MY-ASSIGNMENTEXPERT™可以为您提供ntnu.edu TEP4120 Thermodynamics热力学课程的代写代考和辅导服务！
这是挪威科技大学 热力学课程的代写成功案例。
TEP4120课程简介
Course content
Concepts and definitions; the thermodynamic system, properties, phase equilibrium of pure substances, equations of state for gases, tables of thermodynamic properties, work and heat. First law of thermodynamics; thermodynamic cycles, change of state, internal energy, enthalpy, specific heat; open systems, steady-state and transient processes. Second law of thermodynamics; reversible and irreversible processes, the Carnot cycle, the thermodynamic temperature scale, entropy, the entropy production concept. Thermodynamic power cycles, refrigeration cycles, the Otto cycle and the Diesel cycle, the gas-turbine process. Introduction to exergy analysis.
Prerequisites
LKnowledge: The course provides the student with knowledge about: – Conservation laws for mass and energy (including the 1st law of Thermodynamics). – Forms of energy such as work (power) and heat, internal energy and enthalpy. – Entropy and the 2nd law of thermodynamics. – Reasons for thermodynamic losses in the form of irreversibilities. – The quality of different forms of energy measured as the ability to produce work. – Destruction of energy quality in processes. – Ideal gas model, its assumptions, applications and limitations. – Different cyclic processes such as Carnot, Rankine, Otto, Diesel and Brayton. The course gives the student insight about: – Operation of steam and gas based power stations, internal combustion engines, heat pumps and refrigeration cycles. – The main components of heat & power processes, such as steam and gas turbines, compressors, pumps, fans, heat exchangers and valves. – The ability of fluids to change phase (solid, liquid and gas).
TEP4120 Thermodynamics HELP（EXAM HELP， ONLINE TUTOR）
A manometer of mercury indicates a reading of $1 \mathrm{~m}$ when the fluid pressure, water, $\left(P_1\right)$ at point A is $0.065 \mathrm{~m} \mathrm{Hg}$ vacuum. Determine the fluid pressure, oil, $\left(P_2\right)$ at point B.
$P_1$
$P_2$
0.065
0.035
$0.065 \mathrm{~m} \mathrm{Hg}$
Given:
$$
\begin{aligned}
& h_{\text {water }}=(0.35+1) \mathrm{m}=1.35 \mathrm{~m} \
& h_{\text {oil }}=0.35 \mathrm{~m} \
& h_{H g}=1.00 \mathrm{~m} \
& \gamma_{H g}=133 \mathrm{kN} / \mathrm{m}^3 \
& \gamma_{\text {oil }}=8.95 \mathrm{kN} / \mathrm{m}^3 \
& \gamma_{\text {water }}=9.80 \mathrm{kN} / \mathrm{m}^3
\end{aligned}
$$
Find:
The fluid pressure $\left(P_2\right)$ at point B.
Solution:
$$
\begin{aligned}
& P_2+\gamma_{\text {oil }} h_{\text {oil }}-\gamma_{\text {water }} h_{\text {water }}=P_1 \
& P_1=-\gamma_{H g}(0.065 \mathrm{~m}) \
& P_2=-\gamma_{H g}(0.065 m)-\gamma_{\text {oil } 1.35 m}+\gamma_{H g}(1.00 m)+\gamma_{\text {water }} 0.35 m \
& P_2=115.706 \mathrm{kPa}
\end{aligned}
$$
Water moves inside a pipe at $75 \mathrm{~m} / \mathrm{s}$ with a flow rate of $145 \mathrm{~kg} /$ in in order to generate power by moving a propeller located inside the pipe.
$$
\begin{aligned}
& v=75 \
& \dot{m}=145
\end{aligned}
$$
Given:
Velocity $(\mathrm{v})=75 \mathrm{~m} / \mathrm{s}$
Mass flow rate $(\dot{m})=145 \mathrm{~kg} / \mathrm{s}$
Find:
Determine the power generation at the propeller.
