Which option is an example of a physical model?

Used ______ must be hot drained for 12 hours or crushed before disposal.

WITCHER [35]

Answer:

A

Explanation:

The answer is towels because towels after a little bit of just sitting around have a chemical reaction that cam cause them to spontaneously combust

5 0

3 years ago

The "Crawler" developed to transport the Saturn V launch vehicle from the assembly building to the launch pad is the largest lan

Illusion [34]

Answer:

a) 152000 slugs

b) 2220000 kg or 2220 metric tons

Explanation:

A body with a weight of 4.9*10^6 lbf has a mass of

4.9*10^6 lbm * 1 lbf/lbm = 4.9*10^6 lbm

This mass value can then be converted to other mass values.

1 slug is 32.17 lbm

Therefore:

4.9*10^6 lbm * 1 slug / (32.17 lbm) = 152000 slugs

1 lb is 0.453 kg

Therefore:

4.9*10^6 lbm / (1/0.453) * kg/lbm = 2220000 kg

5 0

3 years ago

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o

Leni [432]

Answer:

(a) The power input to the compressor: $\dot{W}=73.07 kJ/s = 73.07 kW$

(b) The volume flow rate of the refrigerant at the compressor inlet: $\dot{v}=0.209 m^{3}/s$

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

$h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg$

Now we proceed to know the work with the following expression:

$\dot{W}=\dot{m}(h_{2}-h_{1})$

Now we replace values and our result is:

$\dot{W}=73.07 kJ/s = 73.07 kW$

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

$\nu=0.1739m^{3}/kg$

With our specific volume and the mass rate, we can calculate the volume rate:

$\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s$

6 0

3 years ago

A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/d

joja [24]

Solution :

Given :

k = 0.5 per day

$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$

$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$

Volume, V $= 200 \ m^3$

Now, input rate = output rate + KCV ------------- (1)

Input rate $= Q_s C_s + Q_{sw}C_{sw}$

$=(40 \times 10) + (0.5\times 100)$

$= 2 \times 10^5 \ mg/s$

The output rate $= Q_m C_{m}$

= ( 40 + 0.5 ) x C x 1000

$=40.5 \times 10^3 \ C \ mg/s$

Decay rate = KCV

∴ $KCV =\frac{0.5/d \times C \ \times 200 \times 1000}{24 \times 3600}$

= 1.16 C mg/s

Substituting all values in (1)

$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$

C = 4.93 mg/L

4 0

3 years ago

One kilogram of water fills a 0.140 m^3 rigid container at an initial pressure of 1.8 MPa. The container is then cooled to 40°C.

Pavel [41]

Answer:

T1 = 299.18 °C

P2 = 0.00738443 MPa

Explanation:

From the data, we can get two properties for the initial condition. These are pressure and specific volume.

The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

$v=\frac{vol (m^{3})}{mass (kg)} = \frac{0.140 m^{3}}{1 kg} = 0.140 \frac{ m^{3}}{kg}$

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:

$P^{sat} = 1.8 MPa$

$T^{sat} = 207.12 C$

$v_{g} = 0.1103\frac {m^{3}}{kg}$ specific volume for the saturated vapor

$v_{l} = 0.001167 \frac{m^{3}}{kg}$ specific volume for the saturated liquid

Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.

Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of $0.140 \frac{ m^{3}}{kg}$ at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.

Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:

$P^{sat} at 40C = 0.00738443 MPa$

7 0

3 years ago