Numerade

Rohit S.

Books Assigned

Fundamentals of Thermodynamics

Richard E.… 6th Edition

1,400 solutions

Fundamentals of Heat and Mass…

Theodore L.… 7th Edition

1,269 solutions

A Heat Transfer Textbook

John H.… 3rd Edition

1,457 solutions

Viewed Questions

$A Brayton-cycle inlet is at $300 \mathrm{K}$ and $100 \mathrm{kPa}$ and the combustion adds $670 \mathrm{kJ} / \mathrm{kg}$. The maximum temperature is $1200 \mathrm{K}$ due to material considerations. What is the maximum allowed compression ratio? For this ratio, calculate the net work and cycle efficiency assuming variable specific heat for the air (Table A.7).$

A Brayton-cycle inlet is at $300 \mathrm{K}$ and $100 \mathrm{kPa}$ and the combustion adds $670 \mathrm{kJ} / \mathrm{kg}$. The maximum temperature is $1200 \mathrm{K}$ due to material considerations. What is the maximum allowed compression ratio? For this ratio, calculate the net work and cycle efficiency assuming variable specific heat for the air (Table A.7).

A process fluid having a specific heat of $3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ and flowing at $2 \mathrm{~kg} / \mathrm{s}$ is to be cooled from $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ with chilled water, which is supplied at a temperature of $15^{\circ} \mathrm{C}$ and a flow rate of $2.5 \mathrm{~kg} / \mathrm{s}$. Assuming an overall heat transfer coefficient of $2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, calculate the required heat transfer areas for the following exchanger configurations: (a) parallel flow, (b) counterflow, (c) shell-and-tube, one shell pass and two tube passes, and (d) cross-flow, single pass, both fluids unmixed. Compare the results of your analysis. Your work can be reduced by using IHT.

A fin of triangular axial section (cf. Fig. 4.12) $0.1 \mathrm{~m}$ in length and $0.02 \mathrm{~m}$ wide at its base is used to extend the surface area of a $0.5 \%$ carbon steel wall. If the wall is at $40^{\circ} \mathrm{C}$ and heated gas flows past at $200^{\circ} \mathrm{C}\left(\bar{h}=230 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\right)$, compute the heat removed by the fin per meter of breadth, $b$, of the fin. Neglect temperature distortion at the root.

Questions asked

INSTANT ANSWER

2) 0.0 \( \xrightarrow[0.1 \mathrm{~m}]{\stackrel{L}{\longrightarrow}} \) \[ \begin{array}{ll} k=54 \mathrm{w} / \mathrm{m}^{\circ} \mathrm{C} & L=0.1 \mathrm{~m}, \quad t=0.02 \mathrm{~m}, \quad \text { dept } \\ h=200 \mathrm{~W} / \mathrm{m}^{2} \mathrm{C} \quad & T_{\text {wall }}=200^{\circ} \mathrm{C} \quad T_{\text {air }}=10^{\circ} \mathrm{C} \end{array} \] no o \[ A_{f}=2 w\left[L^{2}+\left(\frac{t}{2}\right)^{2}\right]^{1 / 2} \] \[ \begin{array}{l} A_{p}=\left(\frac{t}{2}\right) \times L \\ n_{f}=\frac{1}{m L} \frac{I_{1}(2 m L)}{I_{0}(2 m L)} \\ m=\sqrt{\frac{h p}{k A_{c}}} \text { or } m=\sqrt{\frac{2 h}{k t}} \\ \text { I } A_{f}=2 \times 0.2\left[0.1^{2}+\left(\frac{0.02}{2}\right)^{2}\right]^{1 / 2} \\ A_{f}=0.0402 \mathrm{~m}^{2} \\ \text { Ma } 2000 \\ A_{p}=\left(\frac{t}{2}\right) \times L \\ m=\sqrt{\frac{2 h}{k t}} \\ A_{p}=0.001 \mathrm{~m} \\ m=\sqrt{\frac{2 \times 200}{54 \times 0.02 \mathrm{~m}}} \\ \frac{\theta}{\theta_{b}}=\frac{T-T_{p}}{T_{b}-T_{\infty}}=\Psi_{0} \frac{[m(L-x)]}{I_{0} m L} \\ m=19.24 \\ \text { For } x=0.05 \mathrm{~m} ; T-T_{\infty}=\frac{19.24(0.1-0.05)}{19.24 \times 0.1} \times 190^{\circ} \mathrm{C} \\ T= \end{array} \]

ANSWERED

Andreas Papavassiliou

Numerade educator

A triangular straight fin of 0.1 m in length, 0.02 m thick at the base, and 0.2 m in depth is used to extend the surface of a wall at 200 C. The wall and the fin are made of mild steel (k = 54 W/m C). Air at 10 C (h = 200 W/m2 C) flows over the surface of the fin. Evaluate the temperature at 0.05 m from the base and at the tip of the fin. Determine the rate of heat removal from the fin and the fin efficiency.