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3 years ago

True or false : In improper integrals infinte intervals mean that both of the integration limits are should be infinity

Engineering

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration

Explanation:

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125

Gekata [30.6K]

Answer:

a) $\dot m_{3} = 135\,\frac{lbm}{s}$ , b) $h_{3}=168.965\,\frac{BTU}{lbm}$ , c) $T = 200.829\,^{\textdegree}F$

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

$\dot m_{1} + \dot m_{2} - \dot m_{3} = 0$

The mass flow rate exiting the tank is:

$\dot m_{3} = \dot m_{1} + \dot m_{2}$

$\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}$

$\dot m_{3} = 135\,\frac{lbm}{s}$

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

$\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

$h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}$

Properties of water are obtained from tables:

$h_{1}=180.16\,\frac{BTU}{lbm}$

$h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)$

$h_{2}=29.032\,\frac{BTU}{lbm}$

The specific enthalpy at outlet is:

$h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }$

$h_{3}=168.965\,\frac{BTU}{lbm}$

c) After a quick interpolation from data availables on water tables, the final temperature is:

$T = 200.829\,^{\textdegree}F$

8 0

3 years ago

Read 2 more answers

(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm = 0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

A) would the 1040 steel be a possible candidate for this application

Yield strength of 1040 steel < stress ( in order to be a possible candidate )

stress = p / A0 = ( 18000 ) / ( $\frac{\pi }{4}$ ) * 0.0075^2

= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )

Therefore 1040 steel is not a possible candidate for this application

B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm

Area1 = ( $\frac{\pi }{4}$ ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%

6 0

3 years ago

A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c

Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice $C_p=2.03\ KJ/kg.K$

Latent heat for ice H=336 KJ/kg

Specific heat for ice $C_p=4.187\ KJ/kg.K$

We know that sensible heat given as

$Q=mC_p\Delta T$

Heat for -15F to 32 F:

$Q=mC_p\Delta T$

$Q=9\times 2.03(32+15) KJ$

Q=858.69 KJ

Heat for 32 Fto 200 F:

$Q=mC_p\Delta T$

$Q=9\times 4.187(200-32) KJ$

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So 7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

8 0

3 years ago

Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity

Digiron [165]

Answer: D) All of the above

Explanation:

Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.

4 0

3 years ago

If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab

12345 [234]

Answer:

Explanation:

load = 4500lb lift height= 30 ft

time =15 s

velocity= $\frac{30}{15}$ ft/s