater than the diffusion rate, Scenario 1 occurs

    (B) 

    Since the reaction rate is much greater than the diffusion rate , Scenario 2 occurs

    (C) 

    Since the reaction rate is much lower than the diffusion rate, Scenario 1 occurs

    (D) 

    Since the reaction rate is much lower than the diffusion rate, Scenario 2 occurs

    Marks:2 Explanation
    • Answer:

      The formulae for Thiele modulus is as follow   \(\theta\space=\space L\sqrt{\frac{K}{D}}\)

           Or     \(\theta \space =\space characteristic\space length\times \sqrt{\frac{\space reaction \space rate}{\space pore\space diffusion\space rate}}\)

      So , if \(\theta\) is high then reaction rate is extremely high and pore diffusion rate is extremely low.

      Considering the above facts , we can say that " surface reaction will dominate" and " reactions will be much faster than pore diffusion".

      So most of reaction will takes place on the surface than pores.

      Hence Pt, present only in the immediate vicinity of the external surface of the pellet is used for catalizing the reaction.

  • Q.23

    A CSTR has a long inlet pipe. A tracer is injected at the entrance of the pipe. The E-curve obtained
    at the exit of the CSTR is shown in the figure below.

    Assuming plug flow in the inlet pipe, the ratio (rounded off to the second decimal place) of the
    volume of the pipe to that of the CSTR is _______

     

    Marks:2 Explanation
    • Answer:

      When a tracer is passed through a PFR(pipe) or CSTR in series with PFR(pipe) then the exiy age distribution is shifted by \(\tau_P\) in the time axis , where \(\tau_P\) is the residence time for the pipe. From the figure it can be depicted that curve is shifted by 5 minutes so \(\tau_P\) = 5min.

      Now,         \(E(t)\space= \space \frac{1}{\tau_M}e^{\frac{-(t-\tau_P)}{\tau_M}}\)    , where \(\tau_M\) is the residence time for CSTR.

      At t= 5min , E(t) = 0.05 , putting these value in above formulae we get 

                    \(0.05\space=\space \frac{1}{\tau_M}e^{\frac{-(5-5)}{\tau_M}}\space \implies \space 0.05\space = \space \frac{1}{\tau_M}\space\implies \space \tau_M\space= \frac{1}{0.05}=20min\).

      Thus Volume of Pipe/Volume of CSTR =  \(\frac{\tau_P}{\tau_M}\space = \frac{5}{20}=0.25\).

  • Q.24

    A liquid flows through an “equal percentage” valve at a rate of 2 m\(^3\)/h when the valve is 10% open.
    When the valve opens to 20% the flowrate increases to 3 m\(^3\)/h. Assume that the pressure drop
    across the valve and the density of the liquid remain constant. When the valve opens to 50%, the
    flowrate (in m\(^3\)/h, rounded off to the second decimal place) is_______

    Marks:2 Explanation
    • Answer:

      ..

  • Q.25

    A PI controller with integral time constant of 0.1 min is to be designed to control a process with
    transfer function

    \(G_p(s) = {10\over {s^2\space + \space 2s \space + \space 100}}\)

    Assume the transfer functions of the measuring element and the final control element are both unity
    \((G_m = 1, G_f = 1)\). The gain (rounded off to the first decimal place) of the controller that will
    constitute the critical condition for stability of the PI feedback control system is _______

    Marks:2 Explanation
    • Answer:

      .

  • Q.26

    For a unit step input, the response of a second order system is

    \(y(t) =K_p[1 \space - \space {{1\over \sqrt{1 - \zeta^2}}e^{-\frac{\zeta t}{\tau}}sin(\frac{\sqrt{1\space - \space \zeta^2}}{\tau}t \space + \space \phi)}]\)

    where, \(K_p\) is the steady state gain, \(\zeta\) is the damping coefficient, \(\tau\) is the natural period of ocsillation and \(\phi\) is the phase lag. The overshoot of the system is \(exp(-\frac{\pi \zeta}{\sqrt{1 \space - \space \zeta^2}})\). For a unit step input, the response of the system for an initial steady state condition at t = 0 is shown in the figure below.

    What is the natural period of ocsillation (in seconds) of the system?

     

    (A) 

    15.9

    (B) 

    50

    (C) 

    63.2

    (D) 

    100

    Marks:2 Explanation
    • Answer:

      .

  • Q.27

    A vertical cylindrical tank with a flat roof and bottom is to be constructed for storing 150 m\(^3\) of
    ethylene glycol. The cost of material and fabrication for the tank wall is Rs 6000 per m\(^2\) and the
    same for the roof and the tank bottom are Rs 2000 and Rs 4000 per m\(^2\), respectively. The cost of
    accessories, piping and instruments can be taken as 10% of the cost of the wall. 10% of the volume
    of the tank needs to be kept free as vapour space above the liquid storage. What is the optimum
    diameter (in m) for the tank?

    (A) 

    3.5

    (B) 

    3.9

    (C) 

    7.5

    (D) 

    7.8

    Marks:2 Explanation
    • Answer:

      .

  • Q.28

    A catalytic reforming plant produces hydrogen and benzene from cyclohexane by de-hydro
    aromatisation. In order to increase the production of hydrogen, the owner plans to change the
    process to steam reforming of the same feedstock that produces hydrogen and carbon dioxide.
    Stoichiometrically, what is the maximum ratio of pure hydrogen produced in the proposed process
    to that in the existing process?

    (A) 

    1

    (B) 

    2

    (C) 

    5

    (D) 

    6

    Marks:2 Explanation
    • Answer:

      .

