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lbvjy [14]
3 years ago
7

Can you answer what is attached.

Engineering
1 answer:
kirill [66]3 years ago
3 0
29.4 bro I hope that helps
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A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round
Harlamova29_29 [7]

Answer:

\frac{Q_{square}}{Q_{circle}} =  0.785  

Explanation:

given data

types of drinking straws

  1. square cross-sectional shape
  2. round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square  = perimeter of circle  

4 × S = π × D

here S is length and D is diameter

S = \frac{\pi D}{4}        ....................1

and

ratio of  flow rate through the square and circle is here

\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}  

\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}  

\frac{Q_{square}}{Q_{circle}} =  0.785  

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4 years ago
Define a separate subroutine for each of the following tasks respectively.
Valentin [98]

Answer:

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Explanation:

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5 0
3 years ago
Read 2 more answers
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

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Answer:

Explanation:

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Answer:

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