I need the solution!! Thx so much

Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th

shepuryov [24]

Answer:

$a=38.5 ft/sec^{2}$

Explanation:

Note that acceleration is the rate change of velocity i.e

$acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt} \\$ .

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

$\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\$

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

$a=653*0.1667*0.354\\$

$a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\ a=38.5 ft/sec^{2}$

3 0

3 years ago

The transfer of energy by the movement of particles that are in contact with each other

jeka57 [31]

Answer: I belive that the Answer is C.) Conduction

Explanation:

3 0

3 years ago

Read 2 more answers

Un montañero de 65kg de masa ha ascendido a la cima del Everest, la montaña más alta del mundo de 8848m de altura sobre el nivel

andrew-mc [135]

Answer:

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

Explanation:

Por la definición de trabajo sabemos que el montañero debió contrarrestar trabajo causado por la gravedad terrestre. Si asumimos que el cambio de la altura es muy pequeño en comparación con el radio del planeta (6371 kilómetros vs. 0,5 kilómetros), entonces podemos considerar que la aceleración gravitacional es constante y la ecuación de trabajo ( $\Delta W$ ), medido en joules, que reducida a:

$\Delta W = m\cdot g\cdot \Delta z$ (1)

Donde:

$m$ - Masa del montañero, medido en kilogramos.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$\Delta z$ - Distancia vertical de ascenso del montañero, medida en metros.

Si tenemos que $m = 65\,kg$ , $g = 9,807\,\frac{m}{s^{2}}$ y $\Delta z = 500\,m$ , entonces el trabajo realizado por el montañero para subir es:

$\Delta W = (65\,kg)\cdot \left(9,807\,\frac{m}{s^{2}} \right)\cdot (500\,m)$

$\Delta W = 318727,5\,J$

El trabajo realizado para subir los últimos 500 metros es 318727,5 joules.

7 0

3 years ago

I need help with this please

Firdavs [7]

C is the answer to the question

6 0

3 years ago

Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t

faltersainse [42]

Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

4 0

3 years ago