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Delvig [45]
3 years ago
8

An insulated 40 ft3 rigid tank contains air at 50 lbf/in2 and 120oF. A valve connected to the tank is now opened, and air is all

owed to escape until the pressure inside drops to 25 lbf/in2. The air temperature within the tank and at the exit during this process is kept constant at 120oF by an electric heater placed in the tank. Determine the electrical work [Btu] for the process. Assume that the air behaves like an ideal gas.
Engineering
2 answers:
Elan Coil [88]3 years ago
8 0

Answer:

The electrical work is 184.49 Btu

Explanation:

T=120°F=580R

From table of properties of air the enthalpy of air at 580R is he=138.66 Btu/lbm, internal energy at initial and final state is u=98.9 Btu/lbm

The mass balance is:

me=m1-m2

And the energy balance assuming the tank as the steady flow is:

We=mehe+m2u2-m1u1

The initial mass m1 is:

m_{1}=\frac{P_{1}V }{RT_{1} }

Where P1=50 psig

V=40 ft^3

T1=120°F=580 R

m_{1}=\frac{50*40}{0.3704*580}  =9.3 lbm

The final mass m2 is:

m_{2} =\frac{P_{2}V }{RT_{2} }

P2=25 psig

V=40 ft^3

T2=580 R

m_{2}=\frac{25*40}{0.3704*580}  =4.66 lbm

me=9.3-4.66= 4.64 lbm

W_{e}=(4.64*138.66)+(4.66*98.9)-(9.3*98.9)=184.49 Btu

goldfiish [28.3K]3 years ago
5 0

Answer:

The electrical work for the process is 256.54 Btu.

Explanation:

From the ideal gas equation:

n = PV/RT

n is the number of moles of air in the tank

P is initial pressure of air = 50 lbf/in^2 = 50 lbf/in^2 × 4.4482 N/1 lbf × (1 in/0.0254m)^2 = 344736.2 N/m^2

V is volume of the tank = 40 ft^3 = 40 ft^3 × (1 m/3.2808 ft)^3 = 1.133 m^3

T is initial temperature of air = 120 °F = (120-32)/1.8 + 273 = 321.9 K

R is gas constant = 8.314 J/mol.k

n = 344736.2×1.133/8.314×321.9 = 145.94 mol

The thermodynamic process is an isothermal process because the temperature is kept constant.

W = nRTln(P1/P2) = 145.94×8.314×321.9×ln(50/25) = 145.94×8.314×321.9×0.693 = 270669 J = 270669 J × 1 Btu/1055.06 J = 256.54 Btu

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A strain gage is mounted at an angle of 30° with respect to the longitudinal axis of the cylindrical pressure. The pressure vess
GuDViN [60]

Answer:

1790 μrad.

Explanation:

Young's modulus, E is given as 10000 ksi,

μ is given as 0.33,

Inside diameter, d = 54 in,

Thickness, t = 1 in,

Pressure, p = 794 psi = 0.794 ksi

To determine shear strain, longitudinal strain and circumferential strain will be evaluated,

Longitudinal strain, eL = (pd/4tE)(1 - 2μ)

eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)

eL = 3.64 x 10-⁴ radians

Circumferential strain , eH = (pd/4tE)(2-μ)

eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)

eH = 1.79 x 10-³ radians

The maximum shear strain is 1790 μrad.

4 0
3 years ago
Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i
IceJOKER [234]

Answer:

Gc(s) = \frac{0.1s + 0.28727}{s}

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comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

8 0
4 years ago
A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.
Tems11 [23]

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

S_{ut}=K \left(\cfrac ne \right)^n

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}

And we can replace values

K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87

Thus we get that the strength coefficient is K = 591.87 MPa

6 0
4 years ago
Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.
Sholpan [36]

Answer:

Frequency = 3.5294\times 10^{14}s^{-1}

Wavenumber = 1.1765\times 10^6m^{-1}

Energy = 2.3365\times 10^{-19}J

Energy = 1.4579 eV

Energy = 2.3365\times 10^{-12}erg

Explanation:

As we are given the wavelength = 850 nm

conversion used : (1nm=10^{-9}m)

So, wavelength is  850\times 10^{-9}m

The relation between frequency and wavelength is shown below as:

Frequency=\frac{c}{Wavelength}

Where, c is the speed of light having value = 3\times 10^8m/s

So, Frequency is:

Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}

Frequency=3.5294\times 10^{14}s^{-1}

Wavenumber is the reciprocal of wavelength.  

So,  

Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}

Wavenumber=1.1765\times 10^6m^{-1}

Also,  

Energy=h\times frequency

where, h is Plank's constant having value as 6.62\times 10^{-34}J.s

So,  

Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})

Energy=2.3365\times 10^{-19}J

Also,  

1J=6.24\times 10^{18}eV

So,  

Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)

Energy=1.4579eV

Also,  

1J=10^7erg

So,  

Energy=(2.3365\times 10^{-19})\times 10^7erg

Energy=2.3365\times 10^{-12}erg

5 0
3 years ago
Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid
AnnZ [28]

Answer:

5.328Ibm/hr

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

for this case we can define the following equation for mass flow using the first law of thermodynamics

m=\frac{Q}{h1-h2}

where

Q=capacity of the radiator =5000btu/hr

m = mass flow

then using thermodynamic tables we found entalpy in state 1 and 2

h1(x=0.97, p=16psia)=1123btu/lbm

h2(x=0, p=16psia)=184.5btu/lbm

solving

Q=\frac{5000}{1123-184.5} =5.328Ibm/hr

3 0
3 years ago
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