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2 years ago

Determine how much concrete you will need for a slab which is 50 feet by 30 feet wide and 1 foot thick

Engineering

1 answer:

Levart [38]2 years ago

6 0

Answer:

Amount of concrete need to make slab = 1,500 feet³

Explanation:

Given:

Length of slab = 50 feet

Width of slab = 30 feet

Height of slab = 1 feet

Find:

Amount of concrete need to make slab

Computation;

Amount of concrete need to make slab = Volume of cuboid

Volume of cuboid = (l)(b)(h)

Amount of concrete need to make slab = (50)(30)(1)

Amount of concrete need to make slab = 1,500 feet³

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Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12

Katarina [22]

Answer:

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

Explanation:

The model for the turbine is given by the First Law of Thermodynamics:

$- \dot W_{out} + \dot m \cdot (h_{in} - h_{out}) = 0$

The turbine power output is:

$\dot W_{out} = \dot m\cdot (h_{in}-h_{out})$

The volumetric flow is:

$\dot V = \frac{\pi}{4} \cdot \left( \frac{2}{12}\,ft \right)^{2}\cdot (620\,\frac{ft}{s} )$

$\dot V \approx 13.526\,\frac{ft^{3}}{s}$

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

$\nu = 1.33490\,\frac{ft^{3}}{lbm}$

The mass flow is:

$\dot m = \frac{\dot V}{\nu}$

$\dot m = \frac{13.526\,\frac{ft^{3}}{s} }{1.33490\,\frac{ft^{3}}{lbm} }$

$\dot m = 10.133\,\frac{lbm}{s}$

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

$h = 1479.74\,\frac{BTU}{lbm}$

State 2 (Saturated Vapor)

$h = 1146.1\,\frac{BTU}{lbm}$

The turbine power output is:

$\dot W_{out} = (10.133\,\frac{lbm}{s} )\cdot (1479.1\,\frac{BTU}{lbm}-1146.1\,\frac{BTU}{lbm})$

$\dot W_{out} = 3374.289\,\frac{BTU}{s}$

6 0

3 years ago

A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.

Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

$E_{in}- E_{out}=E_{system,final}- E_{system,initial}$

Where

$E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}$

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

$E_{in}- E_{out}=-Q_{out}+W_{in}$

Regarding the energy system we have the following:

$E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}$

By doing the calculations we have:

$E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ$

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

$W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ$

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

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3 years ago