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Mrac [35]
2 years ago
6

A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a

ir conditioning unit is
Engineering
1 answer:
prohojiy [21]2 years ago
8 0

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds  

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

                          = 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

                              = 2622.6 / 25  = 104.904 ft^3/sec

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2 years ago
Consider the following class definitions: class smart class superSmart: public smart { { public: public: void print() const; voi
arsen [322]

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a) There is no any private member of smart which are public members of superSmart.

Explanation:

5 0
2 years ago
Toeboards are usually ___ inches high and used on landings and balconies.
Ksju [112]

Answer:4 inches

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6 0
2 years ago
Steam enters a turbine operating at steady state at 1 MPa, 200 °C and exits at 40 °C with a quality of 83%. Stray heat transfer
Andrei [34K]

Answer:

(a) Work out put=692.83\frac{KJ}{Kg}

(b) Change in specific entropy=0.0044\frac{KJ}{Kg-K}

Explanation:

Properties of steam at 1 MPa and 200°C

        h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

      h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}

 s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}

h_2=2134.57\frac{KJ}{Kg}

s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}

s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}

s_2=6.6944\frac{KJ}{Kg-K}

(a)

Work out put =h_1-h_2

                      =2827.4-2134.57 \frac{KJ}{Kg}

Work out put =692.83 \frac{KJ}{Kg}

(b) Change in specific entropy

     s_2-s_1=6.6944-6.69\frac{KJ}{Kg-K}

Change in specific entropy =0.0044\frac{KJ}{Kg-K}

3 0
2 years ago
In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y =
monitta

Answer:

Ψ = 10(y^2) + c

<em><u>y = 1.067m</u></em>

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =<em>u/y</em>

<em><u>u = 20y</u></em>

<em><u>Ψ = ∫20ydy</u></em>

<em><u>Ψ = 10(y^2) + c</u></em>

<em><u>(b)</u></em>

<em><u>the flow is half below y = 1.6*(2/3)=1.067 m</u></em>

<em><u>this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m</u></em>

3 0
2 years ago
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