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fiasKO [112]
3 years ago
10

Rain falls on a 1346 acre urban watershed at an intensity of 1.75 in/hr for a duration of 1 hour. The catchment land use is 20%

parkland, 30% residential, and 50% multiple detached residential area. The time to concentration is 1 hr. Determine the peak runoff. Use runoff coefficients published in APWA 5600.

Engineering
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

Detailed solution is given below:

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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
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

3 0
3 years ago
An electric kettle is required to heat 0.64 kg of water from 15.4°C to 98.2°C in six
skelet666 [1.2K]

Answer:

Almost done

Explanation:

I am just finishing up my work

7 0
2 years ago
Select the correct text in the passage.
sineoko [7]
It is habahi Yw with yuuuuuy I am a little more confused about
5 0
3 years ago
Read 2 more answers
Air within a piston cylinder assembly executes a Carnot refrigeration cycle between hot and cold reservoirs at TH=600 K and TC=3
Nataly [62]

Answer:

See explaination

Explanation:

for a reverse carnot cycle T-S diagram is a rectangle which i have shown

net work for a complete cycle must be equal to net heat interaction.

Kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco
allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

d_1=50 mm,f_1=0.0048

For pipe 2

d_2=75 mm,f_2=0.0058

Q=2.8 l/s

Q=2.8\times 10^{-3]

We know that Q=AV

Q=A_1V_1=A_2V_2

A_1=1.95\times 10^{-3}m^2

A_2=4.38\times 10^{-3} m^2

So V_2=0.63 m/s,V_1=1.43 m/s

head loss (h)

h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}

Now putting the all values

h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0
3 years ago
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