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Alenkasestr [34]
2 years ago
14

An ideal Diesel Cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rat

e of heat addition to this cycle when it produces 200 hp of power; the cycle is repeated 1200 times per minute; and the state of the air at the beginning of the compression is 95 kPa and 17oC. Use constant specific heats at room temperature.
Engineering
1 answer:
Mariulka [41]2 years ago
5 0

Answer:

A) Q'_in = 304.66 hp

B) T_max = 1109.29 °C

Explanation:

We are given;

compression ratio; r = 18

cutoff ratio; r_c = 1.5

Power; W' = 200 hp

Temperature; T1 = 17°C = 290 K

A) To find the rate of heat addition, we will use the derived formula;

Q'_in = (W' × r^(k - 1) × k(r_c - 1))/(r^(k - 1) × k(r_c - 1)) - ((r_c)^(k) - 1)

Where k is a constant = 1.4

Thus;

Q'_in = (200 × 18^(1.4 - 1) × 1.4(1.5 - 1))/((18^(1.4 - 1) × 1.4(1.5 - 1)) - ((1.5^(1.4)) - 1)

Q'_in = 304.66 hp

B) To find the maximum temperature, we will use the Formula;

T_max = T1 × r^(k - 1) × r_c

T_max = 290 × 18^(1.4 - 1) × 1.5

T_max = 1382.29 K

Converting to °C gives;

T_max = 1109.29 °C

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mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}

=  2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}

                 = 15 GPa

fracture will occur as the value is less than E/10 = 22.5

7 0
3 years ago
Ok there........................................................................
Juliette [100K]

Answer:

ok THERE

Explanation:

4 0
3 years ago
Chemical engineering got is unofficial start around the time of the __________ __________ ________.
Gre4nikov [31]

Answer:

Option A,  World War II

Explanation:

During the period of industrial revolution around 1915-25, the chemical engineering has taken a new shape. During this period (i.e around the world war I), there was rise in demand for  liquid fuels, synthetic fertilizer, and other chemical products. This lead to development of chemistry centre in Germany . There was rise in use of synthetics fibres and polymers. World war II saw the growth of catalytic cracking, fluidized beds, synthetic rubber, pharmaceuticals production, oil & oil products, etc. and because of rising chemical demand, chemical engineering took a new shape during this period

Hence, option A is the right answer

4 0
2 years ago
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan
hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) =  30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

R = \dfrac{u^2}{15(e+f_s)}

where;

R= radius

f_s = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

f_s = 0.20

So;

R = \dfrac{30^2}{15(0.08+0.20)}

R = \dfrac{900}{15(0.28)}

R = \dfrac{900}{4.2}

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B  in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0
3 years ago
Problem 7.16 (GO Tutorial) A cylindrical bar of steel 10.4 mm (0.4094 in.) in diameter is to be deformed elastically by applicat
shepuryov [24]

Answer:

P = 18035.25 N

Explanation:

Given

D = 10.4 mm

ΔD = 3.2 ×10⁻³ mm

E = 207 GPa

ν = 0.30

If

σ = P/A

A = 0.25*π*D²

σ = E*εx

ν = - εz / εx

εz = ΔD / D

We can get εx as follows

εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴

Now we find εx

ν = - εz / εx   ⇒   εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³

then

σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa

we have to obtain A:

A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²

Finally we apply the following equation in order o get P

σ = P/A   ⇒  P =  σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N

4 0
3 years ago
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