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gogolik [260]
2 years ago
12

Design an algorithm for computing √n

Engineering
1 answer:
a_sh-v [17]2 years ago
8 0
Positive integer n design an algorithm
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You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______
stich3 [128]
I believe it’s D) 20 feet
3 0
3 years ago
Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.
saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 \mathbf{m^{3}}

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; v_1 = vg_1 = 0.0996 \mathbf{m^{3}} / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \mathbf{m^{3}} / kg

From temperature T₂ = 200⁰ C

v_f_2 = 0.0016 \ m^3/kg  

vg_2 = 0.127 \ m^3/kg  

Since  vf_2 < v_2 , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}

x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}

x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}

\mathsf{x_2 =0.78}

At temperature T₂, the specific internal energy u_f_2 = 850.6 \ kJ/kg , also ug_2 = 2594.3 \ kJ/kg

Thus,

u_2 = uf_2 + x_2 (ug_2 -uf_2)

u_2 =850.6  +0.78 (2594.3 -850.6)

u_2 =850.6  +1360.086

u_2 =2210.686 \ kJ/kg

<u>At state 3:</u>

Temperature T_3=T_2 = 200 ^0 C ,

V_3 = 2V_1 = 0.06 \ m^3

Specific volume v_3 = 0.2  \ m^3/kg

Thus; vg_3 =vg_2 = 0.127 \ m^3/kg ,

SInce v_3 > vg_3, therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at v_3 = 0.2  \ m^3/kg and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \ m^3/kg

The specific internal energy u_3 at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

u_2-u_1

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

u_3-u_2

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0
3 years ago
Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu
Sonja [21]

Answer:

a

The rate of work developed is \frac{\r W}{\r m}= 300kJ/kg

b

The rate of entropy produced within the turbine is   \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

Explanation:

     From  the question we are told

          The rate at which heat is transferred is \frac{\r Q}{\r m } = -  30KJ/kg

the negative sign because the heat is transferred from the turbine

          The specific heat capacity of air is c_p = 1.1KJ/kg \cdot K

          The inlet temperature is  T_1 = 970K

          The outlet temperature is T_2 = 670K

           The pressure at the inlet of the turbine is p_1 = 400 kPa

          The pressure at the exist of the turbine is p_2 = 100kPa

           The temperature at outer surface is T_s = 315K

         The individual gas constant of air  R with a constant value R = 0.287kJ/kg \cdot K

The general equation for the turbine operating at steady state is \

               \r Q - \r W + \r m (h_1 - h_2) = 0

h is the enthalpy of the turbine and it is mathematically represented as          

        h = c_p T

The above equation becomes

             \r Q - \r W + \r m c_p(T_1 - T_2) = 0

              \frac{\r W}{\r m}  = \frac{\r Q}{\r m} + c_p (T_1 -T_2)

Where \r Q is the heat transfer from the turbine

           \r W is the work output from the turbine

            \r m is the mass flow rate of air

             \frac{\r W}{\r m} is the rate of work developed

Substituting values

              \frac{\r W}{\r m} =  (-30)+1.1(970-670)

                   \frac{\r W}{\r m}= 300kJ/kg

The general balance  equation for an entropy rate is represented mathematically as

                       \frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma  = 0

          =>          \frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)

    generally (s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]

substituting for (s_1 -s_2)

                      \frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} +  c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]

                      Where \frac{\sigma}{\r m} is the rate of entropy produced within the turbine

 substituting values

                \frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]

                    \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

           

 

                   

   

5 0
3 years ago
What phenomenon allows water to reach the top of a building?
Artemon [7]

Answer:

Option C: water pressure.

Explanation:

Water pressure allows water to reach the top of a building.

6 0
3 years ago
Reception of signals from a radio facility, located off the airway being flown, may be inadequate at the designated mea to ident
lakkis [162]

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

<h3>What is altitude?</h3>

Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.

The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

Therefore, the correct answer is 22 NM of a VOR.

To learn more about altitudes refer to:

brainly.com/question/1159693

#SPJ4

3 0
2 years ago
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