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zhenek [66]
3 years ago
8

Air at 27°C, 1 atm flows parallel to a flat plate, which is electronically heated. The plate is 0.5 m long in the direction of f

low. The uniform heat flux from the plate to the air is 3,000 W/m². A trip wire placed at the leading edge of the plate ensures that the boundary layer is turbulent over the entire plate. A thermocouple mounted at the trailing edge of the plate records a temperature of 127°C. Consider the following:
a) Determine the local Nusselt number Nuy at the trailing edge of the plate.
b) Calculate the Reynolds number Rex at the trailing edge of the plate.
c) Evaluate the air free stream velocity (m/s).

Engineering
1 answer:
dmitriy555 [2]3 years ago
8 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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How will you maintain the orderliness of your storage area 2pts?​
fiasKO [112]

Explanation:

Label and group products. One would think that a general cleanup would be the first step, but no, it's not. ...Clean up the area. ...Put up demarcation lines. ...Stack properly. ...Keep the aisles, paths and ramps clear. ...Have all the safety signs in place.

4 0
2 years ago
What is the difference between an arch and a dome?
bonufazy [111]
This is an arch, its basically a half circle attach to a rectangle, you could also think of it as an upside down U. A dome is a Sphere with the inside hollowed out.

1 difference is a dome is a 3 dimensional shape while an arch is normally not. Or that a dome is the complete shape with a arch act as it’s diameter.

4 0
2 years ago
zener shunt regulator employs a 9.1-V zener diode for which VZ = 9.1 V at IZ = 9 mA, with rz = 40 and IZK= 0.5 mA. The available
gulaghasi [49]

Answer:

V_z=9.1v

V_{zo}=8.74V

I=10mA

R=589 ohms

Explanation:

From the question we are told that:

Zener diode Voltage V_z=9.1-V

Zener diode Current I_z=9 .A

Note

rz = 40\\\\IZK= 0.5 mA

Supply Voltage V_s=15

Reduction Percentage P_r= 50 \%

Generally the equation for Kirchhoff's Voltage Law is mathematically given by

V_z=V_{zo}+I_zr_z

9.1=V_{z0}+9*10^{-3}(40)

V_{zo}=8.74V

Therefore

At I_z-10mA

V_z=V_{z0}+I_zr_z

V_z=8.74+(10*10^{-3}) (40)

V_z=9.1v

Generally the equation for Kirchhoff's Current Law is mathematically given by

-I+I_z+I_l=0

I=10mA+\frac{V_z}{R_l}

I=10mA+\frac{9.1}{0}

I=10mA

Therefore

R=\frac{15V-V_z}{I}

R=\frac{15-9.1}{10*10^{-3}}

R=589 ohms

5 0
3 years ago
134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.
vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = - 10^{\circ}C = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = 60^{\circ}C = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T =60^{\circ}C:

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at  P = 140 kPa, T =- 10^{\circ}C:

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = s'_{s} and h'_{s} = 278.06 kJ/kg.K at 700kPa

(a) Isentropic efficiency of the compressor, \eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}

\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%

(b) The temperature of the environment, T_{e} = 27^{\circ}C = 273 + 27 = 300 K

Availability at state 1, \Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg

Similarly for state 2, \Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg

Now, the efficiency of the compressor as per the second law;

\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757 = 67.57%

4 0
3 years ago
Select the parameters that are included in a baseline performance check. you may select more than one. select one or more: a. co
pishuonlain [190]

Answer:

A F E C D B

Explanation:

6 0
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