Ask question

Ask question

All categories

Lubov Fominskaja [6]

3 years ago

9

A frustum of cone is filled with ice cream such that the portion above the cone is a hemisphere. Define the variables di=1.25 in

., d0=2.25 in., h=2 in., and determine the volume of the ice cream.

1 answer:

LekaFEV [45]3 years ago

4 0

Answer:

The volume of the ice cream is V = 13.9898 in^3

Explanation:

Given

di = 1.25 in

d0 = 2.25 in

h = 2 in

For the calculation

di = 1.25/2

d0 = 2.25/2

h = 2

>> Vcon = $\pi$ * h/3 * (di^2+di*d0+d0^2);

>> Vcon = 5.0437

>> Vtop = 2 * $\pi$ * d0^3;

>>Vtop = 8.9461

>> V = Vcon + Vtop

V = 13.9898 in^3

You might be interested in

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb

Nezavi [6.7K]

Answer:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- $\xi_1$ and $\xi_2$ : extent of the reactions (2).

- $F_{O_2}^2$ , $F_{CH_4}^2$ , $F_{H_2O}^2$ , $F_{HCHO}^2$ and $F_{CO_2}^2$ : Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- $F_{O_2}^2=50mol/s-\xi_1-2\xi_2$ : oxygen mole balance.

- $F_{CH_4}^2=50mol/s-\xi_1-\xi_2$ : methane mole balance.

- $F_{H_2O}^2=\xi_1+2\xi_2$ : water mole balance.

- $F_{HCHO}^2=\xi_1$ : formaldehyde mole balance.

- $F_{CO_2}^2=\xi_2$ : carbon dioxide mole balance.

Thus, the degrees of freedom are:

$DF=7unknowns-5equations=2$

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

$0.900=\frac{\xi_1+\xi_2}{50mol/s}$

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

$0.860=\frac{F_{HCHO}^2}{50mol/s}$

In such a way, one realizes that the output formaldehyde's molar flow is:

$F_{HCHO}^2=0.860*50mol/s=43mol/s$

Which is equal to the first reaction extent $\xi_1$ , therefore, one computes the second one from the fractional conversion of methane as:

$\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s$

Now, one computes the rest of the output flows via:

- $F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s$

- $F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s$

- $F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s$

- $F_{HCHO}^2=43mol/s$

- $F_{CO_2}^2=2mol/s$

The total output molar flow is:

$F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s$

Therefore the output stream composition turns out into:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Best regards.

7 0

2 years ago

The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto

lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0

3 years ago

State five applications of thermochromic materials

rusak2 [61]

Explanation:

The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.

8 0

2 years ago

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities <em> $n_{th}$ </em> therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

5 0

2 years ago

A sum of $500,000 will be invested by a firm two years from now. If money is worth 12%, what will be the worth of this investmen

notka56 [123]

Answer:

investment 10 years from now is $1,238,000 .

Explanation:

given data

sum = $500,000

rate = 12% =0.12

total time = 10 year

solution

as present value After 2 years from now is $500,000

so time period is now = 8 year ( 10 - 2 )

so we apply future value formula that is

Future value = present value × $(1+r)^{t}$ ............1

put here value we get

Future value = $500,000 × $(1+0.12)^{8}$

Future value = $500,000 × 2.476

Future value = $1,238,000

so investment 10 years from now is $1,238,000 .

8 0

3 years ago