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The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the

soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

(dP - y*L*sin(Q))/ L = 2*t / r

dP / L - y*sin(Q) = 2*t / r

Where dP / L = - dP/dx,

dP / dx = -2*t / r - y*sin(Q)

Where r = D /2 ,

dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)

dP / dx = -8 + 0

dP/dx = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

dP / dx = 8 - 62.4

dP/dx = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

dP / dx = -8 + 62.4

dP/dx = 54.4 lb / ft^3

7 0

3 years ago

A 2 m3 rigid tank initially contains air at 100 kPa and 22 degrees C. The tank is connected to a supply line through a valve. Ai

Vaselesa [24]

Answer:

a. 9.58kgs b. 340.32KJ

Explanation:

Volume of tank= 2m³

Initial Pressure Pi= 100KPa

Initial Temperature Ti= 22 C= 295K

Line Pressure P₁= 600 KPa

Line Temperature T= 22 C= 295K

Final Pressure P2= 600 KPa

Final Temperature T2= 77 C= 350K

Use Ideal Gas Equation

PV= mRT

P₁V₁= m₁RT₁

m₁= (100 x 2)/(0.287 x 295) = 2.3622kg

P₂V₂= m₂RT₂

m₂= (600 x 2)/(0.287 x 350) = 11.946 kg

Since valve is closed and no mass leave

m₁ + mi = m₂ + me

as per above condition me= 0

mi= m₂ - m₁ = 11.946 - 2.3622 = 9.5838kg

Applying energy equation

m₁u₁ + mihi + Q = m₂u₂ + mehe + W

me and W=0

m₁u₁ + mihi + Q = m₂u₂

m₁CvT₁ + miCpTi + Q = m₂CvT₂

Q = m₂CvT₂- m₁CvT₁ - miCpTi

Q = (11.946 x 0.717 350) - (2.3622 x 0.717 x 295) - (9.5868 x 1.004 x 295)

Q = -340.321 KJ (Negative sign doesn't matter as energy is not a vector quantity)

4 0

3 years ago

Free pts was my day I was a dog but it would need to be

Ne4ueva [31]

Answer:

hi

Explanation:

8 0

3 years ago

Read 2 more answers

A steam power plant operates on the reheat Rankine cycle. Steam enters the highpressure turbine at 12.5 MPa and 550°C at a rate

gayaneshka [121]

Answer:

A) condenser pressure = 9.73 kPa,

B) 10242 kw

C) 36.9%

Explanation:

given data

entrance pressure of steam = 12.5 MPa

temperature of steam = 550⁰c

flow rate of steam = 7.7 kg/s

outer pressure = 2 MPa

reheated steam temperature = 450⁰c

isentropic efficiency of turbine( nt ) = 85% = 0.85

isentropic efficiency of pump = 90% = 0.90

From steam tables

at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg, S3 = 6.6317 kJ/kgK

also for an Isentropic expansion

S4s = S3 .

therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa

h4s = 2948.1 kJ/kg

nt = 0.85

nt (0.85) = $\frac{h3-h4}{h3-h4s}$ = $\frac{3476.5 - h4}{3476.5 - 2948.1}$

making h4 subject of the equation

h4 = 3476.5 - 0.85 (3476.5 - 2948.1)

h4 = 3027.3 kj/kg

at P5 = 2 MPa and T5 = 450⁰c

h5 = 3358.2 kj/kg, s5 = 7.2815 kj/kgk

at P6 , x6 = 0.95 and s5 = s6

using nt = 0.85 we can calculate for h6 and h6s

from the chart attached below we can see that

p6 = 9.73 kPa, h6 = 2463.3 kj/kg

B) the net power output

solution is attached below

c) thermal efficiency

thermal efficiency = 1 - $\frac{qout}{qin}$ = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%

8 0

3 years ago

A specimen of a 4340-steel alloy with a plane strain fracture toughness of 54.8 MPa sqrt(m) (50 ksi sqrt(in.)) is exposed to a s

Lunna [17]

Answer:

critical stress $\sigma _c$ = 1382.67 MPa

Explanation:

given data

plane strain fracture toughness = 54.8 MP

length of surface creak = 0.5 mm

we take here

parameter Y = 1.0

solution

we apply critical stress formula that is

critical stress $\sigma _c$ = $\frac{K}{Y\sqrt{\pi \times a} }$ .............................1

here K is design stress plane strain fracture toughness and a is length of surface creak so put all these value in equation 1

critical stress $\sigma _c$ = $\frac{54.8 \times 10^6}{1 \sqrt{\pi \times 5 \times 106{-4}}}$

solve it we get

critical stress = 1382.67 MPa

As exposed stress 1030 MPa is less than critical stress 1382 MPa

so that fracture will not be occur here

7 0

3 years ago