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3 years ago

Depending on the economy, financial institution, and your credit history, interest rates commonly range from

Engineering

2 answers:

svp [43]3 years ago

5 0

From what we need a picture for prod

STALIN [3.7K]3 years ago

4 0

Answer:

<em>When interest rates increase, it affects the ways that consumers and businesses can.called the Fed—cut the target range for its benchmark interest rate by 0.25%.Why does the Fed cut interest rates when the economy begins to struggle—or raise.The prime rate represents the credit rate that banks extend to their most.</em>

Explanation:

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In a steam power plant, the temperature of the burning fuel is 1100 °C, and cooling water is available at 15 °C. Steam leaving t

nadya68 [22]

Answer:

b. 1655.7 KJ/kg ( net work produced )

c. 2324.86 KJ/kg

d. 0.25539 --- 25.5%

e. 3980.63 KJ/kg

Explanation:

Given:-

Condenser exit parameters:

P1 = 50 KPa , saturated liquid

Boiler exit / Turbine exit parameters:

P3 = 2 MPa

T3 = 1100°C

Solution:-

- Adiabatic and reversible processes for pump and turbine are to be applied

- Assume changes in elevation heads within the turbomachinery to be negligible.

- Assume steady state conditions for fluid flow and the use of property tables will be employed.

Isentropic compression of water in pump:

Pump inlet conditions : Pump exit to Boiler pressure:

P1 = 50 KPa, sat liquid P2 = P3 = 2 MPa

h1 = 340.54 KJ/kg s2 = s1 = 1.0912 KJ/kg.K

s1 = 1.0912 KJ/kg.K h2 = 908.47 KJ/kg

- Apply energy balance for the pump and determine the work input ( Win ) required by the pump:

Win = h2 - h1

Win = 908.47 - 340.54

Win = 567.93 KJ/kg

Isentropic expansion of steam in turbine:

Turbine inlet conditions : Turbine exit to condenser pressure:

P3 = 2MPa, T3 = 1100°C P4 = P1 = 50 kPa

h3 = 4889.1 KJ/kg s4 = s3 = 8.7842 KJ/kg.K .. superheated

s3 = 8.7842 KJ/kg.K h4 = hg = 2665.4 KJ/kg

- Apply energy balance for the turbine and determine the work output ( Wout ) produced by the turbine:

Wout = h3 - h4

Wout = 4889.1 - 2665.4

Wout = 2223.7 KJ/kg

- The net work-output obtained from the cycle ( W-net ) is governed by the isentropic processes of pump and turbine.

W_net = Wout - Win

W_net = 2223.7 - 567.93

W_net = 1655.77 KJ/kg ... Answer

- The fraction of work generated by turbine is used to operate the pump. The a portion of Wout is used to drive the motor of the pump. The pump draws ( Win ) amount of work from pump. The ratio of work extracted from turbine ( n ) would be:

n = Win / Wout

n = 567.93 / 2223.7

n = 0.25539 ... Answer ( 25.5 % ) of work is used by pump

- The amount of heat loss in the condenser ( consider reversible process ). Apply heat balance for the condenser, using turbine exit and condenser exit conditions:

Ql = h4 - h1

Ql = 2665.4 - 340.54

Ql = 2324.86 KJ/kg ... Answer

- The amount of heat gained by pressurized water in boiler ( consider reversible process ). Apply heat balance for the boiler, using pump exit and boiler exit conditions:

Qh = h3 - h2

Qh = 4889.1 - 908.47

Qh = 3980.63 KJ/kg ... Answer

8 0

3 years ago

A developer has requested permission to build a large retail store at a location adjacent to the intersection of an undivided fo

Sergio [31]

Answer:

676 ft

Explanation:

Minimum sight distance, d_min

d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time

v_max is given as 50 mi/h

t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds

Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft

For practical purposes, this distance is taken as 676 ft

6 0

3 years ago

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

Allisa [31]

Answer:

theoretical fracture strength = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength = $s_{0} \times \sqrt{\frac{L}{r} }$ .............................1

put here value and we get

theoretical fracture strength = $1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }$

theoretical fracture strength = $16919.98 \times 10^6$

theoretical fracture strength = 16919.98 MPa

3 0

3 years ago

Which type of muscle tissue is both voluntary and striated?

Elena L [17]

Answer:

Skeletal muscle

Explanation:

These classifications describe three distinct muscle types: skeletal, cardiac and smooth. Skeletal muscle is voluntary and striated, cardiac muscle is involuntary and striated, and smooth muscle is involuntary and non-striated.

4 0

4 years ago

Read 2 more answers

Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kin

hammer [34]

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

8 0

4 years ago