PROBLEM 1 (13 PTS)

Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

3 years ago

(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 mm in diameter in a steel alloy when a load of 1000

vampirchik [111]

Answer:

(a) We are asked to compute the Brinell hardness for the given indentation. for HB, where P= 1000 kg, d= 2.3 mm, and D= 10 mm.

Thus, the Brinell hardness is computed as

$HB=2P/\pi D{D-\sqrt{D^2-d^2}$

$=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]$

(b) This part of the problem calls for us to determine the indentation diameter d which will yield a 270 HB when P= 500 kg.

$d=\sqrt{D^2-[D-\frac{2P}{(HB)\pi D} } ]^2\\=\sqrt{(10mm)^2-[10mm-\frac{2*500}{450( \pi10mm)} } ]^2$

6 0

4 years ago

Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends

Ierofanga [76]

Answer:

$\t(x)_{max} =\dfrac{p\times L}{2\times \pi}$

Explanation:

Given that

Shear modulus= G

Sectional area = A

Torsional load,

$t(x) = p sin( \frac{2\pi}{ L} x)$

For the maximum value of internal torque

$\dfrac{dt(x)}{dx}=0$

Therefore

$\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}$

Thus the maximum internal torque will be at x= 0.25 L

$t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}$