  • Q.29

    The volume of  a sphere of diameter 1 unit is _________ than the volume of a cube of side 1 unit.

    (A) 

    least

    (B) 

    less

    (C) 

    lesser

    (D) 

    low

    Marks:1 Explanation
    • Answer:

      Let the diameter of the sphere be d and radius be r.

      Let the side of the cube be a.

      d = 1 unit and a = 1 unit;

      Volume of a sphere = \(\frac{4}{3}\pi r^3 = \frac{\pi}{6}d^3\) (since r = d/2) = \(\frac{\pi}{6}\times 1^3\) = 0.5236 cubic unit

      Volume of the cube = \(a^3\) = \(1^3\) = 1 cubic unit

      Here, it can be seen that volume of the cube is greater than that of the sphere.

      So, option (B) is correct.

      The volume of  a sphere of diameter 1 unit is less than the volume of a cube of side 1 unit.

  • Q.30

    The unruly crowd demanded that the accused be ____________ without trial.

    (A) 

    hanged

    (B) 

    hanging

    (C) 

    hankering

    (D) 

    hung

    Marks:1 Explanation
    • Answer:

      Here the four options are:

      (A) hanged - The action of killing someone by hanging a noose arounf the person's neck and then letting them hang by it till his/her death (usually by Law enforcement authorities do it ot a criminal)

      (B) hanging - The present continuous form of the verb 'hang'.

      (C) hankering - present continuous form of hanker (feel a strong desire for or to do something).

      (D) hung - The past/past perfect tense of hang.

      As it can be infered by the usage of the words 'crowd', 'demanded' and 'trial' that the whole context is about a culprit being prosecuted and crowd being so angry they are demanding theeir own judgement to be followed without trial. So, in this context option (A) is suitable.

      The unruly crowd demanded that the accused be hanged without trial.

  • Q.31

    Choose the statement(s) where the underlined word is used correctly:
    (i) A prone is a dried plum.
    (ii) He was lying prone on the floor.
    (iii) People who eat a lot of fat are prone to heart disease.

    (A) 

    (i) and (iii) only

    (B) 

    (iii) only

    (C) 

    (i) and (ii) only

    (D) 

    (ii) and (iii) only

    Marks:1 Explanation
    • Answer:

      (i) A prone is a dried plum. - A prone is not a dried plum. Dried plums are called prune

      (ii) He was lying prone on the floor. -  Prone means lying flat, especially face downwards. So this statement is grammatically correct.

      (iii) People who eat a lot of fat are prone to heart disease. -  Prone also means likely or liable to suffer from, do, or experience something unpleasant or regrettable. So, this statement is also correct in grammatical use.

      As, it can be seen that both statements (ii) and (iii) are true while (i) is incorrect in grammatical use of word 'prone'.

      Hence, option (D) is correct.

  • Q.32

    Fact: If it rains, then the field is wet.
    Read the following statements:
    (i) It rains
    (ii) The field is not wet
    (iii) The field is wet
    (iv) It did not rain
    Which one of the options given below is NOT logically possible, based on the given fact?

    (A) 

    If (iii), then (iv).

    (B) 

    If (i), then (iii).

    (C) 

    If (i), then (ii).

    (D) 

    If (ii), then (iv).

    Marks:1 Explanation
    • Answer:

      Going by options:

      Option (A): If (iii) the field is wet, then (iv) it did not rain. 

      This option is correct since the field can be wet due to mltiple reasons besides rain. So, if the field is wet then it can be even when it did not rain.

      Option (B): If it rains, then (iii) the field is wet. 

      Yes, if it rains the field is definitely going to get wet.

      Option (C): If (i) it rains, then (ii) the field is not wet.

      If it rains then the field is definitely going  to be wet. It is n ot possible for the field to not get wet if it rains.

      Options (D): If the field is wet, then it did not rain.

      If the field is wet then if might had rain as field can be wet because of many reasons. So. it is totally possible that it did not rain even if the field is wet.

      So, only option (C) is not possible.

      Hence, option (C) is correct.

  • Q.33

    A window is made up of a square portion and an equilateral triangle portion above it. The base of
    the triangular portion coincides with the upper side of the square. If the perimeter of the window is
    6 m, the area of the window in m\(^2\) is ___________.

    (A) 

    1.43

    (B) 

    2.06

    (C) 

    2.68

    (D) 

    2.88

    Marks:1 Explanation
    • Answer:

      Let the sides of the square be a m. Then sides of the triangle will also be a m as it is a equilateral triangle mounted on the square matching its side.

      The perimeter of the figure above is = 5a m

      => 5a = 6

      => a = 6/5 = 1.2 m 

      Therefore area of the window is = \(a^2 \space + \space \frac{\sqrt3}{4}a^2\) = \(1.2^2 \space + \space \frac{\sqrt3}{4}1.2^2\) = 2.06 m\(^2\)

      Hence, option (B) is correct.

  • Q.34

    Students taking an exam are divided into two groups, P and Q such that each group has the same
    number of students. The performance of each of the students in a test was evaluated out of 200
    marks. It was observed that the mean of group P was 105, while that of group Q was 85. The
    standard deviation of group P was 25, while that of group Q was 5. Assuming that the marks were
    distributed on a normal distribution, which of the following statements will have the highest
    probability of being TRUE?

    (A) 

    No student in group Q scored less marks than any student in group P.

    (B) 

    No student in group P scored less marks than any student in group Q.

    (C) 

    Most students of group Q scored marks in a narrower range than students in group P.

    (D) 

    The median of the marks of group P is 100.

