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2 years ago

15

Find the rate of heat transfer through a 6 mm thick glass window with a cross-sectional area of 0.8 m2 if the inside temperature

is 310 K and the outside temperature is 265 K. The thermal conductivity of glass is (0.9 J/s m K )

1 answer:

kiruha [24]2 years ago

3 0

Answer:

6.9

Explanation:

I had the same question lol your welcomr if itd not right in sorry

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The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

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Answer:

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An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

$P_{cr} = 13.62 kips$

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2 years ago