Engineering, 22.09.2020 20:01 calraezaj2020

Piessunzed water (pin-10 bar, Tin= 1 10°C) enters the hottom of an L 12-m-long vertical tube of diameter D110 mm at a mass flow rate of 1.5kg/s. The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at o 7 bar, T 600°C. Determine the change in the rate at which the following quantities enter and exit the tube: (a) the combined thermal and Row work, (b) the mechanical energy, and
(c) the total energy of the water. Also, (d) determine the heat trans- fer rate. q. Hint: Relevant properties may be obtained from a thermodynamics text.
Consider the tube and inlet conditions above Heat transfer at a rate of 3.89 MW is delivered to the tube. For an exit pressure of p 8 bar, determine
(a) the temperature of the water at the outlet as well as the change in
(b) combined thermal and flow work,
(c) mechanical energy, and (d) total energy of the water from the inlet to the outlet of the tube. Hint: As a first estimate, neglect the change in mechanical energy in solving part (a). Relevant properties may be obtained from a thermodynamics text

Answers: 2

Show answers

Answers

Answer from: tmcdowell69

(1) $\Delta E_{tw}=4845.43 kW$

(2) $\Delta E_m=6.329 kW$

(3) $\Delta E_t= 4851.759 kW$

(4) q= 4851.759 kW

Explanation:

At the saturation temperature, water starts boiling, and before that heat is added at constant pressure as latent heat.

From the saturated water-pressure table, at the pressure P=10 bar, we have

The saturated temperature of the water, $T_{sw}=179.88^{\circ} C$

The specific volume of water, $v_{ws}=v_f=0.00127 m^3/kg$

Specific enthalpy of water, $h_{ws}=h_f=762.50 kJ/kg$

The given inlet temperature of the water, $T_i=110^{\circ}$ , so, latent heat added to the water to reach the saturation temperature is

$h_l=C_P(T_{sw}-T_i)$

$\Rightarrow h_l=4.187(179.88^{\circ} -110^{\circ}$

$\Rightarrow h_l=292.587 kJ/kg$

Now, specific enthalpy of the water at the inlet (specific enthalpy of the water at the saturation temperature) (Latent heat capacity).

$\Rightarrow h_i=h_{sw}-h_l$

$\Rightarrow h_i=762.50-292.587=469.912kJ/kg$

The specific volume of the water at intel is the same as the specific volume at the saturation temperature as volume remains unchanged on the addition of latent heat.

So, $v_i=v_{ws}=0.00127 m^3/kg$ .

The outlet temperature, $T_o=600^{\circ} C$ and pressure, P_o=7 bar. From the superheated water table, we have

The specific volume of water, v_o=0.5738 m^3/kg

The specific enthalpy of water, $h_{wo}=3700.9 kJ/kg$

The given mass flow rate, $\dot{m} =1.5 kg/s$ .

The inlet radius and outlet diameter are the same, i.e

d_i=d_o=110 mm=0.11m .

So, Inlet and outlet areas, $A_i=A_f=9.5033\times 10^{-3} m^2$ .

Let the inlet and outlet velocities be V_i and V_o respectively.

For the given specific volume, , and mass flow rate, $\dot{m}$ , the velocity, , at any cross-section having an area is

$V=\frac{v\dot{m}}{A}$ .

So, the inlet velocity,

$V_i=\frac{v_i \dot{m}}{A_i}$

$\Rightarrow V_i=\frac{0.00127\times 1.5}{9.5033\times 10^{-3}}$

$\Rightarrow V_i=0.20 m/s$ .

Similarly, the outlet velocity,

$V_o=\frac{v_o \dot{m}}{A_o}$

$\Rightarrow V_o=\frac{0.5738\times 1.5}{9.5033\times 10^{-3}}$

$\Rightarrow V_0=90.57 m/s$ .

(1) The change in combined thermal energy and work flow Change in the thermal energy Change in the flow work

$\Delta E_{tw}= \dot{m}(u_f-u_i)+\dot{m} (P_fv_f-P_iv_i)$

$\Rightarrow \Delta E_{tw}=\dot{m}[(u_f+P_fv_f) - (u_i+P_iv_i)$

$\Rightarrow \Delta E_{tw}=\dot{m} (h_f-h_i)$

$\Rightarrow \Delta E_{tw}=1.5(3700,20-469.912)=4845.43 kW$

(2)The change in mechanical energy

$\Delta E_m=$ Change in kinetic energy + change in potential energy

$\Rightarrow \Delta E_m=\left(\frac 1 2 \dot{m} V_f^2-\frac 1 2 \dot{m} V_i^2\right)+(\dot{m} g z_f-\dot{m} g z_i)$

$\Rightarrow \Delta E_m=\frac 1 2 \dot{m}(V_f^2-V_i^2)+\dot{m} g (z_f-z_i)$

$\Rightarrow \Delta E_m=\frac 1 2\times 1.5((90.57)^2-(0.2)^2)+1.5 \times 9.81 \times12$

$\Rightarrow \Delta E_m=6328.74 J/s=6.329 kW$

(3) The change in the total energy of water,

$\Delta E_t=$ chnge in the thernal energy + change in the flow work + change in the mechanical energy

$\Rightarrow \Delta E_t=4845.43+6.329= 4851.759 kW$ [from part (1) and (2)]

(4) Now, as there is no work done by the water, so, the heat input only caused the change in the total energy.

Hence, the rate of heat transfer, q= 4851.759 kW [from part (3).

Answer from: Quest

explanation:

Answer from: Quest

import math

x=int(input("enter a number: "))

print (math.sqrt(math.fabs(

Another question on Engineering

Engineering, 04.07.2019 18:10

The temperature of air decreases as it is compressed by an adiabatic compressor. a)- true b)- false

Answers: 2

Answer

Engineering, 04.07.2019 18:10

An ideal otto cycle with air as the working fluid has a compression ratio of 8. the minimum and maximum temperatures in the cycle are 300 k and 1340 k. use constant specific heats at room temperature to determine (a) the amount of heat transferred to the air during the heat- addition kj/kg, (b) the thermal efficiency, and (c) the thermal efficiency of a carnot cycle ope limits. process, in rating between the same temperature

Answers: 2

Answer

Engineering, 04.07.2019 18:10

Atmospheric air has a temperature (dry bulb) of 80° f and a wet bulb temperature of 60° f when the barometric pressure is 14.696 psia. determine the specific humidity, grains/lb dry air. a. 11.4 c. 55.8 d. 22.5 b. 44.1

Answers: 1

Answer

Engineering, 04.07.2019 19:10

A)-explain briefly the importance of standards in engineering design. b)- what is patent? c)-explain the relationship between these standards: b.s. and b.s.en d)- in engineering design concepts, types of loads and how they act are important factors. explain.

Answers: 3

Answer

You know the right answer?

Piessunzed water (pin-10 bar, Tin= 1 10°C) enters the hottom of an L 12-m-long vertical tube of diam...

Questions