The component of a fluid system where a fluid is stored, but not under pressure, is called a container.

scoundrel [369]

Answer:

An airfoil creates lift by exerting a downward force on the air as it flows past

3 0

3 years ago

Exercise 5.Water flows in a vertical pipe of 0.15-m diameter at a rate of 0.2 m3/s and a pressureof 275 kPa at an elevation of 2

wariber [46]

Answer:

a. Pressure head: 33.03,

Velocity Head: 6.53

b. Pressure Head: -1.97,

Velocity Head: 6.53

Explanation:

a.

Given

Diameter = 0.15-m, radius = 0.075

rate = 0.2 m3/s

Pressure =275 kPa

elevation =25 m.

We'll consider 3 points as the water flow through the pipe

1. At the entrance

2. Inside the pipe

3. At the exit

At (1), the velocity can be found using continuity equation.

V1 = ∆V/A

Where A = Area = πr² = π(0.075)² = 0.017678571428571m²

V1 = 0.2/0.017678571428571

V1 = 11.32 m/s

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 2:

P1/yH20 + V1²/2g + z1 = P2/yH20 + V2²/2g + z2

Substitute in the values

P1/yH20 + 20 = (275 * 10³Pa)/yH20 + 25

P1/yH20 = (275 * 10³Pa)/yH20 + 25 - 25

=> P1/yH20 = (275/9.81 + 5)

P1/yH20 = 33.03

The velocity head at point one is then given by

V2²/2g = 11.32²/2 * 9.8

V2²/2g = 6.53

b.

The value of pressure at point 1, is given by Bernoulli equation between point 1 and 3:

P1/yH20 + V1²/2g + z1 = P3/yH20 + V3²/2g + z3

Substitute in the values

33.03 + 20 = P3/yH20 + 55

P3/yH20 = 33.03 + 20 - 55

=> P1/yH20 = -1.97

The velocity head at point three is then given by

V2²/2g = V3²/2g = 6.53

4 0

3 years ago

Read 2 more answers

An aluminum alloy [E = 73 GPa] cylinder (2) is held snugly between a rigid plate and a foundation by two steel bolts (1) [E = 18

V125BC [204]

Answer:

$\sigma_A = 58.43 N/mm^2$

Explanation:

Given data:

length of Steel bolt $L_1 = 335 mm$

Length of aluminium cylinder $L_2 = 275 mm$

Pitch of bolt p = 1mm

Modulus of elasticity of steel E = 215 GPa

Modulus of elasticity of aluminium = 74 GP

Area of bolt $= \frac{\pi}{4} 14^2 = 153.93 mm^2$

Area of cylinder = 2300 mm^2

n =1

By equilibrium

$\sum F_y = 0$

$P_A -2P_S = 0$

$P_A =2P_S$

By the compatibility

$\delta _s + \delta_A = nP$

Displacement in steel is $\delta_s = \frac{P_sL_s}{E_sA_s}$

Displacement in Aluminium is $\delta_A = \frac{P_AL_A}{E_AA_A}$

from compatibility equation we have

$\frac{P_sL_s}{E_sA_s} + \frac{P_AL_A}{E_AA_A} = nP$

$\frac{P_s\times 335}{180\times 10^3\times 153.93} + \frac{P_A\times 275}{73\times 10^3\times 2300} = 1\times 1$

$1.20\times 10^[-5} P_s + 1.44\times 10^{-6}P_A = 1$

substitute $P_A =2P_S$

$1.20\times 10^[-5} P_s + 1.44\times 10^{-6} (2\times P_s) = 1$

$1.488\times 10^{-5} P_s = 1$

$P_s = 67204.30 N$

$P_A = 134,408.60 N$

Stress in Aluminium $\sigma = \frac{P_A}{A_A}$

$= \frac{134,408.60}{2300}$

$\sigma_A = 58.43 N/mm^2$

8 0

3 years ago

A(n) _______ is turned by the crankshaft, compresses the air, and forces it into the engine.

zzz [600]

Answer:

Explanation:

internal combustion engine.

3 0

3 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

3 years ago