    Marks:2 Explanation
    • Answer:

      For Group P:

      Mean = 105 and Standard deviation = 25

      Since, the marks are distributed on normal distribution. Median = Mode = Mean = 105

      Range of marks obtained by students of this group is 80 to 130.

      For Group Q:

      Mean = 85 and Standard deviation = 5

      Since, the marks are distributed on normal distribution. Median = Mode = Mean = 85

      Range of marks obtained by students of this group is 80 to 90.

      It can be seen from the range that there are certainly students in Group Q who have scored less than students in Group P and vice-versa. So, options (A) and (B) are not correct.

      Standard deviation of the marks scored by students of Group Q is 5 and Group P is 25 and standard deviation of a distribution curve shows how far away the data are distributed from the mean value. So, it can be concluded that marks data of Group Q isnarrower than that of the Group P. Hence, options (C) is correct.

      Since median of normal distribution is equal to its mean, so median marks of Group P = 105. Hence, options (D) is also incorrect.

      Hence, options (C) is correct.

  • Q.35

    Two paper pulp plants P and Q use the same quality of bamboo as a raw material. The chemicals
    used in their digester are as follows:

      Plant P Plant Q
    NaOH Yes No
    Na\(_2\)S Yes No
    Na\(_2\)CO\(_3\) Yes Yes
    NaHCO\(_3\) No Yes
    Na\(_2\)CO\(_3\) No Yes

    Which one of the following statements is CORRECT?

    (A) 

    Plant P and Plant Q both use the Sulfite process

    (B) 

    Plant P and Plant Q both use the Kraft process

    (C) 

    Plant P uses Sulfite process

    (D) 

    Plant P uses Kraft process

    Marks:1 Explanation
    • Answer:

      Sulfate (Kraft) Pulping Process.

      • This is an alkaline process.
      • \(Na_2SO_4\) is added to the cooking liquor. So its common name is sulfate process.
      • The presence of sodium sulfide makes bleaching of pulp easier and the paper produced has better strength.

      Chemical reactions involved.

              1.Digestion (hydrolysis and solubilization of lignin)

                     \(R-R^{\prime} + NaOH\to R''COONa+ROH\)

                     \(R-R'+Na_2S \to Mercaptans\)

               2.Chemical recovery from black liquor

                 (a) Smelting

                     \(2NaR+air \to Na_2CO_3+CO_2\)

                      \(Na_2SO_4+2C \to Na_2S+2CO_2\)

                   (b) Causticizing

                    \(Na_2CO_3(aq)+Ca(OH)_2(s)\to 2NaOH(aq)+CaCO_3\)

                          \(CaCO_3 \to CaO+CO_2\) , \(CaO+H_2O\to Ca(OH)_2\)

      Thus Plant P uses Kraft process.

  • Q.36

    India has no elemental sulphur deposits that can be economically exploited. In India, which one of
    the following industries produces elemental sulphur as a by-product?

    (A) 

    Coal carbonisation plants

    (B) 

    Petroleum refineries

    (C) 

    Paper and pulp industries

    (D) 

    Iron and steel making plants

    Marks:1 Explanation
    • Answer:

      In India, there are no mineable elemental sulphur reserves. Pyrites were used as a substitute for sulphur in manufacture of sulphuric acid by PPCL but the operations were discontinued in 2003 for production of pyrite. The by-product recovery of sulphur is limited from petroleum refineries,in petroleum refineries, elemental sulphur is recovered from HS2 and other gaseous streams in the Sulphur Recovery Unit (SRU).

  • Q.37

    Match the industrial processes in Group-1, with the catalyst materials in Group-2.

    Group-1 Group-2
    (P) Ethylene polymerisation (I) Nickel
    (Q) Petroleum feedstock cracking (II) Vanadium pentoxide
    (R) Oxidation of \(SO_2 \space to \space SO_3\) (III) Zeolite
    (S) Hydrogenation of oil (IV) Aluminium triethyl with titanium chloride promoter

     

    (A) 

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

    (B) 

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

    (C) 

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

    (D) 

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

    Marks:1 Explanation
    • Answer:

      Ethylene polymerisation uses Zeigler Natta catalyst.ZN catalyst system usually contains two parts: a transition metal (Group IV metals, like Ti, Zr, Hf) compound and an organoaluminum compound (co-catalyst). The common examples of ZN catalyst systems include \(TiCL_4 +Al(Et)_3\).

      In catalytic cracking, the feedstock is gas oil which is vaporised and passed through a zeolite, produced as a fine powder , heated to about 700-800 K in the reactor.

      In oxidation of \(SO_2 \space to\space SO_3\) the sulfur dioxide is mixed with air at pressure slightly higher than atmospheric, and is passed through a catalyst tower, called a converter. The tower contains 3 or 4 layers of vanadium oxide catalyst.

      During hydrogenation, vegetable oils are reacted with hydrogen gas at about 60ºC. A nickel catalyst is used to speed up the reaction. The double bonds are converted to single bonds in the reaction. In this way unsaturated fats can be made into saturated fats – they are hardened.

  • Q.38

    Terms used in engineering economics have standard definitions and interpretations. Which one of
    the following statements is INCORRECT?

    (A) 

    The profitability measure ‘return on investment’ does not consider the time value of money

    (B) 

    A cost index is an index value for a given time showing the cost at that time relative to a
    certain base time

    (C) 

    The ‘six-tenths factor rule’ is used to estimate the cost of an equipment from the cost of a
    similar equipment with a different capacity

    (D) 

    Payback period is calculated based on the payback time for the sum of the fixed and the
    working capital investment

    Marks:1 Explanation
    • Answer:

      Payback period is the time in which the initial cash outflow of an investment is expected to be recovered from the cash inflows generated by the investment. 

