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3 years ago

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"A fluid at a pressure of 7 atm with a specific volume of 0.11 m3/kg is constrained in a cylinder behind a piston. It is allowed

to expand under an isothermal process(T = 23 oC) to a pressure of 0.85 atm according to the ideal gas law.a) Calculate the work done by the fluid in the piston. b) Calculate the specific gas constant of the gas. c) calculate the molecular weight of the gas"

1 answer:

AlekseyPX3 years ago

7 0

Answer:

Work done by the fluid in the piston=164.5kJ/kg

Specific gas constant= 0.263 kJ/kg K

Molecular weight of gas= 31.54 kmol

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An equation used to evaluate vacuum filtration is Q = ΔpA2 α(VRw + ARf) , Where Q ≐ L3/T is the filtrate volume flow rate, Δp ≐

larisa86 [58]

Answer:

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2 years ago

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

8 0

2 years ago

A well-insulated, full, water tank containing 197 kg of water initially at 43.9°C is being supplied with 16.7°C water at a rate

joja [24]

Answer:

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Explanation:

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2 years ago

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

3 0

2 years ago

An open tank in a petroleum company lab contains a layer of oil on top of a layer of water. The water height is 5 times the oil

maks197457 [2]

Answer:

Explanation:

Given

specific gravity of oil=0.79

height of oil column is h

height of water column is 5h

Gauge pressure at bottom is equivalent to $26.9\ mm$

$P_{gauge}=\rho_{Hg} gh_{Hg}$

$P_{gauge}=13.6\times 10^3\times 9.8\times 26.9\times 10^{-3}\ Pa$

Pressure due to oil and water at bottom

$P=\rho _{oil}hg+\rho _{w}5hg$

$P=0.79\rho _whg+5\rho _whg=5.79\rho _{w}hg$

$P_{gauge}=P$

$13.6\times 10^3\times 9.8\times 26.9\times 10^{-3}=5.79\times 10^3\times 9.8\times h$

$h=63.18\ mm$

6 0

2 years ago