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3 years ago

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Water flows around a 6-ft diameter bridge pier with a velocity of 12 ft/s. Estimate the force (per unit length) that the water e

xerts on the pier. Assume that the flow can be approximated as an potential fluid flow around the front half of the cylinder, but due to flow seperation, the average pressure on the rea half is constant and approximately equal to 1/2 the pressure at point A.

1 answer:

jolli1 [7]3 years ago

3 0

Answer: hello the diagram related to your question is missing please the third image is the missing part of the question

Fx = 977.76 Ib/ft

Explanation:

<u>Estimate the force that water exerts on the pier </u>

V = 12 ft/s

D( diameter ) = 6 ft

first express the force on the first half of the cylinder as

Fx1 = - $-2\int\limits^\pi _\frac{\pi }{2} {Ps*cos\beta *a} \, d\beta$ ---------------- ( 1 )

where ; Fy = 0

Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β ) ------------- ( 2 )

Input equation (2) into equation ( 1 ) (note : assuming Po = 0 )

attached below is the remaining part of the solution

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At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

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Answer:

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Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

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$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

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The correct question;

An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?

Answer:

h'_2 = 40 W/K.m²

Explanation:

We are given;

L1 = 1m

L2 = 5m

T_s = 400 K

T_(∞) = 300 K

V = 100 m/s

q = 20,000 W/m²

Both objects have the same shape and density and thus their reynolds number will be the same.

So,

Re_L1 = Re_L2

Thus, V1•L1/v1 = V2•L2/v2

Hence,

(h'_1•L1)/k1 = (h'_2•L2)/k2

Where h'_1 and h'_2 are convection coefficients

Since k1 = k2, thus, we now have;

h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)

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Westkost [7]

Answer:

a) $m_{2} = 0.753\,kg$ , b) $Q_{in} = 2122.963\,kJ$

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.02726\,\frac{m^{3}}{kg}$

$u = 1192.94\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

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$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.06643\,\frac{m^{3}}{kg}$

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$Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}$

Initial and final masses are:

$m_{1} = \frac{V_{1}}{\nu_{1}}$

$m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }$

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$m_{2} = \frac{V_{2}}{\nu_{2}}$

$m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }$

$m_{2} = 0.753\,kg$

a) The final mass within the tank is:

$m_{2} = 0.753\,kg$

b) The total amount of heat transfer is:

$Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}$

$Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )$

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