      Payback Period =\(\frac{Initial\space Investment}{\space Cash\space Inflow\space per\space Period}\).

      Thus statement (D) is incorrect.

  • Q.39

    Standard pipes of different schedule numbers and standard tubes of different BWG numbers are
    available in the market. For a pipe / tube of a given nominal diameter, which one of the following
    statements is TRUE?

    (A) 

    Wall thickness increases with increase in both the schedule number and the BWG number

    (B) 

    Wall thickness increases with increase in the schedule number and decreases with increase in
    the BWG number

    (C) 

    Wall thickness decreases with increase in both the schedule number and the BWG number

    (D) 

    Neither the schedule number, nor the BWG number has any relation to wall thickness

    Marks:1 Explanation
    • Answer:

                               .

  • Q.40

    Two design options for a distillation system are being compared based on the total annual cost.
    Information available is as follows:

      Option P Option Q
    Installed cost of the system (Rs in lakhs) 150 120
    Cost of cooling water for condenser (Rs in lakhs/year) 6 8
    Cost of steam for reboler (Rs in lakhs/year ) 16 20

    The annual fixed charge amounts to 12% of the installed cost. Based on the above information,
    what is the total annual cost (Rs in lakhs /year) of the better option?

    (A) 

    40

    (B) 

    42.4

    (C) 

    92

    (D) 

    128

    Marks:1 Explanation
    • Answer:

      .OPTION P:

      Annual fixed charge = 12% of installed cost = 12% of 150 = Rs 18lakh

      Cost of cooling water = Rs 6lakh , Cost of steam for Reboiler = Rs 16lakh.

      Total cost = Rs (18+6+16) lakh = Rs 40lakh.

      OPTION Q:

      Annual fixed charge = 12% of installed cost = 12% of 120 = Rs 14.4 lakh

      Cost of cooling water = Rs 8 lakh , Cost of steam for Reboiler = Rs 20 lakh.

      Total cost = Rs (14.4+8+20)lakh = Rs 42.4 lakh.

      Thus better option is P for Rs 40 lakh.

  • Q.41

    A system exhibits inverse response for a unit step change in the input. Which one of the following
    statement must necessarily be satisfied?

    (A) 

    The transfer function of the system has at least one negative pole

    (B) 

    The transfer function of the system has at least one positive pole

    (C) 

    The transfer function of the system has at least one negative zero

    (D) 

    The transfer function of the system has at least one positive zero

    Marks:1 Explanation
    • Answer:

      An inverse response occurs when the transfer function has a positive zero i.e., in the right half plane.

      Physical Examples

      Heat exchanger ( Drum Boiler )

      Reboiler section of a distillation column

      Exothermic tubular reactor

  • Q.42

    Match the instruments in Group-1 with process variables in Group-2.

      Group-I   Group-II
    P Conductivity meter I Flow
    Q Turbine meter  II Pressure
    R Piezoresistivity element III Composition

     

     

    (A) 

    P-II, Q-I, R-III

    (B) 

    P-II, Q-III, R-I

    (C) 

    P-III, Q-II, R-I

    (D) 

    P-III, Q-I, R-II

    Marks:1 Explanation
    • Answer:

      A conductivity meter measures an aqueous solution's ability to carry an electrical current.They are used heavily in agriculture to measure the salinity levels of surface water and of soil samples. Their main function is to measure composition.

      Turbine flow meters measure the rate of flow in a pipe or process line via a rotor that spins as the media passes through its blades. The rotational speed is a direct function of flow rate and can be sensed by a magnetic pick-up, photoelectric cell, or tachometer. 

      Piezoresistive pressure sensors are one of the very-first products of MEMS technology. Those products are widely used in biomedical applications, automotive industry and household appliances.

      The sensing material in a piezoresistive pressure sensor is a diaphragm formed on a silicon substrate, which bends with applied pressure. A deformation occurs in the crystal lattice of the diaphragm because of that bending. This deformation causes a change in the band structure of the piezoresistors that are placed on the diaphragm, leading to a change in the resistivity of the material. This change can be an increase or a decrease according to the orientation of the resistors.

  • Q.43

    What is the order of response exhibited by a U-tube manometer?

    (A) 

    Zero order

    (B) 

    First order

    (C) 

    Second Order

    (D) 

    Third order

    Marks:1 Explanation
    • Answer:

      A second order linear instrument is a U-tube manometer for measuring pressure differences. The liquid in the U-tube tends to oscillate from side to side in the tube with a frequency determined by the weight of the liquid. The damping factor is determined by viscosity in the liquid and friction between the liquid and the sides of the tube.

  • Q.44

    Hydrogen iodide decomposes through the reaction 2HI ⇋ H\(_2\) + I\(_2\). The value of the universal gas
    constant R is 8.314 J mol\(^{-1}\)K\(^{-1}\). The activation energy for the forward reaction is 184000 J mol\(^{-1}\).
    The ratio (rounded off to the first decimal place) of the forward reaction rate at 600 K to that at 550
    K is _______.

