Answer: Power
Explanation:
The rate at which work is done or the amount of work done based on a period of time is referred to as power.
Power can also be defined as the amount of energy that is being transferred per unit time. The unit of power is one joule per second or simply called the watt.
Methane (CH4) at 298 K, 1 atm enters a furnace operating at steady state and burns completely with 140% of theoretical air entering at 400 K, 1 atm. The products of combustion exit at 500 K, 1 atm. The flow rate of the methane is 1.4 kg/min. Kinetic and potential energy effects are negligible and air can be modeled as 21% O2 and 79% N2 on a molar basis.
Required:
Determine the dew point temperature of the products, in K.
In a production facility, 3-cm-thick large brass plates (k 5 110 W/m·K, r 5 8530 kg/m3, cp 5 380 J/kg·K, and a 5 33.9 3 1026 m2/s) that are initially at a uniform temperature of 25°C are heated by passing them through an oven maintained at 700°C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h 5 80 W/m2·K, determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be
Answer:
the surface temperature of the plates when they come out of the oven is approximately 445 °C
Explanation:
Given the data in the question;
thickness t = 3 cm = 0.03 m
so half of the thickness L = 0.015 m
thermal conductivity of brass k = 110 W/m°C
Density p = 8530 kg/m³
specific heat [tex]C_p[/tex] = 380 J/kg°C
thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s
Temperature of oven T₀₀ = 700°C
initial temperature T[tex]_i[/tex] = 25°C
time t = 10 min = 600 s
convection heat transfer coefficient h = 80 W/m².K
Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.
So, using analytical one-term approximation method, the Fourier number > 0.2.
now, we determine the Biot number for the process
we know that; Biot number Bi = hL / k
so we substitute
Bi = hL / k
Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109
Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )
The interpolation method used to find the
λ₁ = 0.1039 and A₁ = 1.0018
so
The Fourier number т = ∝t/L²
we substitute
Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²
т = 0.02034 / 0.000225
т = 90.4
As we can see; 90.4 > 0.2
So, analytical one-term approximation can be used.
∴ Temperature at the surface will be;
θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation
θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )
so we substitute
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )
θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998
θ(L,t)[tex]_{wall[/tex] = 0.3775
so we substitute into equation 1
θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)
0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )
0.3775 = ( T(L,t) - 700 ) / ( - 675 )
0.3775 × ( - 675 ) = ( T(L,t) - 700 )
- 254.8125 = T(L,t) - 700
T(L,t) = 700 - 254.8125
T(L,t) = 445.1875 °C ≈ 445 °C
Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 6 when 16.0% of the crack length is removed.
Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
([tex]p_t[/tex])₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2
where ([tex]p_t[/tex])₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
divide both sides by 2(σ₀)₁
[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex] = 36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]
1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂
([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁
([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24
The boundary work is positive during an expansion process.
a.
False
b.
True
Answer:
True
Explanation:
During expansion process, the boundary work is positive while in case of contraction, the boundary work is negative. During expansion process, the work is done by the system while in case of compression process, work in done on the system
Hence, the given statement is true
In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit
Answer: hello your question lacks some data below is the missing data
Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol
H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.
H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg
Answer :
a) 34.98 lit/min
b) 1432.53 m^3/min
Explanation:
a) Calculate how much water is produced
density of water = 1 kg/liter
First we will determine the mass of condensed water using the relation below
inlet mass - outlet vapor mass = 0.0339508 * n * 18/1000 ----- ( 1 )
where : n = 57241.57
hence equation 1 = 34.98 Kg/min
∴ volume of water produced = mass of condensed water / density of water
= 34.98 Kg/min / 1 kg/liter
= 34.98 lit/min
b) calculate the Volumetric flow rate of air entering the unit
applying the relation below
Pv = nRT
101325 *V = 57241.57 * 8.314 * 305
∴ V = 1432.53 m^3/min
hmmmmmmmm i already put the photo as attachment its
Answer:
letse see
Explanation:
The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.
Answer:
the liquid woulriekwvhrnsshsnekwb ndrhwmoadi
If it took a 30m capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) water from the
Complete question:
If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.
Answer:
the flow rate of the water from the tank is 0.05 m³/min
Explanation:
Given;
volume of water in the tank, v = 30 m³
length of the waterproof faucet, L = 2cm = 0.02 m
duration of water flow through the tank, t = 10 hours
The flow rate of the water from the tank is calculated as;
[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]
Therefore, the flow rate of the water from the tank is 0.05 m³/min
1) Each of the following would be considered company-confidential except
A) a contract bid B) employee salaries C) your company's strategic plan D) your company's address
what is the most common type of suspensions system used on body over frame vehicles?
Answer:
Engine
Explanation:
Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.
What is a Semi-independent suspension?Semi-independent suspension give the front wheels some individual movement.
This suspension only used in rear wheels.
Thus, the correct option is Semi-independent suspension
Learn more about Semi-independent suspension
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