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SSSSS [86.1K]
2 years ago
12

Two aerial photographs were taken 30 seconds apart over one east-bound lane of l-80 near Grand Island, NE. The following results

were recorded.
Position from start of road section Vehicle 2000 2300 1700 1200 600 2940 3200 2400 2100 1730 1000 2 3 4
Plot the trajectories of the vehicles on graph paper and compute the average flow (vph), density (veh/mi) and space mean speed (mph) over the 3000 ft length of the lane.
Engineering
1 answer:
NikAS [45]2 years ago
8 0

Answer:

the average flow (vph) = 222.69 veh/hr.  

The average velocity = 111348 ft/hr.

density = 2 × 10⁻³ veh/ft.

Explanation:

The first thing to do in this particular question is to determine the average velocity.

The average velocity = [ ( 2940 - 2000/30) + ( 3200 - 2300/30) + ( 2400 - 1700/30) + ( 2100 - 1200) + ( 1730 - 600/30) + ( 1000 - 0/30).

The average velocity = [31.33 + 30 + 23.33 + 30 + 37.66 + 33.66]/ 6 = 30.99 ft/sec.

Thus, 30.99 ft/sec × 60 × 60 = 111348 ft/hr.

The next thing to do is to determine the density. therefore, the density = 6/ 3000 = 2 × 10⁻³ veh/ft.  

The average flow (vph) =  111348 ft/hr × 2 × 10⁻³ veh/ft.  = 222.69 veh/ hr.

Also, the space mean speed (mph) over the 3000 ft length of the lane = 6/ [ 1/31.33 + 1/30 + 1/23.33 + 1/30 + 1/37.66 + 1/33.66] = 6/ 0.1977 = 30.34 ft/sec.

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A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det
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Answer:

a) T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C), b) \dot Q_{H} = 26.875\,kW

Explanation:

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After some algebraic handling, the temperature of the cold reservoir is determined:

T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}

T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}}  \right)

T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)

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b) The heating load provided by the heat pump is:

\dot Q_{H} = COP_{HP}\cdot \dot W

\dot Q_{H} = (12.5)\cdot (2.15\,kW)

\dot Q_{H} = 26.875\,kW

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Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

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