    Marks:1 Explanation
    • Answer:

      .Arrhenious law ,  \(K\space =\space K_0e^{\frac{-E_A}{RT}}\) ;    \(ln(\frac{K}{K_O})\space=\space \frac{-E_A}{RT}\)

      \(ln(\frac{K_2}{K_O})\space-\space ln(\frac{K_1}{K_O})\space=(\frac{-E_A}{RT_2})-(\frac{-E_A}{RT_1})= \frac{E_A}{R}(\frac{1}{T_1}-\frac{1}{T_2})\)

      \(\implies ln(\frac{K_2}{K_1})= \frac{184000}{8.314}({\frac{1}{550}-\frac{1}{600}}) = 3.353\) ; \((\frac{K_2}{K_1})= e^{3.353}= 28.59\)

  • Q.45

    The variations of the concentrations (C\(_A\), C\(_R\) and C\(_S\)) for three species (A, R and S) with time, in an
    isothermal homogeneous batch reactor are shown in the figure below.

    Select the reaction scheme that correctly represents the above plot. The numbers in the reaction
    schemes shown below, represent the first order rate constants in unit of s\(^{-1}\).

     

    (A) 

    (B) 

    (C) 

    (D) 

    Marks:1 Explanation
    • Answer:

      From the given Concentration vs time graph ; We conclude that \(C_R \space vs \space t\space\&\space C_S \space vs \space t\) graph are identical which leads to the conclusion that rate constant K for reaction \(A\to \space R\space\space\space\space\space\space\&\space\space\space\space A\to S\) are identical. Also the concentration curve of species A is degrading leading to certainty of ir-reversible reaction.

      Hence , option "c" is the correct reaction scheme.

  • Q.46

    For a non-catalytic homogenous reaction \(A \to B\), the rate expressioon at 300 K is \({-r_A (mol m^{-3}s^{-1}) =} {{10C_A}\over {1 + 5C_A}}\), where \(C_A \) is the concentration os A  (in \(mol/m^{3}\)). Theroritically, the upper limit for the magnitude for the reaction rate (\(-r_A in \space mol \space m^{-3} \space s^{-1}\), rounded off to the  first decimal place) at 300 K is _________. 

    Marks:1 Explanation
    • Answer:

      Given: \(-r_A(molm^{-3}s^{-1}) =\space \frac{10C_A}{1+5C_A}\)

       When \(C_A>>>1\space \implies \space(1+5C_A)\space \approx C_A\)

      Thus , \(-r_A\space=\space \frac {10C_A}{5C_A}\space=\space 2molm^{-3}s^{-1}\).

  • Q.47

    For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature are
    nearly equal?

    (A) 

    0.33

    (B) 

    0.5

    (C) 

    1

    (D) 

    2

    Marks:1 Explanation
    • Answer:

      .For Lewis number equal to 1 for air-water system , the wet bulb temperature is equal to adiabatic saturation temperature.

       

  • Q.48

    Match the dimensionless numbers in Group-1 with the ratios in Group-2.

    Group 1 Group 2
    (P) Biot Number (I) \({Buoyancy \space Force} \over {Viscous \space Force}\)
    (Q)Schmidt Number (II) \({iternal \space thermal \space resistannce \space of \space a \space solid} \over {boundary \space layer \space thermal \space resistance}\)
    (R) Grashof Number (III) \({momentum \space diffusivity} \over {mass \space diffusivity}\)

     

    (A) 

    P-II, Q-I, R-III

    (B) 

    P-I, Q-II, R-II

    (C) 

    P-III, Q-I, R-II

    (D) 

    P-II, Q-III, R-I

    Marks:1 Explanation
    • Answer:

      Biot Number = \((\frac{\frac{L_c}{K}}{\frac{1}{h}})​​=​​ \frac{​​​​\space conductive​​​​\space resistance ​​​​\space within ​​​​\space the ​​​​\space body}{​​​​\space convection​​​​\space resistance ​​​​\space at ​​​​\space the​​​​\space surface​​​​\space of​​​​\space the​​​​\space body}\)

      Schmidt Number = \((\frac{\frac{\mu}{\rho}}{D_{AB}})​​​​\space=​​​​\space\frac{​​​​\space Momentum ​​​​\space Diffusivity}{​​​​\space Mass ​​​​\space Diffusivity}\)

      Grashof Number = \(\frac{g\beta.(T_S-T_{\infty})L^3}{\nu^2}​​​​\space=​​​​\space \frac{​​​​\space Buoyancy ​​​​\space forces}{​​​​\space Viscous ​​​​\space forces}\)

  • Q.49

    A binary liquid mixture of benzene and toluene contains 20 mol% of benzene. At 350 K the vapour
    pressures of pure benzene and pure toluene are 92 kPa and 35 kPa, respectively. The mixture
    follows Raoult’s law. The equilibrium vapour phase mole fraction (rounded off to the second
    decimal place) of benzene in contact with this liquid mixture at 350 K is _______

    Marks:1 Explanation
    • Answer:

      .Given: \(x_B=0.2\space ,\space x_A\space=(1-0.2)=0.8\space;\space P_B^{sat}=92KPa,\space P_B^{sat}=35KPa.\)

      Rault's law is : \(y_B.P^t\space=\space x_B.P_B^{sat}\space\space\space\space\space\space\space\space\space\space;\space y_T.P^t \space=\space x_T.P_T^{sat}\)

      \(\implies \space P^t\space =\space x_B.P_B^{sat}\space+\space x_T.P_T^{sat}\space=\space0.2\times92+0.8\times35\space= 46.4KPa\).

      Thus , \(y_B \space=\space\frac{x_B.P_B^{sat}}{P^t}\space=\space\frac{0.2\times92}{46.4}\space=\space0.396\)

  • Q.50

    Steam at 100\(^0\)C is condensing on a vertical steel plate. The condensate flow is laminar. The average
    Nusselt numbers are Nu\(_1\) and Nu\(_2\), when the plate temperatures are 10\(^0\)C and 55\(^0\)C, respectively.
    Assume the physical properties of the fluid and steel to remain constant within the temperature
    range of interest.

