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Dima020 [189]
3 years ago
13

Water flovs in a pipe of diameter 150 mm. The velocity of the water is measured at a certain spot which reflects the average flo

w velocity. A pitot static tube has a meter coefficient of C = 1,05 and is joined to a mercury manometer indicating a reading of 167 mm. Determine the flow rate of the water.
Engineering
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

Q = 0.118 m^{3}/s

Explanation:

Given :

diameter of the pipe, d = 150 mm

                                       = 0.15 m

Pitot tube co efficient, C_{v} = 1.05

manometer reading is given, x = 167 mm

                                                   = 0.167 m

From manometer reading,we can find the difference between the manometer height, h

 h =x\times\left [ \frac{S_{m}}{S_{w}}-1 \right ]

h =0.167\times\left [ \frac{13.6}{1}-1 \right ]

h = 2.1042 m

Now, average velocity is v = C_{v}\sqrt{2.g.h}

                                            = 1.05\times \sqrt{2\times 9.81\times 2.1042}

                                            = 6.74 m/s

Area of the pipe, A = \frac{\pi }{4}\times d^{2}

                                = \frac{\pi }{4}\times 0.15^{2}

                                = 0.0176 m^{2}

Therefore, flow rate is given by, Q = A.v

                                                          = 0.0176 X 6.74

                                                          = 0.118m^{3}/s

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Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

     gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

    gcd(31,415,14,142) =gcd(14,142,3,131)

                                  =gcd=(3,131, 1,618)

                                   =gcd(1,618, 1,513)

                                   =gcd(1,513, 105)

                                   =gcd(105, 43)

                                    =gcd(43, 19)

                                     =gcd(19, 5)

                                      =gcd(5, 4)

                                      =gcd(4, 1)

                                      =gcd(1, 0)

                                      =1

STEP 2

b)  The number of comparison of given input with the algorithm based on  checking consecutive integers and Euclid's algorithm is

     The number of division using Euclid's algorithm =10 from part (a)

      The consecutive integer checking algorithm:

      The number of iterations =14,142 and 1 or 2 division of iteration.

        14,142 ∠= number of division∠ = 2*14,142

         Euclid's algorithm is faster by at least 14,142/10 =1400 times

          At most 2*14,142/10 =2800 times.

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3 years ago
3. What is the mechanical advantage of the pulley system shown below? HII TAI 190 O A1 O E.2 OC.3 OD 4​
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Answer:

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Explanation:

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7 0
2 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

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3 years ago
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DiKsa [7]

Answer:

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6 0
2 years ago
An 1,840 W toaster, a 1,420 W electric frying pan, and a 70 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (T
Viktor [21]

Answer:

A)

Current drawn by toaster = 15.33 A

Current drawn by electric frying pan = 11.83 A

Current drawn by lamp = 0.58 A

B)

The fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

Explanation:

There are three devices plugged into the same outlet.

Toaster = 1840 W

Electric frying pan = 1420 W

Lamp = 70 W

Since the three devices are connected in parallel therefore, the voltage across them will be same but the current drawn by each will be different.

A) What current is drawn by each device?

The current flowing through the device is given by

I = P/V

Where P is the power and V is the voltage.

Current drawn by toaster:

I = 1840/120

I = 15.33 A

Current drawn by electric frying pan:

I = 1420/120

I = 11.83 A

Current drawn by lamp:

I = 70/120

I = 0.58 A

B) Will this combination blow the 15-A fuse?

The total current drawn by all three devices is

Total current = 15.33 + 11.83 + 0.58

Total current = 27.74 A

Therefore, the fuse will definitely blow up since the current drawn by three devices (27.74 A) is way higher than 15 A fuse rating.

5 0
3 years ago
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