1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gwar [14]
3 years ago
6

A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te

mperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa.
Engineering
1 answer:
Artyom0805 [142]3 years ago
3 0

Explanation:

a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures

100% (3 ratings)

A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2

You might be interested in
Which of the following is true about ideal standards? a.Ideal standards provide allowance for normal breakdowns and interruption
marysya [2.9K]

Answer: D. All of the choice A, B and C are correct.

8 0
3 years ago
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m ? K), and the wire/sheath interface
Svet_ta [14]

Question

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.

If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Answer:

a. 4.52W/m

b. 13mm

Explanation:

Given

Diameter of electrical wire = 2mm

Wire Thickness = 2-mm

Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)

Thermal contact resistance = 3E-4m².K/W

Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,

Temperature of the ambient air = 20°C.

Maximum Allowable Sheet Temperature = 50°C.

From the thermal circuit (See attachment), we my write

E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)

= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))

Where r in,i = D/2

= 2mm/2

= 1 mm

= 0.001m

r in,o = r in,i + t = 0.003m

T in, i = Tmax = 50°C

Hence

q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]

= 30/[(Ln3/0.26π) + 1/0.06π)]

= 30/[(1.34) + 5.30)]

= 30/6.64

= 4.52W/m

The critical radius is unaffected by the constant resistance.

Hence

Critical Radius = k/h

= 0.13/10

= 0.013m

= 13mm

5 0
3 years ago
I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have
drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = \lambda e^{-\lambda t}

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [  -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905

(c) The Poisson distribution is presented as follows;

P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!}  = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!}  = 3.915\times 10^{-12}

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}.

4 0
3 years ago
Write a program with total change amount as an integer input, and
AURORKA [14]

Answer:

amount = int(input())

#Check if input is less than 1

if amount<=0:

    print("No change")

else: #If otherwise

    #Convert amount to various coins

    dollar = int(amount/100) #Convert to dollar

    amount = amount % 100 #Get remainder after conversion

    quarter = int(amount/25) #Convert to quarters

    amount = amount % 25 #Get remainder after conversion

    dime = int(amount/10) #Convert to dimes

    amount = amount % 10 #Get remainder after conversion

    nickel = int(amount/5) #Convert to nickel

    penny = amount % 5 #Get remainder after conversion

    #Print results

    if dollar >= 1:

          if dollar == 1:

                print(str(dollar)+" Dollar")

          else:

                print(str(dollar)+" Dollars")

    if quarter >= 1:

          if quarter == 1:

                print(str(quarter)+" Quarter")

          else:

                print(str(quarter)+" Quarters")

    if dime >= 1:

          if dime == 1:

                print(str(dime)+" Dime")

          else:

                print(str(dime)+" Dimes")

    if nickel >= 1:

          if nickel == 1:

                print(str(nickel)+" Nickel")

          else:

                print(str(nickel)+" Nickels")

    if penny >= 1:

          if penny == 1:

                print(str(penny)+" Penny")

          else:

                print(str(penny)+" Pennies")

Explanation:

8 0
3 years ago
The function below takes a single string parameter: input_string. If the input contains the lowercase letter z, return the strin
Whitepunk [10]

Answer:

# string_contains function is defined with input_string

# as arguments

def string_contains(input_string):

   # if-statement that check if letter z is in input_string

   if 'z' in input_string:

       # it print"has the letter z" if it has z

       print("has the letter z")

   else:

       # else it print "not worthwhile"

       print("not worthwhile")

       

# string_contains function is called

string_contains("The animal is zebra")        

string_contains("learning is fun")

Explanation:

The code is written in Python and well commented.

Image of the output when the function is called is attached.

4 0
3 years ago
Other questions:
  • According to the decreasing order of toughness. list the following materials (note: the steels are assumed to have no cold work
    6·1 answer
  • How much work is performed if a 400 lb weight is lifted 10 ft ?
    8·1 answer
  • Am i eating ramon nooddles rn
    10·2 answers
  • The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 40 continuous hours. During that ti
    11·1 answer
  • Steam enters an adiabatic turbine at 8 MPa and 500C with a mass flow rate of 3
    11·1 answer
  • For the system in problem 4, suppose a main memory access requires 30ns, the page fault rate is .01%, it costs 12ms to access a
    14·1 answer
  • A person is planning a bungee jump from a 40 meter high bridge. Under the bridge is a river with crocodiles, so the person does
    7·1 answer
  • How pine are processed ????
    10·1 answer
  • A. The ragion was colonized by European powers
    6·1 answer
  • Explain the S.A. co-ordinate system used in surveying
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!