A spherical balloon is filled with a gas. The outer diameter of the balloon is 20 in. and the thick-ness is 0.012 in. Calculate the maximum permissible pressure in the balloon if the allowable tensile stress and the allowable shear stress in the balloon are 1 ksi and 0.3 ksi, respectively.

Answers

Answer 1

Answer:

1.4 psi

Explanation:

Before diving into the solution to the question above, let's pick out the parameters needed in solving this problem from the question.

=> The measurement for the outer diameter of the balloon = 20 inches, the measurement for the thickness =  0.012 in,  the allowable tensile stress = 1ksi and the allowable shear stress in the balloon =  0.3 ksi.

The first thing to determine is the inner diameter = 20 - 2 ×  0.012 in = 19.976 in.

Therefore, the tensile stress:

1000 = k × [19.976/2]÷ 2 × 0.012 = 2.4 psi.

Also, the sheer stress which is also the maximum permissible pressure in the balloon can be calculated below as:

0.3 × 1000 = k × [19.976/2]/ 4 × 0.012 = 1.4 psi.


Related Questions

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources

Answers

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

You are designing a package for 200 g of snack food that is sensitive to oxygen, and fails when it absorbs 120 ppm of oxygen (by weight). Marketing tells you it wants the snack to be in a plastic pouch measuring 6 inches by 6 inches (ignore seems), so it will have a total surface area available for permeation of 72 in(6" x 6" x 2 sides). You need to recommend an appropriate plastic material for this product, to provide a minimum of 70 days shelf life. Follow these steps:
a. Calculate the allowable oxygen gain, in cm at STP. (5 pts)
b. At this point, you do not know the material you will use, so you do not know the permeability coefficient or the thickness. The better the barrier the plastic you choose, the thinner the material can be to provide the appropriate barrier. Rather than simple trial and error, a sensible approach is to solve for the ratio of P/L that is required. We can solve the basic permeability equation for this ratio: P = 9 At Ap L Use the information you have to determine the required value for P/L, expressing your answer in cm/(100 in? d atm). (5 pts)
c. Use the information in the textbook (chapters 4 and 14 or in another reliable source; provide reference if you have used chapter 4, 14 or any other source) on oxygen permeability coefficients for various polymers to select a polymer that would be suitable, and calculate the required thickness. (Be sure this is reasonable; for example, if the required thickness is more than 20 mils, you need to choose a different polymer!) Note that chapter 4 presents these values in the units you used in (b) while chapter 14 presents values with different units, so unit conversion would be required. In your answer, state the material you have chosen, its oxygen permeability coefficient, and the minimum thickness you recommend. (Be sure to express the thickness with no more than one decimal place.) Obviously, there is more than one solution to this problem, but you only need one. (10 pts)

Answers

Snack food is the sensitivity to auction and avails win the George 120

In a production facility, 3-cm-thick large brass plates (k 5 110 W/m·K, r 5 8530 kg/m3, cp 5 380 J/kg·K, and a 5 33.9 3 1026 m2/s) that are initially at a uniform temperature of 25°C are heated by passing them through an oven maintained at 700°C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h 5 80 W/m2·K, determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be

Answers

Answer:

the surface temperature of the plates when they come out of the oven is approximately 445 °C

Explanation:

Given the data in the question;

thickness t = 3 cm = 0.03 m

so half of the thickness L = 0.015 m

thermal conductivity of brass k = 110 W/m°C

Density p = 8530 kg/m³

specific heat [tex]C_p[/tex] = 380 J/kg°C

thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s

Temperature of oven T₀₀ = 700°C

initial temperature T[tex]_i[/tex] = 25°C

time t = 10 min = 600 s

convection heat transfer coefficient h = 80 W/m².K

Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.

So, using analytical one-term approximation method, the Fourier number > 0.2.

now, we determine the Biot number for the process

we know that; Biot number Bi =  hL / k

so we substitute

Bi =  hL / k

Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109

Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )

The interpolation method used to find the

λ₁ = 0.1039 and A₁ = 1.0018

so

The Fourier number т = ∝t/L²

we substitute

Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²

т = 0.02034 / 0.000225

т = 90.4

As we can see; 90.4 > 0.2

So,  analytical one-term approximation can be used.

∴ Temperature at the surface will be;

θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation

θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )

so we substitute

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )

θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )

θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998

θ(L,t)[tex]_{wall[/tex] = 0.3775

so we substitute into equation 1

θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)

0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )

0.3775 = ( T(L,t) - 700 ) / ( - 675 )

0.3775 × ( - 675 ) = ( T(L,t) - 700 )

- 254.8125 = T(L,t) - 700

T(L,t) = 700 - 254.8125

T(L,t) = 445.1875 °C ≈ 445 °C

Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C

Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.

Answers

Answer:

3  kmol/m^3

Explanation:

Determine the concentration of C in the stream leaving the reactor

Given that the CTSR reaction ; A (k1) → B (k2) → C

K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2

attached below is the detailed solution

concentration of C leaving the reactor= 3 kmol/mol^3

Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3

In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit

Answers

Answer: hello your question lacks some data below is the missing data

Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol

H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.

H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg

Answer :

a) 34.98 lit/min

b) 1432.53 m^3/min

Explanation:

a) Calculate how much water is produced

density of water = 1 kg/liter

First we will determine the mass of condensed water using the relation below

inlet mass - outlet vapor mass =  0.0339508 * n * 18/1000 ----- ( 1 )

where : n = 57241.57

hence equation 1 = 34.98 Kg/min

∴ volume of water produced =  mass of condensed water / density of water

                                                =  34.98 Kg/min / 1 kg/liter

                                                = 34.98 lit/min

b) calculate the Volumetric flow rate of air entering the unit

applying the relation below

Pv = nRT

101325 *V = 57241.57 * 8.314 * 305  

∴ V = 1432.53 m^3/min

The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.

Answers

Answer:

the liquid woulriekwvhrnsshsnekwb ndrhwmoadi

1) Each of the following would be considered company-confidential except
A) a contract bid B) employee salaries C) your company's strategic plan D) your company's address

Answers

Answer is your company’s address

) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).

Answers

Answer:

Va / Vb = 0.5934

Explanation:

First step is to determine total head losses at each pipe

at Pipe A

For 1/4 open gate valve head loss = 17 *Va^2 / 2g

elbow loss = 0.75 Va^2 / 2g

at Pipe B

For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g

elbow loss = 0.75 * Vb^2 / 2g

Given that both pipes are parallel

17 *Va^2/2g +  0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g  + 0.75 * Vb^2 / 2g

∴ Va / Vb = 0.5934

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