Interpretation:
It is impossible that two adiabatic reversible processes can intersect on
Concept Introduction:
From the second law of
Answer to Problem 5.1P
Hence, it is impossible that two adiabatic reversible processes can intersect on
Explanation of Solution
Given information:
It is given that two adiabatic reversible processes do intersect and complete the cycle on
From the given data, we plot a diagram on
Thus, AB is isothermal process; BC and AC are adiabatic reversible processes. This forms a cycle ABC where area inside the cycle is given as work.
Now, for the heat calculations, the adiabatic reversible processes have no heat transfer, so heat given by both process BC and AC are zero. So, heat transfer takes place only by the isothermal process which has only one single temperature
Two thermodynamic curves of same kind (isothermal, adiabatic, polytropic) never intersect each other on
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Chapter 5 Solutions
EBK INTRODUCTION TO CHEMICAL ENGINEERIN
- 4.59 Using the unilateral z-transform, solve the following difference equations with the given initial conditions. (a) y[n]-3y[n-1] = x[n], with x[n] = 4u[n], y[− 1] = 1 (b) y[n]-5y[n-1]+6y[n-2]= x[n], with x[n] = u[n], y[-1] = 3, y[-2]= 2 Ans. (a) y[n] = -2+9(3)", n ≥ -1 (b) y[n]=+8(2)" - (3)", n ≥ -2arrow_forward(30) 6. In a process design, the following process streams must be cooled or heated: Stream No mCp Temperature In Temperature Out °C °C kW/°C 1 5 350 270 2 9 270 120 3 3 100 320 4 5 120 288 Use the MUMNE algorithm for heat exchanger networks with a minimum approach temperature of 20°C. (5) a. Determine the temperature interval diagram. (3) (2) (10) (10) b. Determine the cascade diagram, the pinch temperatures, and the minimum hot and cold utilities. c. Determine the minimum number of heat exchangers above and below the pinch. d. Determine a valid heat exchange network above the pinch. e. Determine a valid heat exchange network below the pinch.arrow_forwardUse this equation to solve it.arrow_forward
- Q1: Consider the following transfer function G(s) 5e-s 15s +1 1. What is the study state gain 2. What is the time constant 3. What is the value of the output at the end if the input is a unit step 4. What is the output value if the input is an impulse function with amplitude equals to 3, at t=7 5. When the output will be 3.5 if the input is a unit steparrow_forwardgive me solution math not explinarrow_forwardgive me solution math not explinarrow_forward
- give me solution math not explinarrow_forwardgive me solution math not explinarrow_forwardExample (6): An evaporator is concentrating F kg/h at 311K of a 20wt% solution of NaOH to 50wt %. The saturated steam used for heating is at 399.3K. The pressure in the vapor space of the evaporator is 13.3 KPa abs. The 5:48 O Transcribed Image Text: Example (7): Determine thearrow_forward
- 14.9. A forward feed double-effect vertical evaporator, with equal heating areas in each effect, is fed with 5 kg/s of a liquor of specific heat capacity of 4.18 kJ/kg K. and with no boiling point rise, so that 50 per cent of the feed liquor is evaporated. The overall heat transfer coefficient in the second effect is 75 per cent of that in the first effect. Steam is fed at 395 K and the boiling point in the second effect is 373 K. The feed is heated by an external heater to the boiling point in the first effect. It is decided to bleed off 0.25 kg/s of vapour from the vapour line to the second effect for use in another process. If the feed is still heated to the boiling point of the first effect by external means, what will be the change in steam consumption of the evaporator unit? For the purpose of calculation, the latent heat of the vapours and of the steam may both be taken as 2230 kJ/kgarrow_forwardExample(3): It is desired to design a double effect evaporator for concentrating a certain caustic soda solution from 12.5wt% to 40wt%. The feed at 50°C enters the first evaporator at a rate of 2500kg/h. Steam at atmospheric pressure is being used for the said purpose. The second effect is operated under 600mmHg vacuum. If the overall heat transfer coefficients of the two stages are 1952 and 1220kcal/ m2.h.°C. respectively, determine the heat transfer area of each effect. The BPR will be considered and present for the both effect 5:49arrow_forwardالعنوان ose only Q Example (7): Determine the heating surface area 개 required for the production of 2.5kg/s of 50wt% NaOH solution from 15 wt% NaOH feed solution which entering at 100 oC to a single effect evaporator. The steam is available as saturated at 451.5K and the boiling point rise (boiling point evaluation) of 50wt% solution is 35K. the overall heat transfer coefficient is 2000 w/m²K. The pressure in the vapor space of the evaporator at atmospheric pressure. The solution has a specific heat of 4.18kJ/ kg.K. The enthalpy of vaporization under these condition is 2257kJ/kg Example (6): 5:48 An evaporator is concentrating F kg/h at 311K of a 20wt% solution of NaOH to 50wt %. The saturated steam used for heating is at 399.3K. The pressure in the vapor space of the evaporator is 13.3 KPa abs. The 5:48 1 J ۲/۱ ostrarrow_forward
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