    Using Nusselt equations for film-type condensation, what is the value of the ratio \(Nu_1 \over Nu_2\)?

     

    (A) 

    0.5

    (B) 

    0.84

    (C) 

    1.19

    (D) 

    1.41

    Marks:1 Explanation
    • Answer:

      .For film type condensation , average heat transfer coefficient over the entire plate is 

      \(h=h_{avg}=\frac{1}{L}\int_0^Lh_xdx=\frac{4}{3}h_{x=L}=0.943[\frac{g\rho_l(\rho_l-\rho_v)h_{fg}K_l^3}{\mu _l(T_{sat}-T_S)L}]^{1/4}\).

      Thus \(h\space \alpha\space[\frac{1}{(T_{sat}-T_S)}]^{1/4}\)

      \( \frac{NU_2}{NU_1}\space=\space \frac{(\frac{hL}{K})_2}{(\frac{hL}{K})_1}\space=\space\frac{h_2}{h_1}\)  \(\implies \space \frac{h_2}{h_1}\space = \space [\frac{(T_{sat}-T_S)_1}{(T_{Sat}-T_S)_2}]^{1/4}\space=\space [\frac{100-10}{100-55}]^{1/4}\space=\space[\frac{90}{45}]^{1/4}\space=\space1.19\)

  • Q.51

    A composite wall is made of four different materials of construction in the fashion shown below.
    The resistance (in K/W) of each of the sections of the wall is indicated in the diagram.

    The overall resistance (in K/W, rounded off to the first decimal place) of the composite wall, in the
    direction of heat flow, is _______

    Marks:1 Explanation
    • Answer:

       

                                        B & C are in parallel , so equivalent = \(E = \frac{0.25}{1+0.25}=\frac{0.25}{1.25}=0.2\)

                                        A,E&D are in parallel , so equivalent = \(F=3+0.2+0.7=3.9 K/W\)

  • Q.52

    In a cyclone separator used for separation of solid particles from a dust laden gas, the separation
    factor is defined as the ratio of the centrifugal force to the gravitational force acting on the particle.
    S\(_r\) denotes the separation factor at a location (near the wall) that is at a radial distance r from the
    centre of the cyclone. Which one of the following statements is INCORRECT?

    (A) 

    S\(_r\) depends on mass of the particle

    (B) 

    S\(_r\) depends on the acceleration due to gravity

    (C) 

    S\(_r\) depends on tangential velocity of the particle

    (D) 

    S\(_r\) depends on the radial location ( r ) of the particle

    Marks:1 Explanation
    • Answer:

      Separation factor \(S_r​​\space=\space \frac{centrifugal\space force}{gravitational\space force}\space = \space \frac{\frac{mv^2}{r}}{mg}=\space \frac{v^2}{rg}\).

      \(\bullet\) Thus \(S_r\) does not depend on the mass of particle.

      So statement (a) is incorrect..

  • Q.53

    For a flow through a smooth pipe, the Fanning friction factor (f) is given by f  = \(mRe^{-0.2}\) in the
    turbulent flow regime, where Re is the Reynolds number and m is a constant. Water flowing
    through a section of this pipe with a velocity 1 m/s results in a frictional pressure drop of 10 kPa.
    What will be the pressure drop across this section (in kPa), when the velocity of water is 2 m/s?

    (A) 

    11.5

    (B) 

    20

    (C) 

    34.8

    (D) 

    40

    Marks:1 Explanation
    • Answer:

      Given: \(f\space=\space mRe^{-0.2}\space\implies f\space \alpha\space v^{-0.2}\).

      We know that \(\frac{\bigtriangleup P}{\rho g}\space=\space \frac{4flv^2}{2gd}\implies \space\bigtriangleup P\space\alpha (fv^2)\implies\space \bigtriangleup P \space \alpha \space v^{1.8}\).

      (\(\because \bigtriangleup P_1=10KPa,v_2=2m/s,v_1=1m/s\))

        \(\frac{\bigtriangleup P_2}{\bigtriangleup P_1}\space=\space(\frac{v_2}{v_1})^{1.8}\) \(\implies \frac{\bigtriangleup P_2}{10}\space=\space(\frac{2}{1})^{1.8}\implies \space\bigtriangleup P_2\space=\space 34.82KPa \).   

  • Q.54

    A vertical cylindrical vessel has a layer of kerosene (of density 800 kg/m\(^3\)) over a layer of water (of
    density 1000 kg/m\(^3\)). L-shaped glass tubes are connected to the column 30 cm apart. The interface
    between the two layers lies between the two points at which the L-tubes are connected. The levels
    (in cm) to which the liquids rise in the respective tubes are shown in the figure below.

     

    The distance (x in cm, rounded off to the first decimal place) of the interface from the point at
    which the lower L-tube is connected is _______

    Marks:1 Explanation
    • Answer:

      .                     

           Since  ;   \(\bigtriangleup P1=\bigtriangleup P2\)

      \(\bigtriangleup P1\space=\space\rho_k\times g \times [20+(30-x)]+\rho_w\times g\times x \space=\space \rho_k\times g\times[50-x]+\rho_w\times g\times x\)

      \(\bigtriangleup P2 = \rho_w\times g\times 42\).

