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charle [14.2K]
3 years ago
14

A rigid, sealed cylinder initially contains 100 lbm of water at 70 °F and atmospheric pressure. Determine: a) the volume of the

tank (ft3 ). Later, a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibriu
Engineering
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

Determine A) The Volume Of The Tank (ft^3) Later A Pump Is Used To Extract ... A rigid, sealed cylinder initially contains 100 lbm of water at 70 degrees F and atmospheric pressure. ... Later a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibrium ...

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Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold
Pavlova-9 [17]

Answer:

Q=127.66W

L=9.2mm

Explanation:

Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.

To solve any problem related to heat transfer, the general equation is used

Q = delta / R

Where

Q = heat

Delta = the temperature difference

R = is the thermal resistance by conduction, convection and radiation

to solve this problem we propose the previous equation

Q = delta / R

later we find R

R=[tex]r=\frac{6L1}{AK1} +\frac{5L2}{AK2}+\frac{1}{Ah}

R=\frac{6(0.0001)}{(1.25)(0.026)} +\frac{5(0.012)}{(1.25)(0.026)}+\frac{1}{(25)(1.25)} =0.235 K/w

Q=(25-(-5))/0.235=127.66W

part b

we use the same ecuation with Q=127.66

Q = delta / R

ΔR=\frac{L}{KA} +\frac{1}{hA} \\R=\frac{L}{(0.035)(1.25)} +\frac{1}{(25)(1.25)}\\ R=22.85L+0.032\\Q=(T1-T2)/R\\\\127.66=(25-(-5))/(22.85L+0.032)\\solving for L\\L=9.2mm

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3 years ago
Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm
igor_vitrenko [27]

Answer:

velocity and pressure in a 2.6-cm:

P2 = 2.53x10^5Pa, v2 = 1.18m/s

Explanation:

Pressure = P, Velocity = v, Height = h, Diameter = d, Radius= r, Area = A

Area = πr^2

From the question:

v1 = 0.5m/s

d1 = 4cm = 0.04m

r1 = d1/2 = 0.04/2 = 0.02m

Since water was pumped from basement, h1 = 0m

P1 = 3.03x10^5 Pa

A1 = π×0.02×0.02

A1 = 0.0004πm^2

v2 = unknown

d2 = 2.6cm = 0.026m

r2 = d2/2 = 0.026/2 = 0.013m

h2 = 5m

P2 = unknown

A2 = π×0.013×0.013

A2 = 0.000169πm^2

Using continuity equation:

A1v1 = A2v2

0.0004π * 0.5 = 0.000169π * v2

v2 = (0.0004π * 0.5)/(0.000169π)

v2 = 1.18m/s

Applying a Bernoulli principle

P + 1/2*density*v^2 + density*g*h =C

C = constant

P1 + 1/2*density*v1^2 + density*g*h1

= P2 + 1/2*density*v2^2 + density*g*h2

Let g = 9.81m/s

density of water = 1000kg/m^3

(P1-P2) = 1/2* density(v2^2 - v1^2) +(density*g*h2) - (density*g*h1)

(P1-P2) = 1/2* density(v2^2 - v1^2) +

density* g(h2-h1)

(3.03x10^5 - P2)= 1/2*1000 (1.18^2-0.5^2) + 1000(9.81(5-0))

(3.03x10^5 - P2) = 500(1.3924-0.25) + 49050

3.03x10^5 - P2 = 571.2 + 49050

3.03x10^5 - P2 = 49621.2

3.03x10^5 - 49621.2 = P2

P2 = 253378.8

P2 = 2.53x10^5Pa

P2 = 2.53x10^5Pa, v2 = 1.18m/s

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