what are some questions about simple machines that don't make me sound like i didn't pay attention in the lesson?

A circular pipe of 25-mm outside diameter is placed in an airstream at 25 °C and 1-atm pressure. The air moves in cross flow ove

Scrat [10]

Answer: a) Fd = 3.24 N/m

b) Q = 520 w/m

Explanation: please find the attached files for the solution

7 0

3 years ago

A bar having a length of 5 in. and cross-sectional area of 0.7 i n . 2 is subjected to an axial force of 8000 lb. If the bar str

barxatty [35]

Answer:

E=1.969 × 10¹¹ Pa

Explanation:

The formula to apply is;

E=F*L/A*ΔL

where

E=Young modulus of elasticity

F=Force in newtons

L=Original length in meters,m

A=area in square meters m²

ΔL= Change in length in meters,m

Given

F= 8000 lb = 8000*4.448 =35584 N

L= 5 in = 0.127 m

A= 0.7 in² =0.0004516 m²

ΔL = 0.002 in = 5.08e-5 m

Applying the formula

E=(35584 * 0.127)/(0.0004516*5.08e-5 )

E=1.969 × 10¹¹ Pa

8 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

Consider an ideal cogeneration steam plant to generate power and process heat. Steam enters the turbine from the boiler at 7 Mpa

igomit [66]

Answer:

1. The diagram T-s or H-s is attached to this answer.

2. The fraction of the steam extracted is 4.088Kg/s

3. The net Power produced per kg of steam exiting the boiler is 1089.5KJ/Kg.

4. The mass flow rate of steam supplied by the boiler is 16.352Kg/s

5. the net power produced by the plant is 11016.2KJ/s.

6. The utilization factor is 0.218.

Explanation:

To analyze this problem we need to find all the thermodynamic coordinates of the system. In the second image attached to this answer, we can see the entire ideal cogeneration steam plant system.

From a water thermodynamic properties chart, we can obtain the information for each point.

+ Steam enters the turbine from the boiler at 7 Mpa and 500 degrees C:

h₆=3410.56KJ/Kg

s₆=6.7993 KJ/Kg

This is an ideal cogeneration steam system, therefore: s₆=s₇=s₈

+One-fourth of the steam is extracted from the turbine at 600-kPa:

h₇(s₇) = 2773.74 KJ/Kg (overheated steam)

+The remainder of the steam continues to expand and exhausts to the condenser at 10 kPa.

h₈(s₈)=2153.58 KJ/Kg (this is wet steam with title X=0.8198)

h₁(P=10Kpa)= 191.83 KJ/Kg (condensed water) s₁=0.64925KJ/Kg

-This flow is pumped to 600KPa, so:

s₂=s₁

h₂(s₂)=192.585KJ/Kg

+The steam extracted for the process heater is condensed in the heater:

h₃(P=600KPa)=670.42KJ/Kg (condensed water)

+The steam extracted for the process heater is condensed in the heater and mixed with the feed-water at 600 kPa:

The mixing process of the flow of point 2 and 3 is an adiabatic process, therefore:

$\dot{Q}_4=\dot{Q}_2+\dot{Q}_3=\dot{m}_2 h_2+\dot{m}_3h_3\\\dot{m}_4 h_4=\dot{m}_2 h_2+\dot{m}_3h_3=\dot{m}_2 0.75h_4+\dot{m}_30.25h_4\\h_4=0.75h_2+0.25h_3=312.043KJ/Kg$

s₄=1.02252

+the mixture is pumped to the boiler pressure of 7 Mpa:

s₅=s₄

h₅(s₅)=323.685KJ/Kg

1)Now we have all the thermodynamic coordinates and we can draw the diagram of the system.

2) To determine the fraction of steam, the mass flow that is extracted from the turbine at state 7, we use the information that this flow is used to generate 8600KJ/s in a process of heat. Therefore:

$P=8600KJ/s=\dot{m}_{3-7}(h_7-h_3)\\\dot{m}_{3-7}=8600KJ/s/(h_7-h_3)=4.088Kg/s$

3)The net power produced per kg of steam exiting the boiler can be obtained as the rest between all the power obtained in the turbine less the power used in the pumps:

$P_{turb}/Kg=(h_6-h_7)+0.75(h_7-h_8)=1101.94KJ/Kg\\P_{pump1}/Kg=h_2-h_1=0.755KJ/kg\\P_{pump2}/kg=h_5-h_4=11.642KJ/Kg\\P_{net}/kg=P_{turb}-P_{pump1}-P_{pump2}=1089.543KJ/Kg$

4) To determine the mass flow rate of steam that must be supplied by the boiler, we only have to remember that the flow used in point 2) is a fourth of the total flow. therefore:

$0.25\dot{m}_{tot}=\dot{m}_{3-7}\\\dot{m}_{tot}=4\dot{m}_{3-7}=16.352Kg/s$