      \(\implies \rho_k\times g\times [50-x]+\rho_w\times g\times x\space=\space \rho_w \times g \times 42\)

      \(\implies x\times g(\rho_w -\rho_k)\space=\space g(\rho_w\times 42 -\rho_k\times 50)\)

      \(\implies \space x\space =\space\frac{(\rho_w\times 42-\rho_k \times 50)} {(\rho_w-\rho_k} \space=\space \frac{(1000\times 42-800\times 50)}{(1000-800)}\space=\space\frac{2000}{200}\space=\space10 cm\)

  • Q.55

    Which of the following curves represents the function \(y = ln(|e^{[|sin(|x|)|]}|)\) ?
    Here, x represents the abscissa and y represents the ordinate.

    (A) 

    (B) 

    (C) 

    (D) 

    Marks:2 Explanation
    • Answer:

      The function given is  \(y = ln(|e^{[|sin(|x|)|]}|)\) .

      Now, \(e^{anything} \) is always positive, so putting exponential function does not alter anything and modulus aroun exponential can be removed.

      => \(y = ln(e^{[|sin(|x|)|]})\)

      => \(y = |sin(|x|)|\)

      When modulus is applied on x then then survr is transformed by by first erasing all the y values for negative domain of x and taking the morror image along the y-axis.

      So, if the function was  y = sin(|x|) then first plot sin(x) which will look like below:

      Now, apply the transformation that is removing all y-value on the negative x-axis and replacing it with the mirror image of the y-values of the positive x-axis along the y-axis. The new plot will look likew as below:

      We know the plot for y = sin(|x|). 

      Now, for y = |sin(|x|)|, in the curve y = sin(|x|), take the mirror image of the negative y-values along the x-axis and then remove all the negative y-values. The new plot will look like below:

       Hence, option (C) is correct.

  • Q.56

    The binary operation □ is defined as a □ b = ab+(a+b), where a and b are any two real numbers.
    The value of the identity element of this operation, defined as the number x such that a □ x = a, for
    any a, is _____.

    (A) 

    0

    (B) 

    1

    (C) 

    2

    (D) 

    10

    Marks:2 Explanation
    • Answer:

      Given: \(a \space \square \space b = ab+(a+b)\)

      Now, \(a \space \square \space x = ax+(a+x)\), also it is given that \(a \space \square \space b = a\)

      \(=> ax+(a+x) = a\)

      Now, putting all the options successively in L.H.S. and equating the result with R.H.S. we get that only for 0 both L.H.S. and R.H.S. are equal.

      Hence, option (A) is correct.

  • Q.57

    Find the missing sequence in the letter series.
    B, FH, LNP, _ _ _ _.

    (A) 

    SUWY

    (B) 

    TUVW

    (C) 

    TVXZ

    (D) 

    TWXZ

    Marks:2 Explanation
    • Answer:

      The first letter of a sequence and last letter of the previous sequence in the series has 3 letters in between them like 'B' and 'F' has 'C', "D' and 'E'. In a sequence letters appearing have a difference of on letter i.e. 'F' and 'H' have 'G' missing between them. Also the number of letters in the successive sequence is in arithmetic progression with common difference as 1 i.e. 1,2,3 so new sequence will have 4 letters.

      Using the pattern deduced the next sequence's last letter and previous sequence's first letter 'P' would have 3 letters between them. Now, after 'P' the four letters are 'Q', 'R', 'S' and 'T'. So, the first letter of the new sequence will be 'T'. 'TUVWXYZ' here all the underlined letters are to be omitted.

      Hence, the sequence is TVXZ' i.e. option (C) is correct.

  • Q.58

    A smart city integrates all modes of transport, uses clean energy and promotes sustainable use of
    resources. It also uses technology to ensure safety and security of the city, something which critics
    argue, will lead to a surveillance state.
    Which of the following can be logically inferred from the above paragraph?

    (i) All smart cities encourage the formation of surveillance states.
    (ii) Surveillance is an integral part of a smart city.
    (iii) Sustainability and surveillance go hand in hand in a smart city.
    (iv) There is a perception that smart cities promote surveillance.

    (A) 

    (i) and (iv) only

    (B) 

    (ii) and (iii) only

    (C) 

    (iv) only

    (D) 

    (i) only

    Marks:2 Explanation
    • Answer:

      The word Surveillance states is used by the critics as suggected by the writer of the given text and is not the central idea suggested by the writer of this paragraph. So, statement (i) cannot be concluded from the given paragraph. 

      It can be seen in the paragraph that the idea of surveillance is introduced with the word 'also' so surveillance is not suggested a primarily important aspect of a smart city. So, statement (ii) is also cannot be logically concluded.

      If two things go hand in hand then it means that these two things are complementary to each other. There isno mention of sustainability and surveillance going hand in hand or being complementary to each other. So, statement (iii) cannot be logically concluded.

      There is a mention of critics's perception which states that smart cities will lead to surveillance states. It can be concured that critics have a perception of cities promoting surveillance. So, statement (iv) can be logically concluded.

      Hence, option (C) is correct.

  • Q.59

    A set of simultaneous linear algebraic equations is represented in a matrix form as shown below.

    \( \left[ {\begin{array}{cc} 0 & 0 & 0 & 4 & 13\\ 2 & 5 & 5 & 2 & 10\\ 0 & 0 & 2 & 5 & 3\\ 0 & 0 & 0 & 4 & 5\\ 2 & 3 & 2 & 1 & 5\\ \end{array} } \right] \)\( \left[ {\begin{array}{cc} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\\ \end{array} } \right] \) = \( \left[ {\begin{array}{cc} 46 \\ 161 \\ 61 \\ 30 \\ 81\\ \end{array} } \right] \)

    The value (rounded off to the nearest integer) of \(x_3\) is _______

    Marks:2 Explanation
    • Answer:

      \(4x_4+13x_5=46\)    _____ (1) By row 1.

