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# Tutor profile: Nishita G.

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Nishita G.
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## Questions

### Subject:Physical Chemistry

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Question:

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled ? (ii) reduced to half ?

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Nishita G.

Let us consider the following reaction: $$aA + bB \Longrightarrow cC + dD$$ Here a moles of compound A and b moles of compound B react to form c moles of compound C and d moles of compound D The rate of any chemical reaction depends on the concentration of the reactants in it. Thus mathematically we can write, Rate = $$k[A]^x [B]^y$$ Where k = rate constant of the reaction A = molar concentration of compound A B = molar concentration of compound B x = order of reaction with respect to A y = order of reaction with respect to B The overall order of the reaction can be given by x+y. Now according to the question, the reaction is 2nd order with respect to one of the reactants. Let that reactant be compound A. Thus x = 2 and the rate of the reaction can now be written as: Rate = $$k[A]^2 [B]^y$$ Let us call this as the original rate of the reaction. i.) If we now double the concentration of reactant A, the new rate of the reaction will be New Rate = $$k [2A]^2[B]^y = 4 k [A]^2[B]$$ = 4 (Rate) Thus the new rate is 4 times the original rate of reaction. Hence the rate of reaction gets quadrupled with the doubling of concentration of reactant A. ii.) If we now halved the concentration of reactant A, the new rate of the reaction will be New Rate = $$k [A/2]^2[B]^y = \frac{k [A]2[B]}{4} = \frac{(Rate)}{4}$$ Thus the new rate is 1/4 times the original rate of reaction. Hence the rate of reaction gets reduced to ¼ of its original rate when the concentration of reactant A is halved.

### Subject:Mechanical Engineering

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Question:

A steam generator (Rankine cycle) comprises a boiler, a super heater, an economizer and an air preheater. The feedwater enters the economizer at 140$$^\circ$$C and leaves as saturated liquid. Air is preheated from a temperature of 25$$^\circ$$C to 250$$^\circ$$C. Steam leaves the boiler drum (evaporator) at 60 bar, 0.98 dry and leaves the super heater at 450$$^\circ$$C. When using coal with a calorific value of 25.2 MJ/kg, the rate of evaporation is 8.5 kg steam per kg coal and the air fuel ratio is 15:1 by mass. Neglecting heat losses and pressure drops, estimate the heat transfer per kg fuel in each component and the efficiency of the steam generator. What are the percentages of the total heat absorption taking place in the economizer, boiler drum and the superheater? Assume $$C_p$$ of air and water as 1.005 and 4.2 kJ/kg K respectively.

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Nishita G.

