Answer:
The distance from the entrance at which the boundary layers meet is 0.516m
The distance from the entrance at which the thermal boundary layers meet is 1.89m
Explanation:
For explanation, look at the attached file
For each topic, find the total number of blurts that were analyzed as being related to the topic. Order the result by topic id.
Your SQL query should print the topic id, topic description and the corresponding count.user (email, password, name, date_of _birth, address, type) primary keyſemail) celebrity (email, website, kind) primary keyemail) blurt (blurtid, email, textlocation, time) primary keyblurtid,email ) foreign key (email) references user(email) hobby (email, hobby) primary keyſ emailhobb, ) foreign key(email) references user(email) . follow (follower, followee) primary key(follower.followee) foreign key follower) references user(email) foreign keyfollowee) references user(email) vendor (id, name rimary keyid) vendorambassador (vendorid. email) primary keyſvendorid) foreign key(email) references userſ email) foreign keyvendorid) references vendor(id) topic (id, description) primary keyſid) vendor_topics (vendorid, topicid) primary keyvendorid, topicid) foreign keyvendorid) references vendor(id) foreign keytopicid) references topic(id)
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Answer:
a) The change in Kinetic energy, KE = -1.95 kJ
b) Power output, W = 10221.72 kW
c) Turbine inlet area,
Explanation:
a) Change in Kinetic Energy
For an adiabatic steady state flow of steam:
.........(1)
Where Inlet velocity, V₁ = 80 m/s
Outlet velocity, V₂ = 50 m/s
Substitute these values into equation (1)
KE = -1950 m²/s²
To convert this to kJ/kg, divide by 1000
KE = -1950/1000
KE = -1.95 kJ/kg
b) The power output, w
The equation below is used to represent a steady state flow.
For an adiabatic process, the rate of heat transfer, q = 0
z₂ = z₁
The equation thus reduces to :
w = h₁ - h₂ - KE...........(2)
Where Power output, ..........(3)
Mass flow rate,
To get the specific enthalpy at the inlet, h₁
At P₁ = 10 MPa, T₁ = 450°C,
h₁ = 3242.4 kJ/kg,
Specific volume, v₁ = 0.029782 m³/kg
At P₂ = 10 kPa, , x₂ = 0.92
specific enthalpy at the outlet, h₂ =
h₂ = 3242.4 + 0.92(2392.1)
h₂ = 2392.54 kJ/kg
Substitute these values into equation (2)
w = 3242.4 - 2392.54 - (-1.95)
w = 851.81 kJ/kg
To get the power output, put the value of w into equation (3)
W = 12 * 851.81
W = 10221.72 kW
c) The turbine inlet area
Answer:
A) W' = 15680 KW
B) W' = 17113.87 KW
Explanation:
We are given;
Temperature at state 1; T1 = 290 K
Temperature at state 3; T3 = 1100 K
Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw
Pressure of air into compressor; P_c = 95 kPa
Pressure of air into turbine; P_t = 760 kPa
A) The power assuming constant specific heats at room temperature is gotten from;
W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in
Now, we don't have T4 and T2 but they can be gotten from;
T4 = [T3 × (r_p)^((1 - k)/k)]
T2 = [T1 × (r_p)^((k - 1)/k)]
r_p = P_t/P_c
r_p = 760/95
r_p = 8
Also,k which is specific heat capacity of air has a constant value of 1.4
Thus;
Plugging in the relevant values, we have;
T4 = [(1100 × (8^((1 - 1.4)/1.4)]
T4 = 607.25 K
T2 = [290 × (8^((1.4 - 1)/1.4)]
T2 = 525.32 K
Thus;
W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000
W' = 0.448 × 35000
W' = 15680 KW
B) The power accounting for the variation of specific heats with temperature is given by;
W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in
From the table attached, we have the following;
At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K
At T3 = 1100 K, h3 = 1161.07 KJ/K
At T1 = 290 K, h1 = 290.16 KJ/K
At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K
Thus;
W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000
W' = 17113.87 KW
Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V =
V = Av =
V =
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = =
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 =
v2 =
v2 = 6.017 m/s