Solution:
$$
\begin{aligned}
& e_{\text {mech }}=k e=\frac{v^2}{2}=\frac{75^2\left(\frac{m}{s}\right)^2}{2}\left(\frac{1\left(\frac{k J}{k g}\right)}{1000\left(\frac{\mathrm{km}}{\mathrm{n}}\right)^2}\right)=2.8125 \mathrm{~kJ} / \mathrm{kg} \
& \dot{W}{\text {max }}=\dot{E}{\text {mech }}=\dot{m} e_{\text {mech }} \
& \dot{W}_{\text {max }}=(120 \mathrm{~kg} / \mathrm{s})(2.8 \mathrm{~kJ} / \mathrm{kg})=406 \mathrm{~kW} \therefore 1 \mathrm{~kJ} / \mathrm{s}=1 \mathrm{~kW}
\end{aligned}
$$
A rigid container has volume of $2 \mathrm{~m}^3$, and holds steam at $260^{\circ} \mathrm{C} .1 / 4$ of the volume is in liquid point and the remaining at vapor form. Determine the pressure of the steam, and quality of the saturated mixture, and density of the mixture.
Given:
Volume $(\mathrm{V})=2 m^3$
Temperature $(\mathrm{T})=260^{\circ} \mathrm{C}$
Find:
- The pressure of the steam.
- The quality of the saturated mixture.
- The density of the mixture.
Solution
From Table A-4, the properties of Temperature(T) at $220^{\circ} \mathrm{C}$
$$
\begin{aligned}
& v_f=0.001276 \mathrm{~m}^3 / \mathrm{kg} \text { and } v_g=0.042175 \mathrm{~m}^3 / \mathrm{kg} \
& P=T_{\text {sat } @ 220^{\circ} \mathrm{C}}=4692.3 \mathrm{kPa}
\end{aligned}
$$
The total mass and the quality are determined as
$$
\begin{aligned}
& m_f=\frac{V_f}{v_f}=\frac{0.25 x 2 \mathrm{~m}^3}{0.001276 \mathrm{~m}^3 / \mathrm{kg}}=391.84 \mathrm{~kg} \
& m_g=\frac{V_g}{v_g}=\frac{0.75 x 2 \mathrm{~m}^3}{0.042175 \mathrm{~m}^3 / \mathrm{kg}}=35.56609 \mathrm{~kg} \
& m_t=m_f+m_g=391.84+35.6=427.44 \mathrm{~kg} \
& m_t=\frac{m_g}{m_t}=\frac{35.6}{427.44}=0.083286
\end{aligned}
$$
The density is determined from
$$
\begin{aligned}
& v=v_f+x\left(v_g-v_f\right)=0.001276+(0.083286)(0.040899)=0.00468231411 \mathrm{~m}^3 / \mathrm{kg} \
& \rho=\frac{1}{v}=\frac{1}{0.00468231411}=213.569 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
$$
A movable cover is located at the top of an opened metal box. The metal box contains steam at $350^{\circ} \mathrm{C}$, and pressure of $2 \mathrm{MPa}$. Currently the steam is about to be cooled. Determine the compression work $W_b$ for the steam and pressure and its final temperature where the mass of the movable box and metal box is $4 \mathrm{~kg}$.
$$
\begin{aligned}
& T=350^{\circ} \mathrm{C} \
& P=2 \
& m=4
\end{aligned}
$$
Final state is cooled to $64.28 \%$
Given:
$$
\begin{aligned}
& T_1=350^{\circ} \mathrm{C} \
& P_1=2 M P a \
& T_2=225^{\circ} \mathrm{C} \
& P_2=2 M P a \
& m=4 k g
\end{aligned}
$$
Find:
Determine the compression work $W_b$ for the steam and pressure and its final temperature.
Solution:
Using table A-6
At First State
$$
T_1=350^{\circ} \mathrm{C}
$$
$$
P_1=2 M P a
$$
$$
v_{f, 1}=0.10381 \mathrm{~m}^3 / \mathrm{kg}
$$
At Final State
$$
\begin{aligned}
& T_2=225^{\circ} \mathrm{C} \
& P_2=2 M P a \
& v_{f, 2}=0.10381 \mathrm{~m}^3 / \mathrm{kg}
\end{aligned}
$$
a) the compression work
$$
\begin{aligned}
& W_b=m P\left(v_1-V_2\right) \
& W_b=4(2000)(0.13860-0.10381)=278.32 k J
\end{aligned}
$$
b) the volume of the metal box at the final state is $64.28 \%$ from its initial volume, so the work will be as
$$
\begin{aligned}
& W_b=m P\left(v_1-0.6428 v_1\right) \
& W_b=4(2000)(0.13860-(0.6428 * 0.13860))=396.06336 \
& P_2=2 * 0.6428=1.2856 \cong 1300 \mathrm{kPa} \
& v_2=0.6428 * 0.13860=0.089092 \mathrm{~m}^3 / \mathrm{kg} \
& T_2=191.60^{\circ} \mathrm{C}
\end{aligned}
$$
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