(This problem requires the knowledge of knowing how to read and extract data from phase diagrams and knowing how to use Steam tables) The problem requires us to find: $$\bullet$$ Efficiency of steam generator $$\bullet$$ % of total heat absorption taking place in each component $$\bullet$$ Heat transfer per kg of fuel in each component This question deals with a steam generator which is a part of a Rankine Cycle. Water is the working fluid in this problem. You can search for diagrams of a standard flow diagram of a steam plant based on Rankine cycle online. It consists of 4 important parts, Pump, Steam generator, Turbine and Condenser in that order. This question asks us to calculate the efficiency of only the steam generator units which are responsible for converting the liquid water into supersaturated steam. Let us understand the function of each of the 3 components involved in the steam generator. The Economiser (1st subcomponent) converts water which is in compressed liquid form to saturated liquid form by heating the working fluid to its Saturated Liquid temperature. An Evaporator (2nd subcomponent) converts this saturated liquid into saturated vapour. During this process the working fluid only undergoes a phase change at its latent heat of vapourization. This means the entire energy provided by the evaporator is used to change the phase of the working fluid. The Superheater (3rd subcomponent) converts this saturated vapour into superheated vapour. Since the steam generator unit changes only the temperature and phase of the working fluid, pressure is constant throughout all the components. But how does the working fluid get heated in these components? The working fluid is heated by burning coal in a furnace. Flue gases from the furnace are provided to these components, where the working fluid extracts energy from it. We can draw a phase diagram based on the data given above. At a pressure of 60 bar, we can get the following data by using the steam tables. Alternatively we can use an online steam table calculator like the one I have used here. $$\bullet$$ Saturated Steam Temperature= 276.681$$^\circ$$C $$\bullet$$ Latent Heat of Steam= 1564.05 KJ/Kg $$\bullet$$ Specific Enthalpy of Saturated Steam= 2783.44 KJ/Kg $$\bullet$$ Specific Enthalpy of Saturated Water= 1219.39 KJ/Kg $$\bullet$$ Specific Volume of Saturated Steam= 0.0318628 m$$^3$$/Kg $$\bullet$$ Specific Volume of Saturated Water=0.00132258 m$$^3$$/Kg When the steam enters the economiser it will need to have $$h_1$$ amount of energy. We know that, $$\Delta h= C_p \times \Delta T$$ The value of $$C_p$$ for water is 4.2 kJ/kg$$^\circ$$C Thus, $$h_1$$ = (140) × 4.2 = 588 kJ/kg. $$h_1$$ can be obtained via the steam table given above. $$h_2$$ represents the specific enthalpy of saturated water, i.e. the energy water has at its boiling point, which occurs inside the boiler. Thus, $$h_2$$ = 1219.39 kJ/kg. $$h_{fg}$$ is the latent heat of vaporization. It is the energy water has between points 2 and 3 when it is undergoing phase change. This value can be obtained directly from the steam table. Thus, $$h_{fg}$$ = 1564.05 kJ/kg. Since steam leaves the superheater at 450$$^\circ$$C, we can again use $$\Delta h= C_p \times \Delta T$$ to calculate $$h_4.$$ Thus $$h_4$$ = 3301.8 kJ/kg. Efficiency of any heat transferring device is given as, Efficiency($$\eta$$) = (Output) ÷ (Input) Output will be energy transferred to the working fluid. Thus Output = $$h_4 - h_1$$ Input will be the energy generated by burning coal. Calorific Value of coal is given as 25.2 MJ/kg. This means on burning 1 kg of coal, 25.2 MJ of energy will be released. According to the question, 8.5 kg of water is evaporated on burning 1 kg coal. Thus on passing 1 kg of working fluid through the steam generator, 1/(8.5) kg of coal will be utilized. Therefore, Input = $$\frac{1}{8.5}$$ x (Calorific Value of coal) Thus, $$\eta _{entire\, boiler} = \frac{w_s(h_4 - h_1)}{w_f\ \times \ C.V} = \frac{8.5\ \times \ 2713.8}{25.2 \ \times \ 1000} = 0.9154\ or\ 91.54 \%$$ This completes part 1 of the question. Next, according to the problem, steam leaves the evaporator 0.98 dry. This value stands for the quality factor of steam. This means that the working fluid exits the evaporator as 98% steam and 2% liquid. We can directly calculate $$h_3$$ from the value of quality with the help of the definition of steam. $$0.98= \frac{h_3-h_2}{h_{fg}}$$, thus $$h_3$$=1213.35 + 0.98 $$\times$$ 1517 = 2752.93 KJ/Kg Now we can find heat transfer in each component, by simply multiplying mass of the working fluid flowing in the component and the enthalpy drop across it. Heat transfer in the economiser = $$\frac{w_s(h_2-h_1)}{w_f} = 8.5 \times 625.35 \times 10^{-3}$$ = 5.3155 MJ/Kg Heat transfer in the Boiler/Evaporator = $$\frac{w_s(h_3-h_2)}{w_f} = 8.5 \times 1539.58 \times 10^{-3}$$ = 13.086 MJ/Kg Heat transfer in the Superheater = $$\frac{w_s(h_4-h_3)}{w_f} = 8.5 \times 548.87 \times 10^{-3}$$ = 4.665 MJ/Kg Thus, % of total heat absorbed in the economiser = $$[(h_2 - h_1) \div (h_4 - h_1)] \times 100 = 23.04 \%$$ % of total heat absorbed in the evaporator = $$[(h_3 - h_2) \div (h_4 - h_1)] \times 100 = 56.73 \%$$ % of total heat absorbed in the superheater = $$[(h_4 - h_3) \div (h_4 - h_1)] \times 100 = 20.23 \%$$ This completes both part 2 and 3 of the question as we have successfully calculated the heat transfer and % heat absorbed in each component. Bonus! We can also find the Heat transfer in the air preheater by using the formula $$\Delta h= C_p \times \Delta T = \frac{w_a c_{p_a}(t_2-t_1)}{w_f} = 15 \times 1.005 \times (250-25) \times 10^{-3}$$ = 3.392 MJ/Kg

### Subject:Physics

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Question:

A hockey puck is traveling at 12.5 m/s on a floor. The coefficient of friction is 0.15. So how far will it travel before coming to a stop ?

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Nishita G.

The work-energy principle states that the work done on an object is equal to the change in the kinetic energy of that object. In this example, the work done on the puck is equal to the change in the kinetic energy of the puck. As a reminder, the formula for kinetic energy is: $$½mv^2$$ The change in kinetic energy of the puck is: Final kinetic energy minus initial kinetic energy Initial Velocity= 12.5 m/s Final Velocity= 0 Let the mass of the puck be $$m$$. Using the formula for kinetic energy, this would be: ½(mass)(final velocity)$$^2$$ - ½(mass)(initial velocity)$$^2$$ = ½$$m$$[(0 m/sec)$$^2$$ - (12.5 m/sec)$$^2$$] = $$0 - 78.125m$$ = $$-78.125m$$ kg*(m/sec)$$^2$$ = $$-78.125m$$ J (where $$m$$ is the mass of the puck.) Now this is equal to the work done on the body by the external frictional force. As a reminder, friction force= $$\mu N$$ (where $$\mu$$ is coefficient of friction and $$N$$ normal reaction on the body) Here the normal reaction on the body will be equal to the weight of the body acting downwards as there is no other force acting in the vertical direction. Normal Reaction=$$mg$$ (where $$m$$ is the mass of the puck and $$g$$ is acceleration due to gravity i.e 9.8 m/s$$^2$$ ) Let the distance traveled by the puck before stopping be $$s$$. As we know work done = $$force \cdot displacement$$ Since displacement and force are in opposite directions with with respect to each other. Work done= $$-fs$$= $$-\mu mgs$$ Now from our previous step we calculated work done on the body= $$-78.125m$$ $$-\mu mgs = -78.125m$$ $$\mu gs = 78.125$$ $$s = \frac{78.125}{\mu g}$$ $$s = 53.15$$ m

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