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3 years ago

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A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 10

1300 Pa. The inlet pure water flow is 68.0 kgmol/h and total inlet gas flow is 57.8 kgmol/h. The tower diameter is 0.747 m. The film mass-transfer coefficients are k’ya=0.0739 kgmol/s.m3 .molfrac and k’xa= 0.169 kgmol/s.m3 .molfrac. Using the design methods for dilute gas mixtures,
(a) Calculate the tower height using k’ya.
(b) Calculate the tower height using K’ya.

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer:

Explanation:

Step by step solved solution is given in the attached document.

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Answer:

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Explanation:

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2 years ago

Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio

SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

a)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

b)

Moment of inertia in the direction of width I = t w³/12

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4 0

3 years ago

A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s

sergiy2304 [10]

Answer:

$D_o=11.9inch$

Explanation:

From the question we are told that:

Thickness $T=0.5$

Internal Pressure $P=2.2Ksi$

Shear stress $\sigma=12ksi$

Elastic modulus $\gamma= 35000$

Generally the equation for shear stress is mathematically given by

$\sigma=\frac{P*r_1}{2*t}$

Where

r_i=internal Radius

Therefore

$12=\frac{2.2*r_1}{2*0.5}$

$r_i=5.45$

Generally

$r_o=r_1+t$

$r_o=5.45+0.5$

$r_o=5.95$

Generally the equation for outer diameter is mathematically given by

$D_o=2r_o$

$D_o=11.9inch$

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

$D_o=11.9inch$

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2 years ago

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

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

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

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

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

$\epsilon =0.799$

5 0

3 years ago