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INPhO 2019, Теория — Задача 6. A Diesel cycle

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Всички материали от INPhO 2019

Indian National Physics Olympiad – 2019 · 3 февруари 2019 г. · 14 т.

Внимание: Бележка към темата

The complete retained question booklet includes summary answer boxes, eighteen blank detailed-answer pages and three blank graph sheets. All matching archived HBCSE key content is retained, including both full handwritten graph sheets and accepted alternatives/ranges. Printed exponent, sign, factor, notation and unit inconsistencies remain literal with labelled source notes.

Native cover and instructions

Indian National Physics Olympiad – 2019

Date: 03 February 2019. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,9\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ . Extra sheets attached: [blank box]. Centre (e.g. Kota): [blank line].

(Do not write below this line)

This booklet consists of 29 pages (excluding this page) and total of 7 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations. Strike out any rough work that you do not want to be considered for evaluation.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Constant	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	9
2	11
3	12
4	7
5	9
6	14
7	13
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key cover

Solutions of Indian National Physics Olympiad – 2019

The native key cover repeats the nine instructions, twelve constant rows, seven-question score table totalling 75 marks, roll-number form and HBCSE address from the question cover. It prints “20 pages (excluding this page)”; the retained key contains twenty-one PDF pages, including two appended handwritten graph sheets.

Условие

Consider $n$ moles of a monoatomic non-ideal (realistic) gas. Its equation of state may be described by the van der Waal’s equation

\left(P+\frac{an^2}{V^2}\right)\left(\frac Vn-b\right)=RT

where $a$ and $b$ are positive constants and other symbols have their usual meanings. The internal energy change of a realistic gas can be given by

dU=C_VdT+\left\{T\left(\frac{dP}{dT}\right)_V-P\right\}dV

As indicated in the above expression, the derivative of pressure is taken at constant volume. We take one mole of the gas ( $n=1$ ) through a Diesel cycle (ABCDA) as shown in the following $P$ - $V$ diagram (diagram is not to scale). During the whole cycle assume that the molar heat capacity at constant volume ( $C_V$ ) remains constant at $3R/2$ . Path AB and CD are reversible adiabats.

Whole native Diesel ABCDA P-V diagram, all arrows, PB=PC, VB,VC,VA=VD and dashed lines.

a) Obtain the temperature at B ( $T_B$ ) in terms of temperature at A ( $T_A$ ), $V_A$ , $V_B$ and constants only.

Summary answer box: $T_B=$ [blank]. [2,5 т.]

b) Let temperature at A to be $T_A=100.00\ \mathrm K$ , $V_A=8.00\ \mathrm l$ , $V_B=1.00\ \mathrm l$ , $V_C=2.00\ \mathrm l$ , $a=1.355\ \mathrm{l^2\cdot atm/mol^2}$ , and $b=0.0313\ \mathrm{l/mol}$ . Calculate the highest temperature reached during the whole cycle.

Summary answer box: Highest temperature = [blank]. [1,5 т.]

c) Calculate the efficiency $\eta$ of the cycle.

Summary answer box: Value of $\eta=$ [blank]. [3 т.]

d) Draw the corresponding $T$ - $S$ (entropy) and $V$ - $T$ diagram for the Diesel cycle. Wherever possible, mention the numerical values of $T$ , $V$ , and $S$ on the diagrams.

Whole original blank T-S axes and answer frame.

Whole original blank V-T axes and answer frame.

Detailed answers can be found on page numbers: [blank]. [7 т.]

Решение

(a)

dU=C_VdT+\left\{\frac{RT}{V-b}-P\right\}dV

AB and CD are reversible adiabats, hence entropy change during these processes are zero.

\Delta S_{\mathrm{AB}}=\Delta S_{\mathrm{CD}}=\int_A^B\frac{dQ}T=\int_C^D\frac{dQ}T=0\qquad (6.1)

\int_A^B\frac{dQ}T=\int\frac{C_V}T\,dT+\int\frac R{V-b}\,dV=0\qquad (6.2)

C_V\ln\frac{T_B}{T_A}=-R\ln\left(\frac{V_B-b}{V_A-b}\right)\qquad (6.3)

T_B=T_A\left(\frac{V_B-b}{V_A-b}\right)^{-R/C_V}=T_A\left(\frac{V_B-b}{V_A-b}\right)^{-2/3}\qquad (6.4)

(b)

Highest temperature during cycle is at $T_C$ .

T_B=407.5\ \mathrm K\qquad (6.5)

which gives, from van der Waal equation

P_B=33.18\ \mathrm{atm}=P_C\qquad (6.6)

\Rightarrow T_C=803.76\ \mathrm K\approx804\ \mathrm K\qquad (6.7)

Different value of $T_C$ (within a range) obtained due to reasonable roundoff in previous step(s) will be credited.

(c)

dQ_{\mathrm{in}}=dQ_{\mathrm{BC}}=dU+P_BdV\qquad (6.8)

=C_VdT+\frac{RT}{V-b}\,dV\qquad (6.9)

=C_VdT+\left(P+\frac a{V^2}\right)dV\qquad (6.10)

Q_{\mathrm{in}}=C_V(T_C-T_B)+P_B(V_C-V_B)-\left.\frac aV\right|_{V_B}^{V_C}\qquad (6.11)

Q_{\mathrm{out}}=Q_{\mathrm{DA}}=C_V(T_A-T_D)\qquad (6.12)

\eta=1-\frac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}}=67.7\%\approx68\%\qquad (6.13)

Different value of $\eta$ (within a range) obtained due to reasonable roundoff in previous step(s) will be credited.

(d)

\Delta S_{\mathrm{BC}}=\int\frac{dQ}{dT}=\int\frac{C_VdT}T+\int\frac R{V-b}\,dV\qquad (6.14)

=C_V\ln\frac{T_C}{T_B}+R\ln\frac{V_C-b}{V_B-b}\qquad (6.15)

=1.73R=\Delta S_{\mathrm{DA}}\qquad (6.16)

Whole original T-S plot, A/B/C/D arrows, temperatures100/408/316/803, entropy axis and1.73R span.

Whole original V-T plot, A/B/C/D arrows, dashed coordinate lines, volumes1/2/8 and temperatures100/316/408/803.

Source note: Eq. (6.12) prints $C_V(T_A-T_D)$ for $Q_{\mathrm{out}}$ , while Eq. (6.13) gives a positive 67.7% efficiency with $1-Q_{\mathrm{out}}/Q_{\mathrm{in}}$ . Eq. (6.14) prints $dQ/dT$ in its first integral rather than the $dQ/T$ used earlier. Eq. (6.16) prints positive $\Delta S_{\mathrm{DA}}=\Delta S_{\mathrm{BC}}$ , whereas the native T-S plot shows DA returning toward smaller entropy. Native equations, values, arrows and graph labels remain literal. Both key plots label $T_C$ as803 K, whereas Eq. (6.7) rounds803.76 K to804 K.

Оригинал в Архива: INPhO2019-Question.pdf · решение от архива: INPhO2019-Solution-20190219.pdf