      \(2x_3+5x_4+3x_3=61\) _____ (2) By row 3.

      \(4x_4+5x_5=30\)      _______ (3) By row 4.

      Solving equation (1) & (3) we get  \(x_4=5,x_5=2\).

      Putting these values in equation (2) we get \(x_3=15\).

  • Q.60

    What is the solution for the second order differential equation \({{d^2y} \over {dx^2}} + y = 0\), with the initial conditions \(y\mid {_{x =0}} = 5 \space and \space {{dy} \over {dx}}\mid {_{x = 0}} = 10\) ?

    (A) 

    y = 5 + 10sin x

    (B) 

    y = 5cos x - 5sin x

    (C) 

    y = 5cos x - 10x

    (D) 

    y = 5cos x + 10sin x

    Marks:2 Explanation
    • Answer:

      A.E is \(D^2+1=0\space\implies D=0\pm i\)

      \(y=e^{0x}[C_1cosx+C_2sinx]\) 

      \(\implies\space5=C_1\space\space\space\space\space \because y|_{x=0}=5\)

      \(\frac{dy}{dx}=-C_1sinx+C_2cosx\)

      \(\implies \space 10=C_2\space\space\space \because \frac{dy}{dx}|_{x=0}=10\)

      Hence \(y=5cosx+10sinx\).

  • Q.61

    Which one of the following is an iterative technique for solving a system of simultaneous linear algebraic equations?

    (A) 

    Gauss Elimination

    (B) 

    Gauss-Jordan

    (C) 

    Gauss-Seidel

    (D) 

    LU Decomposition

    Marks:1 Explanation
    • Answer:

      .Gauss-Seidel

  • Q.62

    The Laplace transform of \(e^{at}sin(bt)\) is

    (A) 

    \(b \over (s - a)^2 + \space b^2 \)

    (B) 

    \((s-a) \over (s-a) ^2 \space+\space b^2\)

    (C) 

    \((s-a) \over (s-a)^2 \space - \space b^2 \)

    (D) 

    \(b \over (s-a)^2 \space - \space b^2\)

    Marks:1 Explanation
    • Answer:

      By shifting property \(L[{e^{at}f(t)}]\space=\bar{f}(s-a)\)

      Here f(t) = sin(bt) ; L[sin(bt)] = \(\frac {b}{s^2+b^2}\)

      Thus \(L[e^{at}f(t)]\space =\space \frac{b}{(s-a)^2+b^2}\) .

  • Q.63

    What are the modulus (r ) and argument (\(\theta\)) of the complex number 3 + 4i ?

    (A) 

    \(r= \sqrt {7}, \space\space\space\space\space \theta = tan^{-1} {({4 \over 3})} \)

    (B) 

    \(r= \sqrt {7}, \space\space\space\space\space \theta = tan^{-1} {({3 \over 4})} \)

    (C) 

    \(r= 5, \space\space\space\space\space \theta = tan^{-1} {({3 \over 4})} \)

    (D) 

    \(r=5, \space\space\space\space\space \theta = tan^{-1} {({4 \over 3})} \)

    Marks:1 Explanation
    • Answer:

      Modulus(r) of a complex number Z = x+iy is \(r\space= \space \sqrt{x^2+y^2}\)

      Thus  r = \(\sqrt{3^2+4^2}\space= \sqrt{25}\space=5\).

      Argument \(\theta\space=\space \tan^{-1}(\frac{y}{x})\space = \tan^{-1}(\frac{4}{3})\).

  • Q.64

    A liquid mixture of ethanol and water is flowing as inlet stream P into a stream splitter. It is split
    into two streams, Q and R, as shown in the figure below.

    The flowrate of P, containing 30 mass% of ethanol, is 100 kg/h. What is the least number of
    additional specification(s) required to determine the mass flowrates and compositions (mass%) of
    the two exit streams?

    (A) 

    0

    (B) 

    1

    (C) 

    2

    (D) 

    3

    Marks:1 Explanation
    • Answer:

      .Since the stream is splitting , composition (mass %) remains the same .

         Overall mass balance :  P = Q  +  R  \(\implies \space 100\space =\space Q+R\)  (1)

        Ethanol balance :    \(P\times30\%\space=\space Q\times 30\%\space+\space R\times30\% \space\implies 100\times 30\%\space = \space Q\times 30\%+R \times30\%\).

      Thus we need either Q or R to determine everything.

      So only one additional requirement is needed.

  • Q.65

    The partial molar enthalpy (in kJ/mol) of species 1 in a binary mixture is given by
    \(\bar h_1 = 2\space - \space 60 x_2^2 \space + \space 100x_1x_2^2\), where \(x_1\) and \(x_2\) are the mole fractions of species 1 and 2,
    respectively. The partial molar enthalpy (in kJ/mol, rounded off to the first decimal place) of
    species 1 at infinite dilution is _______

    Marks:1 Explanation
    • Answer:

      At infinite dilution ,for species 1 , \(x_1\to 0 \space,\space x_2\to 1\).

      Thus, \((\overline h_1^\infty )_ {(x_{1\to 0},x_{2 \to 1})}\space=\space2-60x_2^2+100x_1x_2^2\) ., \(\implies(\overline h_1^\infty )_ {(x_{1\to 0},x_{2 \to 1})}\space=\space2-60\times1^2+100\times 0\times 1^2\space=\space-58(KJ/mol)\)