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Neporo4naja [7]
3 years ago
6

Let r be a bank's interest rate in percent per year (APR). An initial amount of money P, also called as principal, will mature t

o an amount of P(1+r/100)^n after n years have passed. Write a Python program that takes P, r, and n as inputs from the user and provides as output the matured value after n years. For example if the user provides P, r, and n as 1000 0.95 and 5 .Your program must apply the interest-rate formula and provide 1048.4111145526908 as the output.
Engineering
1 answer:
Liono4ka [1.6K]3 years ago
7 0

Answer:

See explanation below.

Explanation:

For this case the program needs to take the inputs as P,r and n and the output would be as A and printed on the system. The code is:

# Inputs

P = float(input("Enter the present value : "))  

r = float(input("Enter your APR : "))  

n = float(input("Enter the number of years : ") )

# Output

A = P*(1 +(r/100))**n

print("The future values is:", A)  

And the result obtained is:

Enter the present value : 1000

Enter your APR : 0.95

Enter the number of years : 5

The future values is: 1048.4111145526908

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You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-
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Explanation:

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Diameter of tank d = 8 m

Length of tank l = 4 m

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For A-36 steel yield stress \sigma = 250 MPa,

Allowable stress \sigma _{allow} = \frac{\sigma}{F.S}

 \sigma _{allow} = \frac{250}{4} = 62.5 MPa

Pressure force is given by,

 P = \rho gh

 P = 1200 \times 9.8 \times 4

P = 47088 Pa

Now for a vertical pipe,

\sigma _{allow} = \frac{Pd}{4t}

Where t = required thickness

 t = \frac{Pd}{4 \sigma _{allow} }

 t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }

t = 1.506 \times 10^{-3} m

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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
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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:

C

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